CHAPTER 1 - THE MOLE

SECTION 1 - RELATIVE WEIGHTS

How do you weigh an atom? In modern jargon, I would suppose that the answer would be "very carefully". An alternate, equally nonsensical answer would be that you could get a very tiny person with very tiny scales (balance, not fish) and very tiny forceps to do it for you. How do you weigh yourself? The answer will probably be with a scales. What good would it do to know that you weighed 135 pounds, for example, if there were no one around for comparison? How would you know if you were at a good weight or not? There would be no insurance charts available. The weights of atoms and molecules also need to be compared to some standard. In fact, the weights found on your periodic chart are comparison or relative weights.

If you have studied any physics or physical science, you will notice that the correct term should be atomic mass rather than atomic weight. Much of the older literature has also used atomic weight rather than mass. Although several incorrect usages do not make a correct usage, the terms will be used interchangeably here.

Objectives of this section:

1. Given the weights (masses) of the elemental components in a decomposition reaction, find the relative (weights) of the elements.

2. Explain what is meant by relative weight (mass).

3. Given a periodic chart, find the atomic weight (mass) of any given element.

4. Given a periodic chart, find the molecular (or formula) weight of any given substance.

5. Describe when the term formula weight is more appropriate than molecular weight.

CHAPTER 1SECTION 1

Pb, as was mentioned earlier in class, is a symbol for the element lead. P4 is a formula for a molecule of the element phosphorus which, in this case, consists of four atoms of phosphorus bonded together. H2O is a formula for a molecule of the compound water. What does the formula H2O tell us? When you read H2O what are you really saying to yourself?

H2O indicates one molecule of water consisting of two atoms of hydrogen bonded to one atom of oxygen. You may want to read this as two parts hydrogen combined chemically with one part oxygen but to do this you must define "parts". It is not true that two grams of hydrogen combine with one gram of oxygen to give three grams of water. The subscripts do not indicate masses. At this point we know nothing of the masses of these atoms. A molecule of water is much too small to see even with the most powerful microscope today. Even if we could see it, it would be so small that we could not measure it directly with any of our instruments.

It is possible to find atomic and molecular masses by indirect means. Look at the electrolysis of water apparatus. Water is added to the reservoir. The stopcocks to the collecting tubes must remain open so that they will be filled with water completely. Pure water does not conduct electricity so a dilute solution of copper (II) sulfate or sulfuric acid must be added. The stopcocks should then be closed and the apparatus is then connected to a battery. The current causes the water to split apart into hydrogen andoxygen. The positive hydrogen ions go to the negative electrode where they combine to form hydrogen gas. The negative oxygen ions move to the positive electrode and form oxygen gas. If the apparatus has been for a time, you should observe that the water no longer fills the collecting tubes completely. You will look at this apparatus again in another section and be able to experiment then. For now, however, notice that one side has less water than the other side. This side has hydrogen gas and the other side has oxygen gas.

A lab worker carefully weighed two empty plastic bags and then collected the gases in the tubes in the bags. To avoid loss, he quickly weighed the bags of gas. Here is the data he collected.

Gas Volume Mass Hydrogen 200 mL 0.0179 gOxygen 100 mL 0.1429 g

Divide 0.1429 g of oxygen gas by 0.0179 g of hydrogen gas to get the relative mass ratio of oxygen to hydrogen.

0.1429 g oxygen gas = _____________

0.0179 g Hydrogen gas

What did you get? (Don't be lazy! Go back and do the problem.) Did you put any units on your answer? For units you should have had only oxygen gas over hydrogen gas since grams divides out in this case.

In 1808 John Dalton published one of the first tables of atomic masses and actually listed the mass ratio of hydrogen to oxygen as 1 to 8 based upon his observations. He thought that the formula for water was HO with one atom of hydrogen to one atom of oxygen. In the above case the weight ratio of the produced gases was 1 to 8 but our sample of hydrogen gas was twice as large as our sample of oxygen gas. The formula for water is H2O so the mass ratio of one hydrogen atom to one oxygen atom is 0.5 to 8 (divide the 1 by 2 since there are two hydrogens) which is also 1 to 16.

The original goal was to find the atomic masses of hydrogen and oxygen but all that we really found was mass ratios of hydrogen and oxygen from a sample of water. Look at the periodic chart. Find the atomic mass of hydrogen. Use the key on the chart if you are not sure which number is the atomic mass. Look up the atomic mass of oxygen also.

Atomic mass of hydrogen (to the nearest whole number) =______Atomic mass of oxygen =______

These are relative masses of the elements compared to some common standard. Each of the values on the chart has been determined experimentally in much the same was as our example above. You can now say one atom of hydrogen weighs 1, one atom of carbon weighs 12, one atom of nitrogen weighs 14, and one atom of oxygen weighs 16. These are the values from the chart. The next question to ask is 1, 12, 14, or 16 what? The answer is that these are just numbers comparing the respective masses to a set standard. Just as a person weighing 150 lbs is 1.5 times heavier than a person weighing 100 lbs, oxygen (16) is 1.33 times heavier than carbon (12). These atomic masses have no real units such as ounces or grams. Arbitrary units of amu (atomic mass units) or au (atomic units) are assigned to these numbers when the need for units arises.

Since it is possible to find relative atomic weights, it should also be possible to compare masses of molecules relative to the same common standard. For water, there are two hydrogens with each weighing 1 for a total of 2 and one oxygen weighing 16 for a total of 16.

Hydrogen 2 atoms x 1.01 = 2.02Oxygen 1 atom x 16.00 = 16.00 Water 1 molecule H2O = 18.02

For most of our problems we will need to keep the number of decimal places that the proper use of significant figures requires. For now, express your answers to the nearest hundredth.

Many textbooks will discuss the term molecular mass and the term formula mass. Although you learned to write the formula of table salt, sodium chloride, as NaCl, it does not really occur as individual NaCl molecules but in a crystalline form as shown.

Sodium atoms (ions actually) are connected to several chloride ions in a network called a crystal lattice. There is usually one chlorine for one sodium so the formula is written as NaCl. It would be correct to say formula mass of NaCl rather than molecular mass. The masses of those compounds that occur in such crystalline structures should be called formula masses. In fact, the term "formula mass" can be used for any compound including those that occur in individual molecular units.

Here are two more examples of finding formula masses. The two are calcium hydroxide, Ca(OH)2, and copper(II) sulfate pentahydrate, CuSO4 5H2O. Each introduces something new.

Formula mass of Ca(OH)2:

Method 1 Method 2

Ca 1 atom x 40.08 = 40.08 Ca 1 atom x 40.08 = 40.08 O 2 atoms x 16.00 = 32.00 OH 1 atoms O x 16.00 H 2 atoms x 1.01 = 2.02 1 atom H x 1.01

74.10 17.01(OH)2 2 OH's x 17.01 = 34.02

74.10

Formula mass of CuSO4 5 H2O:

Method 1 Method 2Cu 1 atom x 63.54 = 63.54 Cu 1 atom x 63.54 = 63.54 S 1 atom x 32.06 = 32.06 S 1 atom x 32.06 = 32.06 O 4 atoms x 16.00 = 64.00 O 4 atoms x 16.00 = 64.00 H 10 atoms x 1.01 = 10.10 159.60 O 5 atoms x 16.00 = 80.00

249.70 H 2 atoms x 1.01 = 2.02 O 1 atom x 16.00 = 16.00 5 H2O molecules x 18.02 =

90.10Total = 249.70

Now try these exercises:

Find the molecular of formula masses:

1. HNO3, nitric acid2. NaOH, sodium hydroxide3. C12H22O11, sucrose or table sugar4. C2H5OH, ethanol or ethyl alcohol5. CCl4, carbon tetrachloride6. Ca3(PO4)2, calcium phosphate7. Na2CO3 10 H2O, sodium carbonate 10 hydrate8. Cl2, chlorine gas9. When is the term formula mass more correct than the term molecular

mass?10. What are the correct units for the mass of one atom of gold (taken

from the periodic table)?11. What does it mean to say that the atomic mass of lithium is 6.94 and

the atomic mass of copper is 63.54?*12. In a chemical reaction 50 pounds of sulfur combined with 50 pounds of

oxygen to form a toxic pungent gas. The formula for the gas must be either SO2 or SO3. Refer back to the electrolysis experiment and periodic chart to discover which must be the correct compound formed.

CHAPTER 1 - THE MOLESECTION 2 - WHAT IS A MOLE?

In the previous section we discovered that the atomic masses on the periodic chart are really relative atomic masses or masses compared to some standard. Nitrogen, for example is approximately 14 times heavier than hydrogen. In this section, we will see how this ratio is very useful for working with very large numbers of molecules.

Objectives:1. Describe why 23 pounds of sodium contain the same number of atoms

as 12 pounds of carbon, 31 pounds of phosphorus, 27 pounds of aluminum, etc.

2. Repeat the numerical value of Avogadro's Number.

3. Define the term "mole".4. Given the number of particles of a substance, calculate the number

of moles of particles.5. Given the number of moles of particles of a substance, calculate

the number of particles.6. Differentiate between the units used for a particle of a substance

and a mole of particles of that substance.7. Given the number of moles of particles of a substance, calculate

the mass of the particles.8. Given the mass of a substance present, calculate the moles of

particles of that substance.9. Given the mass and number of moles of a substance, calculate the

molecular or formula mass. 10. Given the mass present of a compound, calculate the number of

atoms of an element within that compound.

In the electrolysis of water example and sulfur dioxide exercise we were able to deduce the mass ratios of the atoms. It did not matter that the first masses obtained were in grams and the second were in pounds. The relative atomic and molecular masses are independent of the units used as long as the units of the two substances being compared are the same.

The relative masses of some elements are hydrogen, 1; carbon, 12; nitrogen, 14; and oxygen, 16. The masses of 10 atoms would be 10, 120, 149, and 160 respectively. The mass of each atom was multiplied by 10. Divide each mass by the greatest common divisor for the set to find that the relative masses are still 1:12:14:16. This time we shall take 1000 atoms of each. The masses would be: hydrogen,_______; carbon,______; nitrogen,______; and oxygen,______. What is the greatest common divisor for

this set of numbers?___________ What are the relative weights when divided by the g.c.d.?

We will look at one more example. This time we will take ______ atom. The masses will be: carbon, 9000; fluorine, 14,250; and sodium 17,250. The relative masses are 12 for C, 19 for F, and 23 for Na. If you have trouble with this one, be sure to ask for help.

In the above example with carbon, fluorine, and sodium, the number of atoms came out the same since each of the masses given was a multiple of 750. One gram of hydrogen should therefore contain the same number of atoms as nineteen grams of fluorine atoms or twenty-three grams of sodium atoms. Twelve ton of carbon should contain the same number of atoms as 197 tons of gold.

By now you may be saying, "Fine, but how many atoms are in twenty-three grams of sodium?". It was difficult enough to get 750 with simple math. It has been experimentally determined that there are 6.02 x 10 23 atoms in 12 grams of carbon-12 (a particular isotope of carbon). There are three questions that may come to mind concerning this. The answer to one is that the methods used to find this number, except for an indirect approximate method, are rather too complicated to be discussed her. A second question might be that 6.02 x 1023 contains quite a bit of inaccuracy. You are right. Chemists have actually been able to calculate the number more precisely, but our calculations will not demand this.

A third question could be, "How large is 6.02 x 1023 ?". 602,000,000,000,000,000,000,000 certainly takes up more paper space than does one million, 1,000,000 (1 x 106). If we were to add up all of the people who are now alive and have ever lived, we could not even come close to that number. Only about ten times that number of sand grains exist on earth! Indiana is one of the world's largest popcorn producers. If you were to pop 6.02 x 1023 kernels of popcorn, you could make a popcorn ball about one-tenth the size of the moon. The assumption is made, of course, that there are no "old-maids" among the bunch.

How many carbons atoms in a two-ton load of coal? Remember that 12 grams of carbon (about the mass of four pennies) contain 6.02 x 1023 atoms of carbon. A two-ton load of soft Indiana coal is about 75% pure carbon so 2 ton x 0.75 = 1.50 ton of carbon from a two-ton load. Change the 1.50 ton of carbon to grams of carbon:

1.50 tn x 2000 lb x 1 kg x 1000 g = 1.36 x 106g of carbon 1 tn 2.205 lb 1 kg

Now we know that 12 grams of carbon contains 6.02 x 1023 carbon atoms. Either 12 grams of carbon 6.02 x 1023 carbon atoms

or

6.02 x 10 23 carbon atoms 12 grams of carbon

can be used in dimensional analysis problems just as the weight conversions in the problem above. So far we have calculated that two ton of soft coal contains about 1.36 x 106g of carbon. Since every 12 grams contains 6.02 x 1023 carbon atoms, dimensional analysis allows us to set up the problem as follows:

1.36 x 106g of carbon x 6.02 x 10 23 carbon atoms = 6.83 x1023 carbon atoms 12 g of carbon

That is a very large number of carbon atoms. As you do more calculations with extremely large numbers, keep in mind the comparisons on this page. The size of a single atom must indeed be very small.

How many atoms are in 23 grams of sodium?_________________

Does this seem like a mystery question? After all that discussion on large numbers, you may have forgotten that 23 grams of sodium should contain the same number of atoms as 12 grams of carbon. (You may wish to review the first five paragraphs of this section.) Did you say 6.02 x 1023 sodium atoms? Good. If you are not sure, be sure to ask for help before you go on.

Here is a new question for you. How many molecules are in 342 grams of sucrose, C12H22O11?________________________

Up to this point the discussion has been about atoms rather than molecules. In the previous section we were able to compare the weight of one atom to the weight of one molecule. For example, the weight of one atom of oxygen is 16 amu and the weight of one molecule of sucrose is 342 amu. 6.02 x 1023

is the number of oxygen atoms in 16 grams of oxygen so 6.02 x 1023 should be the number of molecules in 342 grams of sucrose. 6.02 x 1023 molecules of water should weigh __________ grams.

The number 6.02 x 10 23 is called Avogadro's Number after Amedeo Avogadro, an Italian scientist who formulated some very important ideas about matter in the 19th century. By now you should have read 6.02 x 1023

enough times to know it and probably think that there should be some shorter notation for it. The term mole is used for this purpose. One mole is defined as 6.02 x 1023 of anything.

_______________________________ONE MOLE = 6.02 X 1023 ANYTHING_______________________________

One mole of carbon atoms (or oxygen atoms or hydrogen atoms) is 6.02 x 1023 carbon atoms. One mole of moles is 6.02 x 1023 moles. Can you imagine what your garden would be like with a mole of moles?

Just a note of interest, the word mole comes from the Latin word "moles" which means a mass. The word molecule is a diminutive of "moles" and means "a small mass". It is helpful to remember the "cule" portion means small so that you do not confuse these two. The correct abbreviation for mole is mol. There is no abbreviation for molecule.

Since one mole is equivalent to 6.03 x 1023, it can be used in dimensional analysis. Always be sure to put the proper units on mole and 6.02 x 1023. Here are a few problems relating moles and molecules.

Example 1:3.54 x 1023 molecules of C2H5OH, ethanol, is equal to how many moles of ethanol?

3.54 x 1023 molecules C2H5OH x 1 mole C2H5OH = 6.02 x 1023 molecules C2H5OH

0.588 moles C2H5OH molecules

Example 2:7.35 moles of sodium atoms is equal to how many sodium atoms?

7.35 moles Na x 6.02 x 10 23 Na atoms = 4.42 x 10 24 Na atoms 1 mole Na

The number of atoms or molecules is a huge number compared to the number of moles. Many students try to put a number such as 1023 on the term mole when they are still attempting to understand the concept. This really makes no sense if you think about the relative sizes of these particles.

Exercises

Work the following exercises. Be sure to use the correct units.

1. Calculate the number of molecules in 3.50 mole of hydrogen molecules (H2).

2. Calculate the number of atoms in 3.50 moles of hydrogen atoms (H).

3. Calculate the number of moles of sodium atoms in 2.50 x 1023 sodium atoms.

4. Calculate the number of moles of methane, CH4, molecules in 2.10 x 1021 molecules of methane.

5. Calculate the number of moles of carbon tetrachloride molecules (CCl4) in 2.10 x 1025 molecules of carbon tetrachloride.

How much does one mole of gold atoms weigh?__________________

To answer this problem you must recall that one mole of gold is equivalent to 6.02 x 1023 gold atoms. From the periodic chart it can be seen that the atomic mass of gold is 196.97 amu. The mass of one mole of gold atoms is therefore 196.97 grams. The relative mass of a carbon atoms is 12 amu. The relative mass of an oxygen atom is 16 amu. Twelve grams of carbon contain 6.02 x 1023 atoms which is one mole. Sixteen grams of oxygen atoms contain 6.02 x 1023 atoms which is one mole.

12 g of C == 1 mole of C== 6.02 x 1023 atoms of C16 g of O == 1 mole of O== 6.02 x 1023 atoms of O196.97 g of Au == 1 mole of Au == 6.02 x 1023 atoms of Au

but

12 amu of C == 1 atom of C16 amu of O == 1 atom of O

What is the mass of one mole of sodium hydroxide,NaOH?______

Here you would just find the molecular mass (actually formula mass) which has either no units or units of amu's. The numerical value of the molecular mass is the mass of one mole with grams as the unit.

Exercises

Work the following to check your understanding. Express youranswers to the nearest hundredth.

1. The mass of one mole of NaCl is ________________.2. The molecular mass of water is _________________.3. The atomic mass of iodine is ___________________.4. The mass of one mole of fluorine atoms is _____________.5. A chemist said that the molecular weight of sulfur dioxide was 64.06

grams. Is that correct? Explain your answer.

The idea of the mole is one of the most used ideas in chemistry calculations. Dimensional analysis will again be very handy for this work. You will need to practice a few problems using grams and moles.

Example 1:A new lab assistant who had never studied high school chemistry was told to weigh out two and a half moles of salt for an experiment. After an initial few minutes of panic, he grabbed a shovel, a large knife, and asked another worker for a map of the nearby salt mines. After recovering from a long bout of hysterical laughter the worker informed him that a mole of salt is a number and not an animal that burrows in salt mines. The embarrassed new lab assistant then looked for a scales to weigh out 2.5 moles, but all he found were gram scales. He debated for a while whether he should ask the other workers for help again or apply for unemployment benefits. Desperately he ran out into the hallway and grabbed the first person he saw - you. Since his name was mugger, everyone ignored your screams. Your only choice was to help him solve the problem. Thankfully, the lab did have a periodic chart.

Solution: 2.500 mole NaCl = ______________g NaCl

The molecular weight of NaCl = 22.99 + 35.45 = 58.44 amu.The weight of one mole of NaCl = 58.44g

2.500 mol NaCl x 58.44 g NaCl = 146.10 g NaCl 1 mole NaCl

Example 2: The next day the lab assistant weighed out 45.53 grams of zinc and wondered how many moles he had. From the periodic table he discovered that zinc had an atomic mass of 65.37 grams. Right away (thanks to your good instruction from the previous day) he knew that he had less than one mole but did not know how to proceed from there. Help him out.

Solution:

45.53 g Zn x 1 mole Zn = 0.6965 moles Zn 65.37 g Zn _______________

Exercises

Work the following exercises before proceeding to the next section.

1. Calculate the number of moles of oxygen atoms in 48.0 g of oxygen atoms (O).

2. Calculate the number of moles of sulfuric acid in 0.200 grams of sulfuric acid, H2SO4.

3. Calculate the grams of nitrogen in 2.50 moles of nitrogen molecules, N2.

4. Calculate the pounds of potassium chloride in 10.0 mole of potassium chloride, KCl.

*5. Given that a sample of a compound contains 2.45 moles and weighs 74.5 grams, calculate the molecular mass of the compound.

Moles, molecules, and grams can all be interrelated with moles serving as the key.

In example 1, our desperate lab assistant could have been told to weigh out 1.50 x 1024 molecules of salt. Since NaCl comes in crystals rather than individual molecules, he would have had a very difficult time in deciding exactly what a molecule was. Of course, he could have tried counting little salt crystals. After the first thousand, he probably would have considered jumping out the window. It is possible to weigh out 1.50 x 1024 NaCl units, however, if this value is converted to grams.

1.50 x 1024 NaCl molecules x1 mole NaCl molecules x 58.44 g 6.02 x 1023 NaCl molecules 1 mole NaCl

molecules= 146 g NaCl

__________

Exercises 1. Calculate the number of grams of methane, CH4, in 1.20 x 1021 molecules of methane.

2. Calculate the number of molecules in 12.0 g of hydrogen gas, H2.

3. Calculate the number of milligrams of potassium chloride, KCl, molecules in 1.50 x 1023 potassium chloride molecules.

4. Calculate the number of molecules of sulfuric acid, H2SO4, in 60.0 pounds of sulfuric acid.

*5. What is the mass in grams of one molecule of H20?

How many oxygen atoms are in 75.0 grams of sucrose, C12H22O11?

Everything needed to solve this problem has already been covered in this unit. Give it a try on your own before you look at the answer.

One way to attack an unfamiliar problem is to list first what is given and then what is to be found. In this problem begin with 75.0 grams of

sucrose. The is no direct relationship between grams and the number of atoms in the sucrose molecule. List the relationships that you do know.

Known (or calculatable) relations:75.0 grams of C12H22O11 =___________oxygen atoms

342 g C12H22O11 = 1 mole C12H22O111 mole C12H22O11 = 6.02 x 1023 molecules C12H22O111 molecule C12H22O11 contains 11 oxygen atoms

This last relationship is the step that you may have overlooked but is very important. A very common mistake on this problem is to attempt to proceed directly from mole to atom. The moles you calculate are moles of molecules and should be labelled as such to prevent confusion. The steps in solving the problem are shown below.

75.0 g C12H22O11x 1 mole C12H22O11 x 6.02 x 10 23 molecules C 12H22O11 342 g C12H22O11 1 mole C12H22O11

11 oxygen atoms = 1.45 x 10 23 oxygen atoms 1 molecule C12H22O11 ________________________

Exercises 1. Give the number of hydrogen atoms in one molecule of cortisone, C21H28O5.

2. Calculate the number of oxygen atoms in 50.0 grams of carbon dioxide, CO2.

*3. Given that there are 2.03 x 1024 oxygen atoms present in a sample of nitric acid, HNO3, calculate the number of grams of nitric acid present.

4. Calculate the number of hydrogen atoms present in 2.50 moles of sulfuric acid, H2SO4.

*5. Calculate the number of millimoles present in a sample of sulfuric acid which contains 7.35 x 1022 atoms of hydrogen.

This chapter began with the question, "what does the formula H2O tell us?" At that time the answer was that H2O represented one molecule of water that contained two hydrogen atoms and one oxygen atom chemically combined. 2 H2O would then represent two molecules of water with each containing two hydrogen atoms and one oxygen atom for a total of four hydrogen atoms and two oxygen atoms. 1/2 H2O would represent what?

Is it a molecule of water that has been cut in half? How do we slice it? That is really a pretty silly question since a molecule is already the smallest unit of compound water. Is 1/2 H2 O illogical or is it possible?

The formula H2O can represent one molecule of water but it can also represent one mole of water. Much of the work in chemistry uses moles because they can be weighed. 2 H2O would then represent two moles of water or two molecules of water depending upon the context of the problem. 1/2 H2O could not represent half of a molecule but it could easily represent half a mole or 3.01 x 1023 molecules of water. How many grams of water would it represent?_________ How many atoms of hydrogen and how many atoms of oxygen?_____________________

If H2O represents one mole of water, what do the subscripts represent? For a molecule they may represent atoms but you cannot say that one mole (6.02 x 1023 molecules) of water contains two atoms of hydrogen and one of oxygen. The size of those three atoms would have to be tremendous and would cause all kinds of plumbing problems. One mole of H2O molecules does contain two moles of hydrogen atoms and one mole of oxygen atoms.

Exercises

1. One molecule of H2SO4 contains____________ atoms.2. One mole of H2SO4 contains _________ molecules of H2SO4.3. One mole of H2SO4 contains _________ atoms.4. One mole of H2SO4 contains _________ H atoms.5. One mole of H2SO4 contains _________ S atoms.6. One mole of H2SO4 contains _________ moles of H atoms.

Section Exercises Weigh out 342.30 g of sucrose, C12H22O11, and 58.44 g of sodium chloride,

NaCl. Use the specifically marked containers in the lab. Do not mix the two and, unlike most lab activities, put the substances back when you are finished with the following questions.

1. Which of these substances contains the most formula units? __________________________________________________________2. How many formula units are in the sucrose?________________3. How many moles are in the sucrose sample?_________________4. How many oxygen atoms are in the sucrose sample?__________5. How many moles of oxygen atoms are in the sucrose?________*6. What is the percentage by weight of oxygen in sucrose?_______

7. How many formula units are in the weighed salt?___________8. How many moles are in the salt sample?____________________9. How many moles of chloride ions are in the weighed salt?_______*10. What is the percentage by weight of the metallic ion in the salt sample?____________

CHAPTER 1 - THE MOLE

SECTION 3 - PERCENT COMPOSITION AND EMPIRICAL FORMULAS

Now that it is possible to interconvert grams, molecules, and atoms through the use of moles, of what use is this skill? This next section shows how chemical formulas and moles can be useful in real lab problems. All of the basic ideas have already been covered. This section will help you put them together.

Objectives:1. State the law of definite composition.2. Given a formula of a compound, calculate the percent composition by weight of each elemental species.3. Differentiate empirical (simplest) formula from molecular formula.4. Given the percent composition by weight of the elemental species, determine the empirical formula.5. Given the gram composition of the elemental species, determine the molecular formula.6. Given either the percent composition or gram composition and the molecular weight of a compound, determine the molecular formula.

This laboratory experiment should proceed Chapter 1, Section 3.

An entire class period may be needed to perform this experiment.

Partial Weight Analysis of a Compound

The blue copper (II) sulfate found in the lab is an example of hydrate. Hydrates have definite amounts of water incorporated chemically within the crystalline structure. The five moles of water molecules per mole of copper sulfate can be removed by heating. This will enable you to determine the percentage of water of hydration present in a sample.

1. Weigh a clean, dry crucible to the nearest O.O1g. ________g2. Grind up crystalline hydrated copper (II) sulfate in a mortar. Weigh out between 2.50 and 3.50 grams of the salt in the crucible. Be sure to clean up any spilled copper sulfate from the balance pan before you record the weight. Weight of sample and crucible _________g3. Weight of sample, calculated _________g4. Place the crucible and its contents on a pipestem triangle set on a ring clamped to a ring stand.

Heat the system gently for 3 to 5 minutes to avoid splattering. Continue to heat but more strongly for 10 more minutes. Record any observations during this time.

Observations:

Obtain a desiccator from the lab or make one from a large jar. The

bottom of the jar should have 1.2 inch solid anhydrous calcium chloride. A

commercial laboratory desiccator may have the lid coated with petroleum jelly so do not set

the lid down flat on the table. The top should be placed upon the jar as

soon as it is filled and not removed until necessary. The crucible will

rest on a wire screen or plate placed above the calcium chloride.

At the end of the 10-minute heating, allow the crucible to cool on the stand for 2 minutes and then use tongs to transfer it to the desiccator.

5. After the crucible and contents have cooled to room temperature in

the closed desiccator, weigh the crucible and solid. Weight of crucible and residue. ______________g6. Calculate the weight of the residue ______________g7. Calculate the weight of the water loss by the original sample ______________g8. Calculate the percentage of water of hydration in the sample ______________g9. Place a few drops of water on the cooled, weighed residue. Feel the bottom of the crucible before and after doing this. Record any observations.

Observations:

Questions:

1. If the hydrated copper (II) sulfate were heated strongly for a longer period of time, it would lose its water of hydration and also begin to decompose into black copper (II) oxide, CuO, and colorless, noxious, pungent sulfur trioxide. Would this cause your results to appear higher than expected or lower than expected? Explain your answer.

2. The lid of a commercial desiccator is not fastened to the top. After it has been on a prepared desiccator for a few minutes, it is often difficult to lift the lid. Why might this happen?

3. Use a dictionary to find the meanings of hydrate, desiccate, and anhydrous.

4. What color is the hydrated copper (II) sulfate?What color is the anhydrous copper (II) sulfate?

5. Hydrated sodium carbonate (washing soda) is often found in powdered laundry detergents. An opened, unused box of the detergent may weigh less than the shipping weight on the box. Account for this discrepancy.

"A rose is a rose is a rose. . ." is a famous quote from Gertrude Stein. It would also be correct to say H2O is H2O is H2O no matter where it is found. Water in the southwestern part of the United States may have a bitter alkaline taste and sea water may have a salty taste, but if all of the impurities were removed, there would be no difference. Salt found in a mine in Michigan has exactly the same composition as salt found in a mine in Poland. Commercially produced Vitamin C (ascorbic acid) has the same composition as Vitamin C found in orange juice. The observation that a pure compound always has the same elemental composition has long been known. It has been called the law of constant composition or law of definite proportions. When you learned to write formulas, you had to assume that all table salt combined to from NaCl rather than NaxCly with x and y representing numerical subscripts.

In the experiment with copper (II) sulfate five hydrate, you found the percent by weight of water in the compound. The weight loss was divided by the total weight of the compound. It is possible to find the percent composition by weight from the formula above. To find the percentage of hydrogen in water, take the weight of hydrogen (2 x 1.01) and divide by the total molecular weight of water ((2 x 1.01 ) + (16.00)).

2.02 x 100 = 11.2% of hydrogen in water by weight 18.02

What does 11.2% hydrogen in water indicate? If 100 grams of water were present, 11.2 grams of compound would be from hydrogen and 88.8 grams of the total weight would be from oxygen.

Exercises 1. Calculate the percent composition of carbon in ethyl chloride,

C2H5Cl.2. Calculate the percent composition of sulfur in aluminum sulfite, Al2(SO3)3.3. Calculate the percent composition of each elemental species in acetic acid, CH3COOH.*4. Calculate the percent of metal ion in a compound where 0.350 g of metal combines with 0.255 g of oxygen.*5. From the formula for copper(II) sulfate 5 hydrate, CuSO4 5

H2O, calculate the percent water found in the hydrated compound.

An entire class period may be needed to perform this experiment.

Experiment: The formula of a Compound

Each compound has its own distinct formula, as was just seen in the

previous section, has its own constant composition. In this experiment you are going to make a compound and then calculate its

simplest formula.

1. Weigh a clean, dry crucible and record its weight to 0.01g ________

2. Obtain a clean piece of magnesium ribbon about 35 cm in length. Use

a fine piece of steel wool to clean the ribbon. Cut the ribbon into

pieces approximately 1 cm and add to the crucible. Weigh the crucible and contents. _______

3. Weight of magnesium ribbon by difference. _______

4. Calculate the moles of magnesium (M.W of Mg = 24.31)_______

5. Set up the ring, triangle, and ring stand as in the previous experiment. Heat the covered crucible (picture 1) gently at first and then gradually increase the heat intensity for 2 minutes.

With your tongs, carefully tilt thelid (picture 2) and continue strongheating of the partially uncovered crucible for an additional 10 minutes.

At the end of this time, allow thecovered crucible and contents to cool.Check to see that all of the magnesiumhas reacted before covering. If thereis still unreacted magnesium, heatstrongly again.

Since the magnesium reacted with both the oxygen and nitrogen in the air, it is necessary to convert the mixture of compounds to only the oxide compound.

Use a medicine dropper to add only enough distilled water to cover the contents of the crucible. Carefully heat the system. Observe the odor of any vapor that has been produced. When the water has evaporated, add

a little more and carefully heat again. Continue to do this until there is no longer a distinct odor produced upon heating.

When this has happened, continue careful heating until all of the water has evaporated. Strongly heat the uncovered crucible for five more minutes. Allow the crucible and its product to cool. Weigh the crucible and its contents.

Weight of crucible and contents________

6. Determine the weight of the magnesium oxide by difference_____

7. Determine the weight of oxygen that combined with the original weight of magnesium(You know how much Mg is present in the compound.) ________

8. Calculate the number of moles of combined oxygen ________

9. Determine the simplest formula for magnesium oxide. (If you are not sure what to do here, read ahead in section 3.)

When samples of unknowns are sent in for analyses, the results often come back expressed as percentages. An example might be that an unknown nitrogen - oxygen compound was found to contain 63.6% N and 36.4% O weight. From this information it is possible to calculate a simplest or empirical formula. Remember that 63.6% N means that out of every hundred grams, there are 63.6 g of nitrogen present. How does this help with the problem? Keep in mind that we are trying to find the subscripts x and y in the formula NxOy. It was mentioned in the last section that the subscripts can represent not only the number or atoms of each elemental species per mole of molecules. If the grams of each species could be changed to moles, the moles could serve as subscripts for the formula.

Example 1: A compound consists of 63.6% N and 36.4% O. What is the empirical formula of the compound?

Solution: 63.6% N and 36.4% O by weight means 63.6 g N and 36.4 g O out of 100 g of

compound

Assume that we have 100 g of compound. We would have 63.6 g N and 36.4 g O in our

sample. Change these gram values to moles.

63.6 g N x 1 mole N = 4.54 mole N 14.0 g N

36.4 g O x 1 mole O = 2.28 mole O 16.0 g O

This would give an empirical formula of N4.5402.28. Having fractional subscripts would be

no problem if the subscripts represented only moles. Since they can also represent

atoms, it is necessary that they be whole numbers. To find the whole number ratio represented by these fractional subscripts, divide both of the numbers by the smaller number to make the smaller value equal to zero.

In this case N4.54/2.28)2.28/2.28 nicely works out to N2O1 which is N2O or the simplest formula of the compound.

If this problem had state that 3.18 g of N and 1.82 g of O were present in a nitrogen - oxygen compound, the moles of nitrogen and oxygen could be calculate directly without using percents and assuming that 100 g sample was present.

The subscripts in Example 1 worked out very nicely to a 2 to 1 ratio of N to O. What do you do if the subscript ratios work out to something like 1 to 1.25? The decimal portion of 0.25 is 1/4 in fractions. If 1.25 is multiplied by 4, a whole number will result. For a ratio such as 1 to 1.25, when one number is multiplied by 4 the other number must also be multiplied by 4. The ratio is now 4 to 5.

Example 2: A compound of arsenic and oxygen was found to contain 75.8% arsenic. What is the empirical formula of the compound?

Solution: Work out the first part of this problem yourself? The ratio of moles of atoms comes out approximately

1 As to 1.5 O. Multiply both numbers by 2 to get the final answer of As2O3 for the empirical formula.

When is the empirical or simplest formula not the actual formula of a compound? At first two phosphorus compounds were thought to have formulas of P2O3 and P2O5, but it was later discovered that the real formulas were P4O6 and P4O10. Many organic compounds have multiples of the simplest formula for their real formulas. Ethane, C2H6, has an empirical formula of CH3. There is no compound with the actual formula of CH3. How can we tell what the real formula for a compound is? If the molecular weight as well as the percent composition by weight of a compound is known, the actual formula can be found.

Example 3:A hydrocarbon compound with a calculated molecular weight of 78.8 was found to consist of 92.3% carbon of

7.7% hydrogen. Calculate the actual formula of the compound.

Solution: To solve this, initially ignore the 78.8 and use the percentage data as before. Calculate the empirical

formula.

The empirical formula should be C1H1. The real formula must be a multiple of C1H1 so the actual molecular weight must be a multiple of the weight of C1H1.

The weight of C1H1 is approximately 13. Divide 78.8 by 13 to get 6.06. This number is close enough to 6 by significant digits. The actual molecular formula is

C6H6. This compound is called benzene.

*Just a word of warning. 6 C1H1 is not the same as C6H6 (CH)6 also represents something different. Sometimes it

is very difficult to know whether the correct solution is C6H6 or has units such as Ca(OH)2 included in it.

6 CH is never correct.

Exercises 1. The analysis of a compound shows that the compound contains 26.67% S, 53.33% O, and 20.00% Mg. Calculate its simplest formula.

2. A sample of an oxide of iron contains 27.59% O and 72.45% Fe. Calculate its empirical formula.

3. Upon analysis, 51.0 grams of aluminum oxide are found to contain 27.0 grams of aluminum. Calculate the simplest formula of the aluminum oxide.

4. A compound has the following composition by weight:C, 57.1% H, 4.8% O, 38.!%

a. If the atomic weights of carbon, hydrogen, and oxygen are respectively 12.0, 1.01, and 16.0, calculate the simplest formula for this compound.

b. If the true molecular weight is shown to be 126, what must be the true molecular formula?

Summary for Chapter 1:

Many new ideas that are extremely important have been covered in this chapter. We saw that although we cannot weigh atoms and molecules directly, we can obtain their weights in comparison to each other. These relative weights (both atomic and molecular or formula) hold between atoms and

molecules whether we are talking about one or ten million atoms or molecules. This enables us to compare the weights of the same number of atoms in common units. One gram of hydrogen atoms, 12 grams of carbon atoms, and 16 grams of oxygen atoms should contain the same number of atoms.

It has been experimentally determined that 0.012 kg of carbon 12 contains Avogadro's Number (6.02 x 1023) of carbon atoms. The mole (mol) is 6.02 x 10 23 of anything. The weights on the periodic chart can represent the weights of the individual atoms in amu's or the weights of moles of atoms in grams. Weights of molecules can be calculated by adding up the atomic weights. The weights of moles of molecules can be found by the same method.

It is possible to interconvert numbers of atoms, molecules, moles, and grams using moles as the key. The number of moles is really the number of particles we are dealing with.

6.02 x 1023 M.W. (atoms)<------>Molecules<------------->MOLES<---->grams

The law of definite composition has long been known. It would be impossible to do much chemistry if the formulas of compounds kept changing. Finding empirical or simplest formulas depended upon having constant composition as well as having subscripts that could represent atoms in one case and moles of atoms in another. This was one important use of mole that we saw in this chapter. In the next chapter we will look at how it is used in chemical reactions.

To be sure that you really understand the material in this chapter, you need to do some review exercises. When you were learning to spell in elementary school, you learned the list of words but were rather upset when your mother gave them to you out of order. As you know now, you do not spell in that order in real life. In chemistry you do not have problems presented to you in a definite order either, so you need to recognize what the problems are all about to do them better. Try to work as many of these as you can without looking back through this chapter.

Chapter Review Problems 1. How much does 5.22 x 1023 atoms of nitrogen weigh?2. How many molecules are present in 48.5 g of HBrO3?3. Calculate the mass in grams of 1.5 moles Ca(OH)2.4. Calculate the number of moles in 52.5 g of Zn(HCO3)2.5. Calculate the number of molecules in 52.5 g of Zn(HCO3)2.6. Calculate the number of moles of atoms in 52.5 g of Zn(HCO3)2 7. Calculate the percent by weight of hydrogen in Zn(HCO3)2.8. Calculate the total number of atoms in 52.5 g of Zn(HCO3)2.9. Calculate the number of zinc atoms present in 52.5 g of Zn(HCO3)2.10. Find the percentage of nitrogen in ammonium nitrate, NH4NO3.11. A photography developing compound was found to contain 36.5% sodium, 25.4% sulfur, and 38.1% oxygen. Calculate its simplest

formula.12. Analysis of a compound gives 90.6% Pb and 9.4% O. What is the empirical formula of this compound?13. Calculate the mass of the metal ion in 50.0 g of MgS.14. Calculate the moles in 1.505 x 1023 molecules of NaOH.15. Calculate the grams in 1.505 x 1023 molecules of NaOH.16. What is the molecular weight of CO2?17. If the molecular mass of an oxide of nitrogen is 108, and 4.02g of N combine with 11.48 g of O, what is the molecular formula of this compound?18. What is the weight of one mole of CO2?19. How many atoms does one molecule of H2O contain?20. How many moles of atoms does one mole of H2O molecules contain?21. What is the mass in grams of an atom of carbon?22. Define mole.23. Which is larger, Avogadro's Number or the number of molecules in 66 g of CO2?24. Calculate the mass in grams of one mole of Na2SO4 10 H2O.25. Blue CuSO4 5 H2O and NaSO4 10 H2O are known as hydrates. What are they called when they no longer contain H2O molecules?

ChemistryChapter 1

Alternate Exercises

Show the calculation setups and answers for all problems.

Find the formula weight (or molecular weight) of:1. Phosphoric acid, H3PO42. Sodium bisulfate, NaHSO43. Bismuth(III) nitrate, Bi(NO3)4. A sample of mercury(II) iodide, HgI2, weighs 9.42 grams. How many moles are in this sample?5. What is the weight of 0.l20 mole of ammonium carbonate, (NH4) CO3?6. How many molecules are contained in 3.65 moles of chlorine, Cl2?Calculate the percent composition by weight of barium sulfate, BaSO4 7. Ba8. S9. O

10. How many moles are present in 8.42 grams of sodium chlorate, NaClO3?A sample of oxygen gas, O2, weighs 30.0 grams. How many molecules and how many atoms are present in this sample?11. Molecules?12. Atoms?

*13. A mixture of sand and salt is found to be 42 percent NaCl by weight. How many moles of NaCl are in 36 grams of this mixture?14. What is the weight of 8.5 x 1022 molecules of carbon dioxide, CO2?15. A sample of the mineral crysoberyl contains 2.25 g Be, 13.49g of Al, and 4.00 g O. What is its empirical formula?16. What is the empirical formula of a compound that is 47.7% K, 13.2% B, and 39.1% O?17. If the compound in problem 16 has a molecular weight of 246, what is its molecular formula?*18. Pewter is 85% tin and 15% antimony. How many moles of antimony are contained in 50 g of pewter?*19 What mass of silicon is combined with 6.40 g of oxygen in the

compound SiO2?*20. If one atom of an element has a mass of 3.330 x 10-22g, what is that element's atomic weight?

Name______________________Date______________________

Chemistry QuizChapter 1 - Section 3

"Percent Composition and Empirical Formulas"

Show your set-up on any calculation problem. You may use a Periodic Chart and a calculator.

1. Five samples of mercury(II) oxide (HgO) were found to contain 92.6% mercury and 7.4% oxygen. How does this illustrate the Law of Definite Composition?

2. Find the percent composition by weight of aluminum in aluminum hydroxide {Al (OH)3}.

3. Find the percent composition by weight of oxygen in sulfuric acid

(H2SO4).

4. Identify each of the following as simplest formula, molecular formula, or both.a. HOb. P4O10c. H2Od. C12H22O11e. N2O4

5. A compound is found by analysis to contain 75.0% carbon and 25.0% hydrogen. What is the empirical formula?

6. A compound is found by analysis to contain 19.57 g iron and 8.42 g oxygen. What is the simplest formula?

7. The analysis of a gas reveals that it consists of 92.3% carbon and 7.7% hydrogen. Its molecular weight is 26. What is the molecular formula?

Name____________________

Date____________________

Chemistry QuizChapter 1 - Section 3

"Percent Composition and Empirical Formulas"

Show the set-up in each calculation problem to receive credit. Label each answer. You may use a Periodic Chart and a calculator.

1. Identify each of the following as simplest formula, molecular formula or both.

a. C12H22O11b. C2H6c. H2Od. COe. HO

2. The molecular weight of hemoglobin is 68,000. Each molecule of hemoglobin contains 4 atoms of iron. What is the percent by weight of iron in hemoglobin.

3. Benzene has an empirical formula of CH and a molecular weight of 78. What is the molecular formula for benzene?

4. Determine the simplest formula for a compound with 36.5 percent sodium, 0.8 percent hydrogen, and 24.6 percent phosphorus, and 38.1 percent oxygen.

5. Cyclopropane contains only two elements, carbon and hydrogen. If a sample is found to contain 3.00 grams of carbon and 0.250 grams of hydrogen, what is the simplest formula?

6. If the molecular weight of cyclopropane is found to be 39, what is its molecular formula?

7. Every sample of pure limestone is found to contain 40.0 percent calcium, 12.0 percent carbon, and 48.0 percent oxygen. Explain why one would not expect these values to change from sample to sample.

ANSWERS TO EXERCISES

Chapter 1 - Section 1

1. HNO3 MW = 63.01 The units and either amu's or nothing2. NaOH MW = 40.003. C12H22O11 MW = 342.294. C2H5OH MW = 46.065. CCl4 MW = 153.826. Ca3(PO4)2 MW = 310.197. Na2CO3 10H2O MW = 286.138. Cl2 MW = 70.919. The term formula weight is a more correct term for compounds that occur as crystal lattices.10. The units for relative weight of one atom of gold could be amu's, au's, or nothing. The mass of one atom of gold in grams will be calculated later.11. This means that copper is 63.54 times heavier than lithium. 6.4412. According to the electrolysis problem, we can divide the weight of sulfur by the weight of oxygen. 50lb sulfur = 1 sulfur 50lb oxygen 1 oxygenThe unit of pounds divides our. This means that the relative weight of sulfur is equal to the relative weight of oxygen. Sulfur weighs 32.06 and oxygen weighs 16.00. In SO2, there are two oxygens which would weigh 32.00 together. In SO3, the three oxygens would weigh 48.00. SO2 is the compound with sulfur and oxygen weighing the same so SO2 is the answer.

Section 2

page 121. 3.50 moles H2 x 6.02 x 10 23 molecules = 2.11 x 10 24 H 2 molecules 1 mole H22. 3.50 moles H atoms x 6.02 x 10 23 H atoms = 2.11 x 1024 H atoms

1 mole H atoms3. 2.50 x 1023 Na atoms x 1 mole Na atoms = 0.415 mole of Na atoms 6.02 x 1023 Na atoms4. 2.10 x 1021 molecules CH4 x 1 mole CH4___________ =

6.02 x 1023 molecules CH4

.00349 moles CH4 molecules 3.49 x 10-3 moles CH4

5. 2.10 x 1025 molecules CCl4 x 1 mole CCl4 =6.02 x 1023 molecules CCl4

34.9 moles CCl4 molecules

Page 14

1. 1 mole NaCl has a mass of 58.44 g 2. 18.01 or 18.01 amu but not 18.01 grams. This is for moles only.3. AW of I = 126.90 4. 1 mole of F = 19.00g 5. No. The weight of a mole of sulfur dioxide is 64.06 grams but molecular weight has units of amu or nothing. Molecular weight is a companion, not an actual weight.

Page 15

1. 48.0g 0 atoms x 1 mole 0 atoms = 3.00 mole 0 atoms 16.0g 0 atoms

2. 0.200g H2SO4 x 1 mole H2SO4 = 0.00204 mole H2SO4 98.08g H2SO4 3. 2.50 moles of N2 x 28.01g N2 = 70.03g N2 1 mole N2 My calculation may round off numbers differently than yours. Do not be concerned if your last digit and mine do not agree.4. 10.0 mole K Cl x 74.56g KCl x 1 lb = 1.64 lb KCl 1 mole KCl 454g This last step is a little different. If you do not know the English to metric conversions, look them up in your text when you need them.5. One mole of a compound weighs a certain number of grams. It can be expressed as ? grams which can equal ? grams. ? moles 1 mole 73.5 grams = 30.0 grams. The mass of one mole is identical 2.45 mole 1 mole to the molecular weight in amu's. The answer is 30.0 amu.

Page 16

1. 1.20 x 1021 CH4 molecules x 1 mole CH4___ x 16.04g CH4 = 6.02 x 1023 CH4 1 mole CH4 0.320g CH4

2. 12.0g H2 x 1 mole H2 x 6.02x10 23 H 2 = 3.58 x 10 23 H 2 molecules 2.02g H2 1 mole H2

3. 1.50x1023 KCl molecules x 1 mole KCl x 74.55g KCl x 1000mg 6.02x1023 KCl molecules 1 mole KCl 1g = 1.86 x 10 4 mg KCl This involves a metric-to-metric conversion.

4. 60.0 lb H2SO4 x 454g x 1 mole H2SO4 x 6.02 x 10 23 H 2SO4 molecules= 1 lb 98.08g H2SO4 1 mole H2SO4

1.67 x 10 26 H 2SO4 molecules

5. 1 molecule H2O x 1 mole H2O x 18.01g = 2.99x10 -23 g H 2O 6.02 x 1023 molecules H2O 1 mole H2O

One molecule of H2O is not very large, is it?

Page 17

1. One molecule of cortisone has 28 H atoms.

2. 50.0g CO2 x 1 mole CO2 x 6.02 x 10 23 CO 2 molecules x 2 0 atoms 44.0lg CO2 1 mole CO2 1 CO2 molecule

= 1.37 x 10 24 atoms of O

3. One molecule of HNO3 has 3 atoms of O 2.03 x 1024 O atoms x 1 HNO3 molecules 1 mole HNO3 x 3 atoms of O 6.02 x 1023 molecules HNO3

63.01g HNO3 = 70.8g HNO3 1 mole HNO3 This was a different problem.

4. 2.50 moles H2SO4 x 6.02x10 23 H 2SO4 molecules x 2 atoms H 1 mole H2SO4 1 molecule H2SO4

= 3.01 x 10 24 atoms H

Did you get confused and try to use grams? Read each problem and think.

5. 7.35 x 1022 H atoms x 1 H2SO4 molecules x 1 mole H2SO4 x 2 H atoms 6.02 x 1023 molecules

1000 millimole H2SO4 = 1 mole H2SO4

61.0 millimoles H2SO4

This is also a difficult problem. The problem, itself, did not state that the millimole were to be of H SO . That was my mistake and I hope it did not confuse you too much.

Page 19 - top of page

1. 1 H2SO4 molecule contains 7 atoms2. 1 mole H2SO4 contains 6.02 x 10 23 molecules H2SO43. 1 mole H2SO4 contains 6.02 x 10 23 x 7 = 4.21 x 10 24 atoms 4. 1 mole H2SO4 contains 6.02 x 10 23 x 2 = 1.20 x 10 24 H atoms

5. 1 mole H2SO4 contains 6.02 x 10 23 x 1 = 6.02 x 10 23 S atoms6. 1 mole H2SO4 contains 2 moles of H atoms

P 19 - Exercise

1. MW of sucrose = 342.20 MW of sodium chloride = 58.44 342.30g of C12H22O11 and 58.44g of NaCl are both one mole. Each would have 6.02 x 1023 molecules.

2. 6.02 x 1023 molecules

3. 1 mole

4. 1 mole C12H22O11 x 6.02 x 10 23 Molecules C12H22O11 x 11 O atoms 1 mole C12H22O11 1 mole C12H22O11

= 6.62 x 10 23 O atoms

5. There are 11 moles of oxygen atoms.

6. 11 O atoms x 16.00g = 176.00g atoms

176.00g x 100 = 51.43% O Do not worry about this problem. 342.20g You will look at this in the next section.

7. 6.02 x 10 molecules

8. 1 mole of NaCl

9. 1 mole of chloride ions

10. 22.99g x 100 = 39.34% Na Do not worry about this problem yet. 58.44g

Section 3

P.24

1. MW of C2H5Cl = 64.51 24.02g C x 100 = 37.23% C 64.51g C2H5CL

2. MW of Al2(SO3)3 = 294.15 96.19g S x 100 = 32.70% S 294.15g Al2(SO3)33. MW of CH3COOH = 60.05 24.02g C x 100 = 40.00% C 60.05g CH3COOH

4. 0.350g of metal 0.350g metal x 100 = 57.9% metal

+0.255g of oxygen 0.605g compound .605g of metal oxide

The method for find percent by weight is still identical.

5. CuSO4 5H2O 180.14g H2O x 100 = 53.02% H2O MW = 339.74 339.74g CuSO4 5H2O

P. 29 Exercises

1. Assume 100g of compound is present There is a 26.67g S x 1 mole S atoms = 0.8319 moles a mistake in 32.06g the problem. Hg should be 53.33g O x 1 mole O atoms = 3.333 moles O Mg. 16.00g

20.00g Mg x 1 mole Mg = 0.8227 moles Mg 24.31 g

Mg .8227 S 0.8319 O 3.333 <-- 1 st in order of metallic character Mg .8227 S 0.8319 O 3.333 --> Mg S 1.011 O 4.051 --> MgSO4 .8227 .8227 .8227

2. Assume 100g of compound - 27.59g O and 72.41g Fe 27.59g O x 1 mole O = 1.724 mole O atoms 16.00g O 72.41g Fe x 1 mole Fe = 1.296 mole Fe atoms 55.85g Fe Fe 1.296 O 1.724 --> Fe 1.296 O 1.724 --> Fe1 O 1.330 --> Fe3O4

3. 51.0g of aluminum oxide - 27.0g aluminum = 24.0g oxygen 27.0g x 1 mole Al = 1.00 mole 24.0g O atoms x 1 mole O atoms= 26.98g Al Al 16.0g

1.50 mole O Al1O 1.50 --> Al2O3

4. (a) Assume 100g of compound57.1g C x 1 mole C atoms = 4.76 mole C

12.0gC4.8g H x 1mole H atoms = 4.75 mole H

1.01g H38.1g O x 1 mole O atoms - 2.38 mole O

16.0gOC 4.76 H 4.75 O 2.38 --> C 4.76 H 4.75 O 2.38 --> C2H2O1

2.38 2.38 2.38

(b) C2H2O1 's weight adds up to 42. The real weight is 126.126 = 3 so 3 units of C2H2O1 are needed.

42The true formula is C6H6O3. {Note: 3C2H2O1 is wrong.}

Page 31 - Review problems

1. 5.22 x 1023 N atoms x 1 mole N atoms x 14.01g N = 12.1gN atoms 6.02 x 1023 atoms N 1 mole N atoms

2. 48.5g HBrO3 x 1 mole HBrO3 x 6.02 x 10 23 HBrO3 molecules = 128.92g HBrO3 1 mole HBrO3

2.26 x 10 23 HBrO 3

3. 1.5 moles Ca(OH)2 x 74.09g Ca(OH)2 = 111g but with only 2 sig. 1 mole Ca(OH)2 figs. The answer must be 110g Ca(OH)2

4. 52.5g Zn(HCO3)2 x 1 mole Zn(HCO3)2 = 0.280 mole Zn(HCO3)2 187.41g Zn(HCO3)2

5. From #4 0.280 moles Zn(HCO3)2 x 6.02 x 10 23 molecules = 1 mole Zn(HCO3)2 1.69 x 10 23 Zn(HCO 3 ) 2 molecules

6. 0.280 moles Zn(HCO3)2 x 11 moles atoms = 3.08 moles of atoms 1 mole Zn(HCO3)2

1 molecule of Zn(HCO3)2 contains 11 atoms 1 mole of Zn (HCO3)2 molecules contains 11 moles of atoms This is a new type of problem but uses old ideas. There are other ways of solving it also.

7. 2.02g H x 100 = 1.08%H 187.41g Zn(HCO3)2

8. 52.5gZn(HCO312 x 1 mole Zn(HCO3)2x6.02x10 23 Zn (HCO3)2 moleculesx 187.41gZn(HCO3)2 1 moleZn(HCO3)2

11 atoms = 1.86 x 10 24 atoms 1 molecule Zn(HCO3)2

We could have started the answer to #6 also. 3.08 moles of atoms x 6.02 x 10 23 atoms = 1.85 x 10 24 atoms 1 mole atoms

Notice the difference of 1 in the final digit. The last sig. digit is always uncertain and can vary.

9. 52.5gZn(HCO3)2 x 1 mole Zn(HCO3)2x6.02x10 23 Zn(HCO 3)2moleculesx 187.41gZn(HCO3)2 1 mole Zn(HCO3)2

1 atom Zn 1.69 x 10 23 Zn atoms 1 molecule Zn(HCO3)2 = 10. 28.01g N x 100 = 35.0% N 80.04g NH4NO3

11. Assume 100g of compound 36.5g Na x 1 mole Na atoms = 1.59 mole Na atoms 22.99g Na 25.4g S x 1 mole S atoms = 0.792 mole S atoms 32.06g S 38.1g O x 1 mole O atoms = 2.38 mole O atoms 16.00g O

Na2SO3 <--- Na2.01 S1 O3.01

12. Assume 100g90.6g Pb x 1 mole Pb atoms = 0.437 mole Pb

207.19g9.4g O x 1 mole O atoms = 0.59 mole O

16.00gO

Pb3O4 <-- multiply x 3 <--- Pb1O1.35

13. Calculate the percent of metal first in MgS.24.31gMg x 100 = 43.13% 43.13% of 50.0g of MgS must be

56.37gMg5 Mg.

43.13gMg x 50.0g MgS = 21.6g Mg 100g MgS

14. 1.505 x 1023 moleculse NaOH x 1 mole NaOH x 40.00g NaOH = 6.02 x 1023 molecules 1 mole NaOH

10.0g NaOH Of course, you could have begun with the answer to #14.

16. 44.01 amu's - Did you say grams? Review that vocabulary.

17. This is a different problem to read. Molecular mass is the same as molecules weight. The problem is an empirical formula problem.

4.02gN x 1 mole N atoms = 0.287 mole N atoms 14.01gN

11.48gO x 1 mole O atoms = 0.718 mole O atoms 16.00 O

N .287 O .718 --> N1O2.5 --> multiply by 2 --> N2O5 ---

N2O5 has a weighted 108 so the actual formula is N2O5.

18. 44.01g

19. 3 atoms - This problem asked for atoms in a molecule, not mole.

20. 3 moles of atoms - See your teacher if 19 and 20 confuse you.

21. 1 C atom x 1 mole C atoms x 12.01g C atoms = 2.00x10 -23 g C 6.02x1023 atoms of C 1 mole C atoms

This is backward from many of the problems that we have done.

22. A mole is 6.02x10 23 of anything. A mole, by deifintion, is a number and not a weight.

23. Avogadro's Number is 6.02 x 1023 or 1 mole.6 6g CO2 x 1 mole = 1.50 mole (or 9.03x1023 molecules)

44gCO2 S or the number of molecules in 66g of CO2 is larger.

24. 322.18g

25. They are called anhydrons copper (II) sulfate and anhydrous sodium sulfate instead of copper (II) sulfate pentahydrate (or copper (II) sulfate 5 hydrate) and sodium sulfate decalydrate. Anhydrous means "without water."

CHEMICAL FORMULASAND

NOMENCLATURE

CHEMICAL FORMULASand

NOMENCLATURE

One of the basic skills that must be developed by beginning cheistry students is that of writing and naming formulas. Many of the laboratory reagent bottles will have either names or formulas of compounds while the laboratory experiments will list the compound in the other mode. Just as you learned teh alphabet to put words together, you must first be able to recognize symbols before you can learn formulas. Learning to write formulas is much easier than learning to spell English words, however, since the rules are straight forward. The mastery of formula writing as the mastery of spelling requires practice. This is one of the few sections of chemistry where moemorization is a must so clear the cobwebs out of your memory banks.

Objective:

1. When given the name of an ion group, write its formula.

2. When given the formula of an ion group, write its name.

3. When given a Periodic Chart, identify the most common combining number of representative group elemental atoms that can be used in metal-nonmetal compounds.

4. When given the formula of a compound, write its name.

5. When given the name of a compound, write its formula.

6. When presented with a list of problems to name or write formulas, identify and correctly perform the naming or formula writing for nonmetal-nonmetal compounds.

7. Define the words ion, anion, and cation as they were used in this section.

Formulas for compounds were not standarized until the middle to late nineteenth century. This is not say that water was H2O, H3O, and H3O2 before that but that man had to discover that water was H2O. The formulas of compounds are invariant. Water is H2O. It just took a long time to discover that.

Before John Dalton's model of the atom in 1802 to 1808, many scientists thought that matter was infinitely divisible. There was no smallest particle such as the atom. Dalton's model stated that there was a smallest particle for each element, and the particles of one type of element were similar in weight to each other but different from the particles of other elements. Dalton also made an initial chart of atomic weights. In 1828 Berzelius published a more accurate table of atomic weights. Later, other scientists found even more accuarte values.

Another contribution of Berzelius was that of a standard symbol and nomenclature system. The ancient alchemists used assorted symbols such as moon slivers, triangles, and circles to represent elements. Dalton proposed that circles with various shadings and other delineations be adopted. Berzellius chose the first letter of the name of the element where possible or the first and second letter of the elemental name for the symbol. The symbol often represented the Latin name of the element rather than the English name. An example is that of iron and ferrum, Fe. Gold has the symbol Au from "aurum", silver, Ag form "argentum", and lead, Pb from "plumbum".

Once the system for symbols and formulas were standarized and tables of atomic weights were established, it became a matter of time before chemists succeeded in deducing formulas for compounds. They did this by either forming or breaking apart the compound to see how much of each element it contained. Since it was (and is) not possible to see the atoms, it was necessary to use indirest means to discover weight relationships. You will look at weight relationships in the next unit.

As formulas became to be known, the idea of combining number or combining power was advanced. In metal-nonmetal compounds, it can be observed that elements have certain tendencies for combination. With the exception of metal to hydrogen combinations called hydrides, hydrogen can be thought of as a metal. It will combine to form H2O, HCl, H2S, HBr, and HF. Sodium can form NaCl, Na2S, NaBr, and NaF. Sodium and hydrogen have combining numbers of one. Magnesium forms compounds of MgO, MgCl2, MgBr2, and MgI2. Its combining number is two as are those of beryllium, calcium, strontium, barium and radium. In metal-nonmetal compounds, chlorine, bromine, fluorine, and iodine all have combining numbers of one with oxygen and sulfur having two. By looking at a Periodic Chart, the most common combining numbers of many elements can be deduced.

The combining numbers of middle and the bottom groups cannot be deduced in this manner but must be learned.

Binary metal-nonmetal compounds are formed by making the combining numbers of the metal and nonmetal equal. In magnesium chloride, the magnesium has a combining number of two and the chlorine has a combining number of one. It is necessary to have two chlorines to add up to the magnesium combining number. The formula has one magnesium and two chlorines or MgCl2. The idea of combining numbers is sufficient to form most binary compounds. You may have, however, seen bottles in the laboratory with more than two elements in the formula. Common battery acid, or sulfuric acid, is H2SO4. How do we obtain a formula such as this?

The ancient Greeks had known that amber when rubbed attracted light objects just as bits of paper will be attracted to a comb after it has been put through dry hair in the winter. A French chemist, du Fay, discovered in 1733 that there were two kinds of electrical charge. One could be put on glass and the other on amber. Particles containing one of the charges would attract particles containing the other charge, but particles of like charges repelled each other. In the 1740's, Benjamin Franklin suggested that on electrical charge resuleted from a substance which contained more than a greater amount of electrical fluid; and the other charge resulted from a substance which contained a lesser amount of electrical fluid. He called the glass charge positive and the amber charge negative.

Volta, Carlisle, Nicholson, and Davy worked with chemistry and electricity in the late 1700's and early 1800's. Volta found that a current could be produced by two different metals, brine soaked blotting paper in between, and wire connecting them. This voltaic pile (battery) enabled the other men to use electrolysis as a means of separating certain compounds. Davy, in particular, observed that certain substances such as hydrogen and metals would go to the negative pole and oxygen would go to the positive pole.

An atom of sodium is electrically neutral. Sodium metal reacts readily, vigorously, and sometimes violently with water. When Davy isolated the metal sodium at the negative electrode, he observed this phenomena. The particles of sodium that was attracted to the negative electrode had to be positively charged. It did not react with the water. This particle is called a sodium ion. An ion is a charged particle of an atom.

If a salt water solution is used in this electrolysis, chlorine as well as sodium will be present since ordinary table salt is sodium chloride. The chlorine will be formed as diatomic molecules in normal conditions and is a poisonous pale green gas. Chlorine is electrically neutral. The particle that migrates to the positive electrode must be negatively charged. This charged particle is called a chloride ion. Table salt is made up of sodium ions and chloride ions. You would definitely not want to eat sodium atoms or chlorine molecules but eat their ions everyday. Ions, or charged particles, are chemically different from their original species.

How does this help us to deduce a formula such as H2SO4? Some groups of atoms tend to combine together to form charged species that act as if they were single charged atoms. Ions are charged atoms or groups of atoms. It is know that sodium and chlorine both have combining numbers of one but that their ions have opposite charges which cause them to attract. Na1+ and Cl1- can form NaCl. The hydrogen ion has a combining number of one and is positively charged, H+, except when it combines with metals. The one for a combining number is sometimes omitted but any other number must be included. The sulfate group is SO4

2-. This means that it is negatively charged with a combining number of two. The two ions attract each other because they have opposite charges. It requires two hydrogens to equal the combining number of one sulfate so the formula is H2SO4.

The next page has a chart of common ions and ion groups. It will be necessary to learn these. In this class we will use the modern notation of iron (III) for Fe3+ and iron (II) for Fe2+ instead of the older ferric and ferrous. You may occasionally find these older names so be aware that they do exist. Rules for writing and naming formulas follow the ion chart. Compounds can also be made from nonmetals combining with other nonmetals. These compounds have an entirely different set of rules involved with their formulas and names. This will follow the other set of rules.

TABLE OF COMMON IONS AND THEIR CHARGES

_________________________________________________________________NAMESYMBOL CHARGE NAME SYMBOL CHARGE

-----------------------------------------------------------------AluminumAl3+ +3 Lead (II) Pb2+ +2

-----------------------------------------------------------------AmmoniumNH4 2+ +1

-----------------------------------------------------------------Barium

Ba2+ +2 Magnesium Mg2+ +2-----------------------------------------------------------------Calcium

Ca3+ +2 Mercury (I) Hg2 +2-----------------------------------------------------------------Chromiuim (III) Cr +3 mercurous ----------------------------------------------------------------- chromic

Mercury (II) Hg2+ +2-----------------------------------------------------------------Cobalt (II)

Co2+ +2 mercuric-----------------------------------------------------------------

Nickel (II) Ni2++2-----------------------------------------------------------------Copper (I)

Cu+ +1----------------------------------------------------------------- cuprous

Potassium K+ +1-----------------------------------------------------------------Copper (II)

Cu2+ +2 Silver Ag+ +1----------------------------------------------------------------- cupric

Sodium Na2+ +1-----------------------------------------------------------------Hydronium

H3O +1 Tin (II) Sn +2-----------------------------------------------------------------

stannous-----------------------------------------------------------------Iron (II)

Fe2+ +2 Tin (IV) Sn4+ +4----------------------------------------------------------------- ferrous

stannic-----------------------------------------------------------------Iron (III)

Fe3+ +3 Zinc Zn2+ +2----------------------------------------------------------------- Ferric----------------------------------------------------------------------------------------------------------------------------------Acetate C2H3O2 -1 Hydroxide OH- -1-----------------------------------------------------------------Arsenate

AsO43- -3 Hypochlorite C10- -1-----------------------------------------------------------------Bromide

Br- -1 Iodide I- -1-----------------------------------------------------------------Carbonate

CO32- -2 Nitrate NO3- -1-----------------------------------------------------------------Chlorate

C103- -1 Nitrate NO2- -1-----------------------------------------------------------------Chloride

Cl- -1 Oxalate C2O42- -2

-----------------------------------------------------------------ChromateCrO4

2- -2 Oxide O2- -2-----------------------------------------------------------------Cyanide

CN- -1 Permanganate MnO4 -1-----------------------------------------------------------------Dichromate

Cr2O72- -2 Peroxide O22- -2

-----------------------------------------------------------------DihydrogenPhospate PO4

3- -3----------------------------------------------------------------- phosphate H2PO4- -1 Sulfate SO4

2- -2-----------------------------------------------------------------Fluoride

F- -1 Sulfide S2- -2-----------------------------------------------------------------Hydride

H- -1 Sulfite SO32- -2

-----------------------------------------------------------------Hydrogen carbonate HCO3- -1-----------------------------------------------------------------Hydrogen sulfate HSO4- -1-----------------------------------------------------------------

FORMULA WRITINGfor

METAL-NONMETAL COMPOUNDS

1. All compounds are electrically neutral and have balanced combining numbers.

2. An atom is electrically neutral; an ion is a charged species. A molecule is electrically neutral; an ion group is a charged species.

3. A negative ion is called an anion; a positive ion is a cation.

4. A cation is always written first in a formula.In table salt, sodium is first.In ammonium chloride, the ammonium ion is first.

5. The total combining number of the cation species must equal the total combining number of the anion species.

In sodium chloride, Na+ and Cl- ----> Na Cl.In magnesium chloride, Mg2+ and Cl- requires two Cl- toequal one Mg2+ -----> MgCl2.The one is understood.

In magnesium oxide, Mg2+ and O2- ---> MgO.

6. When the cation or anion is an ion group, parenthesis are used when more than one group in needed.

In magnesium sulfate, Mg2+ and SO42- -----> MgSO4.

Only one sulfate ion is needed.

In sodium sulfate, Na+ and SO42- -----> Na2SO4.

Two sodium ions are needed.

In calcium hydroxide, Ca2+ and OH- ----> Ca(OH)2. Two

hydroxides are needed. This is written Ca(OH)2 and not CaO2H2. It is really the calcium ion with two hydroxide ion groups attached.

In magnesium nitrate, Mg2+ and NO3- ---> Mg(NO3)2.

Exercises:

Write the formula for each of the following compounds:

1. Calcium phosphate 6. Lithium carbonate2. Sodium oxide 7. Sodiuim hydrogen carbonate3. Potassium chloride 8. Sodium sulfite4. Sodium carbonate 9. Sodium sulfide5. Calcium sulfate 10. Ammonium sulfate

7. Metals with combining numbers that vary are written so that the combining number is given in the name.

In iron (III) chrloride, Fe3+ and Cl- ----> FeCl3.

In iron (II) bromide, Fe2+ and Br- ----> FeBr2.

Exercises

Write the formula for each of the following compounds:

1. Tin (IV) oxide 4. Copper (II) sulfate2. Tin (II) fluoride 5. Copper (I) sulfate3. Lead (II) chloride 6. Iron (II) hydroxide

NOMENCLATURE RULESfor

METAL-NONMETAL COMPOUNDS

1. The compound has a cation (+) and an anion (-).

2. The cation is listed first in a compound.

3. In most cases the cation has no prefix. It nevr has a suffix (unless the old system for multivalent compounds is used, I.e. ferric for Fe3+.)

4. Some cations have more than one common combining number and this number must be identified by the use of Roman numerals in parentheses after the elemental name.

IUPAC (modern) rules: Fe3+ ----> iron (III) Fe2+ ----> iron (II)

Old system Fe3+ ----> ferric Fe2+ ----> ferrous

The combining number must be deduced from the formula of a compound.

FeS would be iron (II) sulfide since sulfur in the ion form is S2-.

Fe2(SO4)3 would be iron (III) sulfate since the sulfate ion is SO4

2- and there are three of them to make a total combining number of six. Since there are two iron ions present, each iron must have a combining number of three to also make a total of six.

Exercises

What is the combining number of the metal in each of the following?

1. FePO4 6. HgCl22. Fe3(PO4)2 7. CuSO43. SnO2 8. CuCl24. SnO 9. CuI5. PbCl2 10. SnF2

5. All anions have suffixes. The elemental ion will end in -ide. O, oxygen ---> ion is oxide, O2-

Cl, chlorine ---> ion is choride Cl-N, nitrogen ---> ion is nitride, N3-

Exercises

Name the anions of the following atoms:

1. bromine 4. iodine2. phosphorus 5. selenium3. sulfur 6. hydrogen (as an

anion)

The anion groups have suffixes that are already incorporated into their names. This has nothing to do with the number of oxygens attached (except more oxygens ate, fewer ite) or the combining number and must be memorized.

sulfate SO42- nitrate NO3-

Sulfite SO32- nitrate NO2-

6. An aqueous (aq) solution of a compound of hydrogen and an anion may form an acid. The same substance in gaseous (g) form may not be considered to be an acid.

The hydrogen is shortened to hydro- and the anion ending is changed from -ide to -ic. Acid is added to the name.

HCl (g) is hydrogen chloride. HCl (aq) is hydrocloric acid. HF(g) is hydrogen fluoride. HF(aq) is hydrofluoric acid.

An aqueous solution of a compound of hydrogen and an anion group may

also form an acid. The hydrogen part of the name is dropped entirely. The -ate suffix becomes an -ic while the -ite suffix becomes an -ous.

HNO3(g) is hydrogen nitrate. HNO3(aq) is nitric acid.H2SO3(g) is hydrogen sulfite. H2SO3(aq) is sulfurous

acid.

Exercises

Name the following:

1. HBr(g) 6. H2CO3(aq)

2. H2SO4(aq) 7. HCl(aq)

3. HI(aq) 8. HCN(aq)

4. H3PO4(aq) 9. HCl(g)

5. HC2H3O2(aq) 10. H2CrO4(aq)

7. For acids and salts which have several anion groups of a nonmetal and oxygen, a special nomenclature is used. A prefix may be necessary.

ClO4- perchlorate ionClO3- chlorate ionClO2- chlorite ionClO hypochlorite ion

Examples:

KBrO3 Potassium bromate

HClO2(aq) Chlorous acid

Exercises

Name the following:

1. NaClO 4. HClO4(aq)

2. HClO(aq) 5. HClO3(aq)

3. NaClO4 6. KClO2

BINARY NONMETAL-NONMETAL COMPOUNDS

1. The first species in the name or formula is always (almost) to the left or below in the Periodic Chart.

In a compound of sulfur and oxygen, sulfur is first.In a compound of nitrogen and oxygen, nitrogen is first.

2. The first species will have no suffix. It may have a prefix if more than one atom is present.

In CO, the first species is carbon-.In N2O4, the first species in dinitrogen-.

3. The second species will have a number prefix and an -ide ending.In CO, the second species is -monoxide.In N2O4, the second species is tetroxide.

4. The prefixes will indicate the number of atoms of that element present. They are mono, di, tri, tetra, penta, hexa, hepta, octo, and

so on. The last letter may be omitted to facilitate naming.In CO, the second species is -monoxide rather than

monoxide.In N2O4, the second species is -tetroxide rather than

tetraoxide.5. The rules for metal-nonmetal compounds do not hold here. Ions are not formed and combining numbers can vary. The formula is given by the name.

Carbon dioxide ---> CO2

This must have only one carbon (no prefix) and two oxygens (di prefix). The name can be taken directly from the formula according

to the above rules.SO3 ----> sulfur trioxide

This has only one sulfur so no prefix is needed. It has three oxygens so a tri- prefix is needed as well as an -ide suffix.

Exercises

Name or write the formula for each of the following:

1. CO2 4. dichlorine monoxide

2. sulfur dioxide 5. P4O10

3. carbon tetrachloride 6. NO2

Exercises

In order to master a skill such as formula writing and naming, it is necessary to practice. The following exercises are divided into groups. Do the first group, check your answers, and review any types that you missed. Do the same with the next group. Be sure that your spelling is correct.

1. Potassium bromide 6. Calcium nitrate2. SF6 7. Boron nitride3. Ag3PO4 8. Zinc cyanide4. Chromium (III) iodate 9. Sn3(PO4)25. SnCl4 10. Al(ClO4)3

1. CsClO3 6. (NH4)3N2. Phosphorus pentafluoride 7. Lead (II) sulfide3. Zinc hydrogen carbonate 8. Bromic acid4. Aluminum sulfate 9. Cd(IO3)25. Pb(MnO4)2 10. Cl2O

1. Iron (III) chloride 6. Phosphorus trichloride2. KClO3 7. Cu2CO33. Ba(HSO4)2 8. Calcium potassium

phosphate 4. Aluminum chromate 9. HClO5. Hg2(NO3)2 10. Cl2O7

1. MgC2O4 6. KMnO42. Beryllium nitrite 7. Aluminum acetate3. Aluminum sulfide 8. Ammonium nitrite4. Al2(SO3)2 9. CoCl35. nitrogen tri-iodide 10. N2O5

1. Barium hypobromite 6. AuHSO32. CaH2 7. Cesium hypoiodite3. Arsenic (III) sulfate 8. Hg(CN)24. Bismuth (III) oxide 9. Co(MnO4)25. PbCO3 10. IO5

Exercises

If astatine (At) is similar in properties to chlorine (Cl) and gallium (Ga) is similar to aluminum (Al), write the formulas for the

following compounds. 1. potassium astatate2. barium astatide3. gallium sulfate4. hydrastatic acid5. gallium cyanide6. gallium hypoastatite

If selenium (Se) is similar in properties to sulfur (S) and francium (Fr) is similiar to sodium (Na), write the formulas for the following compounds: 1. zinc selenide2. francium phosphate3. cobalt (II) selenite4. selenium dioxide5. francium selenate6. selenium hexafluoride7. francium hydride

Answers to Chemical Formulas and Nomenclature

P.8 Exercises

1. Ca3(PO4)2 6. Li2CO32. Na2O 7. NaHCO33. KCl 8. Na2SO34. Na2CO3 9. Na2S5. CaSO4 10. (NH4)2SO4

P.9 Exercises - top

1. SnO2 4. CuSO42. SuF2 5. Cu2SO43. PbCl2 6. Fe(OH)2

P.9 Exercises - bottom

1. Fe(III) 6. Hg (II)2. Fe (II) 7. Cu (II)3. Sn (IV) 8. Cu (II)4. Sn (II) 9. Cu (I)5. Pb (II) 10. Su (II)

P. 10 Exercises - top

1. bromide 4. iodide2. phosphide 5. selenide3. sulfide 6. hydride

P. 10 Exercises - bottom

1. hydrogen bromide gas 6. carbonic acid2. sulfuric acid - dilate 7. hydrochloric acid (conc. sulfuric acid is 95% 8. hydrocyanic acid sulfuric acid. It is a liquid) 9. hydrogenchloride gas3. hydroiodic acid 10. chromic acid4. phosphoric acid

5. acetic acid

Exercises p. 11

1. sodium hypochlorite 4. perchloric acid2. hypochlorous acid 5. chloric acid3. sodium perchlorate 6. potassium chlorite

Exercises p. 12

1. carbon dioxide 4. Cl2O Carbon (IV) oxide 5. P4O10 - tetraphosphorus2. SO2 decoxide 3. CCl4 (phosphorus (V) oxide)

6. nitrogen dioxide (nitrogen (IV) oxide

Exercies p. 13

1. KBr 6. Ca(NO3)22. Sulfur hexafluoride 7. BN (sulfur (VI) fluoride 8. Zn(CN)23. silver phosphate 9. Sn (II) phosphate4. Cr I3 10. aluminum perchlorate5. Tin (IV) chloride

1. Cesium chlorate 6. ammonium nitride2. PF5 7. PbS3. Zn(HCO3)2 8. HBrO34. Al2(SO4)3 9. Cadmium (II) iodate5. lead (II) permangamate 10. dichlorine monoxide

1. FeCl3 6. PCl32. potassium chlorate 7. Copper (I) carbonate3. barium hydrogen sulfate 8. CaKPO4 barium bisulfate 9. hypochlorous acid4. Al2(CrO3)3 10. dichlorine heptoxide5. mercury (I) nitrate

Exercises p. 13 continued

1. magnesium oxalate 6. potassium permanganate2. Be3N2 7. Al(C2H3O2)33. Al2S3 8. NH4NO24. aluminum sulfite 9. cobalt (III) chloride5. NI3 10. dinitrogen pentoxide

1. Ba (BrO)2 6. gold (I) hydrogen sulfite gold (I)

brsulfite 2. calcium hydride 7. Cs IO3. As2(SO4)3 8. mercury (II) cyanide4. Bi2O3 9. cobalt (II) per

manganate5. lead (II) carbonate 10. iodine pentoxide

Exercises p. 14

1. KAt2. BaAt23. Ga2(SO4)34. HAt5. Ga(CN)36. Ga (AtO)3

1. ZnSe2. Fr3PO43. Co SeO34. Se O25. Fr2SeO46. SeF67. FrH

CHAPTER 2 -- STOICHIOMETRY

SECTION 1 -- CHEMICAL REACTIONS

Chemistry evolved from alchemy in which the aim was tranmutation of metals to gold. The early alchemists did make some important observations about chemical reactions as they tried to convert lead to gold but these were usually not quantitative in nature. One question about chemical reactions is what happens if something is mixed with something else. A modern industrially important question is how much of a chemical is needed to put into a reaction to get the desired amount of product. This chapter will attempt to look at these questions although they involve many more factors than can be addressed here. We will begin by investigating chemical reactions.

Objectives:

1. Given a chemical equation, form pictorial representations of the reactants and products.

2. Define chemical equation, products, and reactants.

3. Illustration of the Law of Conservation of Matter using a chemical reaction.

4. Given a formula equation, balance it.

5. Given a word equation, transform it to a formula equation and balance it.

6. Given a list of equations of the five given types, classify according to type.

7. Given a partial single replacement, metathesis, combustion reaction, complete and balance the equation.

8. Given a solubility chart, determine if a metathesis reaction will occur.

CHAPTER 2

SECTION 1

Modern chemisty, as compared to alchemist chemistry, began in 1860. The period from 1750 to 1860 was the time of changeover. The ideas of Antoine Lavoiser (1743 - 1794), a wealthy French scientist, helped to guide chemistry in the right direction. Lavoisier kept careful records (or rather had his wife do so) of his experiments and extensively used the balances to weigh both reactants and products. Although the idea of conservation of matter had been observed earlier by Joseph Black, Henry Cavandick, and a Russian, Mikhail Lamonosov, it was up to Lavoisier to publicize the idea.

The law of Conservation of Mass of Conservationof Matter states that the weight of the productsin a chemical reaction should be equal to theweight of the reactants.

The writing of modern chemical equations had to follow the introduction and popularization of several theories. In the period from 1802 to 1808, John Dalton, a poor English school teacher, formulated a very important theory of the atom. In it he stated that all matter is composed of tiny particles called atoms, atoms of one element have the same weight, atoms are indivisible, and chemical changes are changes in the combinations of atoms with each other. Although some ancient Greeks such as Leucippets and his student Democritus theorized that there was a smallest particle of nature

such as an atom, it remained just a philosophical entity with the idea that matter was infinitely divisible being more accepted. The idea that there are finite particles of definite masses for the elements paved the way for later ideas. We often accept this idea and the important idea that chemical changes are changes in the combinations of atoms with one another without second thought.

Another aid to the writing of equations was the standardization of chemical symbols. Many new elements were discovered after the 1750 time period

mentioned above. The symbols used for the elements were left over from the alchemist days. The Wizard of Id's hat, in the comics, contains many of those symbols. Some of the

chemistry texts in your classroom should have other examples. These were cumbersome to remember and to use. Dalton made an attempt at a more modern system of symbol writing by using circles with various dots and lines to indicate the different elements but this system was also tedious to use. Berzellius (1779-1848) proposed in 1813 that letters and subscripts (raised numbers) be used to indicate compounds. Water would be H2O. With the exception of subscripts instead of superscripts, this is the system used today.

What else had to preceed equation-writing? It was necessary to know the formula of a compound first. Some compounds such as water, H2O, and hydrogen peroxide, H2O2 consist of the same type elements. Dalton observed that when two elements combine to form more than one compound, the masses of one element that combine with a given mass of the other element are in the ratio of small whole numbers. This is known as the Law of Multiple Proportions. In the water and hydrogen peroxide case set the mass of hydrogen at 2.0 grams. The mass of oxygen in water would then be 16.0 grams. The mass of oxygen in hydrogen peroxide would be 32.0 grams. The ratio of the mass of oxygen in water to mass of oxygen in hydrogen

Hydrogen (set) Oxygen (found) Ratio of O masses-----------------------------------------------------------------water

2.0g 16.0g 1hydrogen peroxide 2.0g 32.0g 2

peroxide would be one to two. According to Dalton's ideas, the formula for water should then be HO and the formula for hydrogen peroxide should be HO2. The true formulas were deduced later when better analytical methods were developed and more accuarate atomic weights were found.

Gay-Lussac's Law of Combining Volumes both confirmed the Dalton model of the atom and illustrated numerical observations that are very close to our modern equations. He observed that two volumes of hydrogen gas and one volume of oxygen would produce two

2 vol. hydrogen + 1 vol. oxygen gives 2 vol water vaper

volumes of water vaper.Dalton disputed this observation by saying that it was the same as two atoms of hydrogen and one atom of oxygen giving two molecules

of water (HO). For this to happen, it would be necessary to cut an atom in half. Even if he had known that water was really H2O, only one molecule could have been predicted. Gay-Lussac also observed that one volume of hydrogen and one of chlorine would give two volumes of hydrogen chloride gas. Three volumes of hydrogen and one volume of nitrogen gs would yield two volumes of ammonia gas (NH3). Look at the pictures below.

Gay-Lussac's observations:

one vol. hydrogen + one vol. chlorine gives two vol. hydrogen chloride

three vol. hydrogen + one vol. nitrogen gives two vol. ammonia

Dalton's view

one atom hydrogen + one atom chlorine two molecules hydrogen chloride

three atoms hydrogen one atom nitrogen gives two molecules ammonia

It remained fo rthe Italian, Amedo Avogadro (1776-1856) to publish a paper in 1811 to explain the problem in Dalton's idea. He stated that the gases were diatomic in nature. Diatomic means that hydrogen gas occurs not as the atomic hydrogen but as a molecule with two hydrogen atoms chemically combined. Those common elements that exist naturally as diatomic molecules are hydrogen, nitrogen, oxygen, fluorine, chlorine, bromine, and iodine. All but the last two are gases.

The pictorial explanation for the three reactions above would be:

two molecules of + one molecule gives two molecules of water hydrogen of oxygen

one molecule of + one molcule of gives two molecules of hydrogen chlorine hydrogen chloride

------------ ------------- ----------------

Draw in the pictures for the third reaction yourself and write the statement for what is happening as in the first two examples.

We would write these reactions as:

2H2 + O2 -----> 2H2O

H2 + Cl2 -----> 2HCl

---H2 + ---N2 ----> ---NH3

The statements above are symbolic representations, of what happens in a chemical reaction. They are called chemical equations. The substances on the left side of the equation are called reactants and those on the right side are called products. According to the Law of Conversation of Matter, we should have the same amount of oxygen and hydrogen on each side of the equation. Nothing can be lost or added during a chemical reaction. The law can also be called the Law of Conservation of Atoms. This means that the right and left side should have the same number of atoms.

How would you go about making the number of atoms the same on each side of the ammonia equation above? This is called balancing the equation.

------H2 + ------N2 ------> ----NH3

What we cannot do is change any subscript. The formula for ammonia is NH3 and nothing else. N2H6 and NH3 are very different formulas. From the preceeding discussion we now know that hydrogen and nitrogen occur as diatomics. The only part that appears to be adjustable is the number of molecules that we bring to the reaction and the number that we get out. When balancing an equation, adjust only the coefficients. The method we will use here is called "balancing by inspection" or "trial and error".

Begin by counting the number of atoms on each side.

----H2 + ----N2 ---------> ----NH3

H's - 2 H's - 3

N's - 2 N's - 1

Next change the coefficients from the understood 1 to whatever is needed. In this case, adjust the product nitrogen (which, of course, also adjusts the product hydrogen).

______H2 + ______N2 ---------> __2___NH3

Recount the atoms: H's - 2 H's -6 N's - 2 N's - 2

Readjust the coefficients if needed.

3 H2 + N2 -------> 2 NH3

Count atoms: H's - 6 H's - 6 N's - 2 N's - 2

The equation is balanced: 3H2 + N2 ------> 2NH3

Try the next example yourself.

Ca(OH)2 + H3PO4 -------> Ca3(PO4)2 + H2O

There are some hints that make working with an equation such as this a little easier. In equations such as this, it is better to save the hydrogens and oxygens for last. Adjust the calcium and phosphorus first. In some cases it is easier if you rewrite water as NOH to show that it can represent a hydroxide ion and a hydrogen ion. If the ion groups such as OH-

and PO43- do not break up the reaction, it is possible to count groups. The

answer to the equation is

3Ca(OH)2 + 2H3PO4 --------> Ca3(PO4)2 + 6H2O

After the experiment you will have plenty of practice with balancing equations.

Experiment: Balancing Chemical Equations

In chapter one we looked at the electrolysis of water apparatus. Here we will use it again along with molecular models to aid in balancing equations according to the Law of Conservation of Atoms.

If your lab has several electrolysisapparatuses, you will need to set up theapparatus. If there is only one, skip parts1, 2, and 3.

1. Set up the electrolysis apparatus. The apparatus should be connected to some D-C 6 volt source. If a rectifier is used, be sure that it is off before you connect the apparatus. If batteries are used, include a knife switch and keep it open until you are ready to begin.

2. Slowly, with stirring, add 10 ml of concentrated sulfuric acid ` to 600 ml of water in a 1000-ml breaker. Open the stopcocks to the collecting tubes. Add the water-acid mixture to the apparatus with a small excess collects in the reservoir.

3. Close the switch and allow the electrolysis to proceed until one of the tubes is two-thirds full of gas. Record both quantitative and qualitative observations.

Observations:

4. Collect a test tube full of gas from the tube with the larger volume by holding an inverted test tube over the stopcock, opening

the stopcock, and then closing the stopcock. Keep the test tube inverted. Hold a blazing splint to the mouth of the test tube. This is a test for which gas? Collect the second gas in the same manner but thrust a glowing splint into this test tube. For which gas is this a test?

Observations:

5. If it is marked on the rectifier or battery, follown the connecting wires to identify which is the positive electrode and which is the negative. Which gas collected at each electrode?

Reverse the electrical connections and close the switch. Which gas is collecting at each electrode here.

6. Write a word equation to describe what is happening.

7. Write a formula equation (not balanced) to describe what is happening.

8. Use the molcular models available to represent the above equation. Construct any additional models needed to conform to the Law of Conservation of Atoms.

Draw pictures of the models or use circle models.

9. Write the complete balanced equation.

Questions:

1. How does the volume of the hydrogen produced compare to the volume of the oxygen produced?

2. At which electrode does the hydrogen collect?

3. Does the oxygen always collect at a particular electrode?

4. What relationship do you find between the relative number of molecules represented by the balanced formula equation and the experimentally measured volumes of the gases produced?

5. The relative volumes of gases collected in the tubes are not exactly two to one, although exactly twice the volume of one gas is set free compared to the other gas. How do you account for the difference?

(Hint: Why is it possible to collect the gases in this manner(displacement of water)?)

Now it is time for a little practice work. As with the skill of formula writing, the only way in which you become a good equation balancer is practice. Remember that practice makes perfect:

1. _____BaCl2 + _____(NH4)2 CO3 -----> _____BaCO3 + ____NH4Cl

2. _____KClO3 -----> _____KCl + _____O2

3. _____Al(OH)3 + ___NaOH ----> ___Na AlO2 + _____H2O

4. _____Fe(OH)3 + ___H2SO4 ____> ___Fe2(SO4)3 + ___H2O

5. _____Na + ___H2O ----> ___NaOH + ___H2

6. _____Mg + ____N2 ----> _____Mg3N2

7. _____Mg + ____O2 -----> ___MgO

8. _____AgNO3 + ___CuCl2 -----> ___AgCl + ___Cu(NO3)2

9. _____C2H6O + ___O2 ---> ___CO2 + ___H2O

10. _____FeCl2 + ___Na3PO4 ----> ___Fe3(PO4)2 + ____NaCl

11. _____HNO3 + ___Sn ----> ___SnO2 + ___NO2 + ___H2O

12. ____NH4OH + ___AgNO3 ----> ___NH4NO3 + ___Ag2O + ___H2O

13. ____NaOH + ___NH4Cl -----> ___NaCl + ___NH3 + ___H2

14. ____NaOH + ___AgNO3 ----> ___NaNO3 + ___Ag2O + ___H2O

15. ____CuSO4 5H2O ----> ____CuSO4 + ___H2O

Formula Writing and Equations

Now that you are an expert in balancing equations and should have already been an expert in formula writing, you are going to be a double expert. You will get some practice in writing and balancing equations. Look at the following example:

Borium chloride + ammonium carbonate -----> borium carbonate + ammonium chloride

You must write each formula first. Much of what I will write down here should be done in your head. To write down the formulas, think of the ion symbols and their combining numbers.

Ba2+ Cl1- + NH41+ CO32- ----> Ba2+ CO3

2- + NH41+ Cl1-

Ba Cl2 + (NH4)2 CO3 ---> BaCO3 + NH4Cl

After you write the formulas, you may not change the subscripts when

balancing the equation. Balance the equation next.

BaCl2 + (NH4)2 CO3 ---> BaCO3 + 2 NH4Cl

If you encounter an equation which apparently cannot be balanced, check your formula writing for mistakes first.

In the following exercises, write and balance the equations for the first ten. Do not forget those elements that are diatomic. In the second ten, write the names of the elements and compounds represented by the formulas.

Exercises:

1. Sodium chloride + silver nitrate ---> silver chloride + sodiumnitrate

2. Zinc + copper (II) sulfate ---> zinc sulfate + copper

3. Silver nitrate + copper ---> copper (II) nitrate + silver

4. Barium chloride + sodium sulfate ---> sodium chloride + bariumsulfate

5. Zinc chloride + ammonium sulfide ---> zinc sulfide + ammonium chloride

6. Nickel + hydrochloric acid ---> nickel (II) chloride + hydrogen

7. Calcium carbonate ---> calcium oxide + carbon dioxide

8. Iron (III) chloride + sodium carbonate ---> sodium chloride +iron (III) carbonate

9. Calcium carbonate + hydrochloric acid ---> calcium chloride +water + carbon dioxide

10. Copper + sulfuric ace ---> copper (II) sulfate + water + sulfur dioxide

11. MgBr2 + Cl2 ---> MgCl2 + Br2

12. Ca(OH)2 + CO2 ---> CaCO3 + H2O

13. Nc + 2HCl ---> NiCl2 + H2

14. Zn + Pb(C2H3O2)2 ---> Pb + Zn(C2H3O2)2

15. NH4NO2 ---> N2 + 2H2O

16. 2AgNO3 + Cu ---> Cu(NO3)2 + 2 Ag

17. Fe S + 2 H Cl ---> H2 S + Fe Cl2

18. CaCO3 + 2 HCl ---> CaCl2 + H2O + CO2

19. 2 Al(OH)3 + 3 H2SO4 ---> Al2 (SO4)3 + 6 H2O

20. Ca(OH)2 + (NH4)2 SO4 ---> CaSO4 + 2NH3 + 2H2O

Types of Chemical Reactions

If sodium chloride and silver nitrate solutions are combined, what are the products?

Now that you are doubly expert on writing formulas and balancing equations, you are going to be triply expert. This new skill does not refer to being a hamburger stand expert but being able to write and balance equations as in the question above.

Many, but not all, chemical reactions can be divided into categories. These categories make it possible for us to write equations. In some reactions, however, the only way we know what will happen is to do the reaction and analyze the products. Here are five types of chemical reactions and examples of each.

1. Combustion Reactions

When a hydrocarbon (hydrogen, carbon compound) is burned,the products are carbon dioxide and water.

Example:CH4 + 2O2 ---> CO2 + 2H2OSome other organic compounds such as ethanol, C2H5OH,

also form carbon dioxide upon combustion.Example:

C2H5OH + 3O2 ---> 2CO2 + 3H2O

2. Combination Reactions

When two or more substances (elements or compounds) combine to form one substance, the reaction is called a combination reaction.

Example:2H2 + O2 ---> H2O

S + O2 ---> SO2

2Na + Cl2 ---> 2NaCl

When it rains over an area heavily polluted with sulfur dioxide, a combination reaction often happens.

Example:H2O + SO2 ---> H2SO3

3. Decomposition Reactions:When one substance breaks up to form two or more substances, the

reaction is called decomposition.

The electrolysis of water gives a good example.

Example:2H2O ---> 2H2 + O2

When baking soda (sodium bicarbonate) is heated, it breaks up into three compounds: sodium carbonate (which must be neutralized or the food will taste bitter), water, and carbon dioxide (which makes a cake rise).

Example:2Na HCO3 ---> Na2 CO3 + H2O + CO2

When Epson Salts, a hydrate of magnesium sulfate, are heated, the water comes off.

Example:MgSO4 7H2O ---> MgSO4 + 7H2O

4. Single Replacement Reaction

When one element replaces another element in a compound, the reaction is called a single replacement, or displacement reaction.

Example:Fe + CuSO4 ---> FeSO4 + Cu2K + 2H2O ---> 2KOH + H2Cl2 + 2NaBr ---> 2NaCl + Br2

In the first of the two experiments accompanying Types of Chemical

Reactions, you will observe and write many single replacement reactions. Just because you can write a reactoin on paper does not mean that it will happen. Some elements are not capable of replacing others as you will see.

Perform the experiment Single Replacement Reactions before proceeding to Metathesis Reactions.

Experiment: Single Replacement Reactions

Some elements are so reactive that they are never found in the free or uncombined state. Other elements are nearly inert. The reactivity of an element is related to its tendency to lose or gain elections. A list or series can be made which indicates which elements are capable of replacing other elements from their compounds. In the general example: A + BC ---> B + AC. A is the more active element since it replaces element B from the compound BC. In a second general reaction X + YZ ---> Y + XZ whre ---> means the reaction does not go, element Y is more active than element X so element X cannot replace it.

Procedure:

1. In some cases you will notice no immediate reaction. Set aside the test tube and observe it again after ten minutes.

Evidence of a reaction will be either evolution of a gas or the appearance of a deposit on the surface of the metal strip.

2. Label six clean test tubes numbers 1 through 6 and place them in a rack. Add the following to the test tubes:

Tube 1 = Copper strip, about 1 X 2 cm, and about 4 ml of 0.1 M silver nitrate.Tube 2 = Lead strip and about 4 ml of 0.1 M copper (II) nitrateTube 3 = Zinc strip and about 4 ml of 0.1 M lead (II) nitrateTube 4 = Zinc strip and about 4 ml of 0.1 M magnesium sulfateTube 5 = Copper strip and about 4 ml of dilute (3M) sulfuric acidTube 6 = Zinc strip and about 4 ml of dilute (3M) sulfuric acid

Record your observations on the data sheet.

_________________________________________________________________Evidence of Reaction Equation (to be completed)

_________________________________________________________________Describe any evidence of reaction Write "No reaction," if noif no reaction was observed, write reaction was observed"None"_________________________________________________________________1.

Cu + Ag NO3 --->__________________________________________________________________2. Pb + Cu(NO3)2 --->_________________________________________________________________3.

Zn + Pb(NO3)2 --->_________________________________________________________________4.

Zn + MgSO4 ---->_________________________________________________________________5.

Cu + A2SO4 ---->_________________________________________________________________6.

Zn + H2SO4 --->_________________________________________________________________

Questions:

1. Complete the following table by writing the symbols of the two elements whose reactivities are being compared in each test:

Tube Number1 2 3 4 5 6

_________________________________________________________________Greater Activity_________________________________________________________________Lesser Activity_________________________________________________________________

2. Arrange Pb, Mg, and Zn in order of their activities, listing the most active first.

(1)_______(2)_______(3)_______

3. Arrange Cu, Ag, and Zn in order of their activities, listing the most active first.

(1)_______(2)_______(3)_______

4. Arrange Mg, H and Ag in order of their activities, listing the most active first.

(1)_______(2)_______(3)_______

5. Arrange all five of the metals (excluding hydrogen) in an activity series, listing the most active first.

(1)_______(2)_______(3)_______(4)_______(5)_______

6. On the basis of the reactions observed in the six test tubes, explain why the position of hydrogen cannot be fixed exactly with

respect to all of the other elements listed in the activity series in Question 5.

7. What additional reaction(s) would be needed to establish the exact position of hydrogen in the activity series of the elements listed in Question 5?

8. On the basis of the evidence developed in this experiment:(a) Would silver react with dilute sulfuric acid? Why or whynot?

(b) Would magnesiuim react with dilute sulfuric acid? Why or why not?

5. Metathesis Reactions

When two positive ions exchange negative ions (or partners), a metathesis or double replacement reaction has happened.

As above, it is often possible to write reactions that are not observed. Most metathesis reactions are done in aqueous solution (the solid dissolved in water). If the possible products of the reaction are also soluble in water, there iis no driving force to make the reaction happen. If one of the possible products is insoluble or only slightly

soluble in water, the solid product is observed.

+

Ax Bz If the possible (mg) (aq) products AZ and

AX + BZ in water are really BX are soluble A + X- + B + Z-. in water, the

ions have no driving force to combine.

(Note: the subscripts (s) for solid, (l) for liquid, (g) for gas, and (aq) for aqueous are often found in reactions to indicate physical states.)

If AZ or BX is either insoluble or slightly

soluble in water, it will form a

solid called a precipitate.

The B+ and X- ions will still be in solution.

To write these reactions you must have a solubility chart. One follows this discussion.

Examples:

AgNO3 + H Cl ----> AgCl + HNO3Ni(NO3)2 + 2NaOH -----> Ni(OH)2 + 2NaNO3The arrow going down indicates a precipitate.

If a gaseous product is formed during a reaction such as this, the reaction can also be written.

Example:Mg CO3 + 2HCl ---> MgCl2 + H2O + CO2

You would have predicted that the product was MgCl2 + H2CO3, H2CO3, carbonic acid, will decompose to form H2O

and CO2. Other examples are in the second experiment.

A Neutrolization reaction is a type of metathesis reaction. A salt and water are formed as well as heat energy.

Example:HCL + NaOH ---> NaCl + H2O2HCL + Mg(OH)2 ---> MgCl2 + 2H2O

Solubility Rules

1. Compounds of the Group IA metals and NH4+ are soluble.

2. Binary compounds of Cl-, Br-, and I- are soluble, except for AgCl, Hg2

Cl2, and PbCl2. (PbCl2 is soluble in hot water.)

3. Compounds containing the NO3- group are all soluble.

4. Compound containing SO4 2- are soluble, except BaSO4, SrSO4 , and PbSO4. CaSO4 and Ag2SO4 are only slightly soluble.

5. Binary compounds of S2- are insoluble except fo rthe sulfide of the metals of Group IA, IIA, and NH4+.

6. Compounds containing the OH- group are insoluble except those of Group IA and Sr, Ba, and Ra.

7. Compounds of metals containing CO32-, PO4

3-, and SO32- are insoluble

except those of Group IA and NH4+.

Insoluble = less than lg of salt 100gH2O dissolves

Soluble = dissolves in water to a greater extent than above.

Use this chart in the next experiment on Double Replacement Reactions. This experiment will give you plenty of practice on writing reactions. It is not necessary to memorize the above chart for work in this chapter. The lab portion of the experiment will take one period, but the questions and problems will take longer.

Experiment: Double Replacement Reactions

A general metathesis or double replacement reaction follows:

AB + CO -----> AD + CD

In solution we really have ions of A+, B-, C+, and D- as is seen in these diagrams.

AB(aq) + CD -------> AD + CB

no reaction

1. If the four ions can exist mutually in solution, they do not recombine to form AD + CB. An example of this case is

NaCl + KNO3 -----> KCl + NaNO3

According to the solubility rules the products are both soluble so no

evidence of a reaction is present. This would best be written as

NaCl + KNO3 ----->

2. A second situation involves the formation of a solid or precipitate.

NaCl + AgNO3 -----> NaNO3 + AgCl

According to the solubility rules NaNO3 is soluble and will stay in solution. AgCl, however, is insoluble and will form a precipitate. This is indicated by a downward arrow .

NaCl + AgNO3 ----> NaNO3 + AgCl

3. A third situation involves the formation of a gaseous product. An unstable product is formed which will decompse to form water and a gas.

Na2CO3 + 2HCl ---> 2NaCl + H2CO3

unstable

H2CO3 will decompose to form H2O and CO2. The entire reaction is correctly written as:

Na2CO3 + 2 H Cl ----> 2 NaCl + H2O + CO2

Other examples of metathesis reaction products that decompose are H2SO3 (H2O and SO2 ) and NH4OH(H2O and NH3 ). The upward arrow indicates the formation of a gas.

4. The fourth type involves the formation of heat in a neutralization reaction.

NaOH + HCl ------> NaCl + H2O

Water is formed in the reaction so its ions, H+ and OH-, are no longer free in the solution. The formation of any slightly ionized compound (water, H2O; acetic acid, HC2H3O2; oxalic acid, H2C2O4; and phosphoric acid, H3PO4) causes the reaction to occur and liberate heat.

Double Replacement Reactions will occur if at least one of the following classes of substances is formed by the reaction:

1. A precipitate2. A gas3. A slightly ionized compound

Procedure:

Measure out 3 ml of distilled water into a test tube. Observe this level. This is the approximate level to which each solution should be added to obtain 3 ml. It is not necessary to measure each volume accurately. Record your observations at the time of mixing. If there is no visible reaction, feel the test tube to determine if heat has been evolved. Complete and balance the equation using for gases and for precipitates where appropriate. If there is no evidence of a reaction, write "No reaction".

Follow the following instructions and be sure to use the proper concentrations of solutions. In 1, for example, 3 ml 0.1M sodium chloride and 3 ml 0.1M potassium nitrate solutions are mixed together in a test tube.

1. Mix 0.1M sodium chloride and 0.1M potassium nitrate solutions.2. Mix 0.1M sodium chloride and 0.1M silver nitrate solutions.3. Mix 0.1M sodium carbonate and conc. (12M) hydrochloric acid solutions.4. Mix 10 percent sodium hydroxide and dil. (6M) hydrochloric acid solutions.5. Mix 0.1M barium chloride and dil. (3M) sulfuric acid solutions.6. Mix dil. (6M) ammonium hydroxide an dil (3M) sulfuric acid solutions.7. Mix 0.1M copper (II) sulfate and 0.1M zinc nitrate solutions.8. Mix 0.1M sodium carbonate and 0.1M calcium chloride solutions.9. Mix 0.1M copper (II) sulfate and 0.1M ammonium chloride solutions.10. Mix 10 percent sodium hydroxide and dil. (6M) nitric acid solutions.11. Mix 0.1M iron (III) chloride and dil. (6M) ammonium hydroxide

solutions.12. Add lg of solid sodium sulfite to 3 ml of water and shake to dissolve. Add about 1 ml of conc. (12M) hydrochloric acid solution, dropwise, using a medicine dropper.

Data for Double Replacement Reactions

_________________________________________________________________ Evidence of Reaction Equation-----------------------------------------------------------------1.

NaCl + KNO3 ------>-----------------------------------------------------------------2.

NaCl + AgNO3 ------>-----------------------------------------------------------------3.

Na2CO3 + HCl ------>-----------------------------------------------------------------4.

NaOH + HCl ------>-----------------------------------------------------------------5.

BaCl2 + H2SO4 ------>-----------------------------------------------------------------6.

NH4OH + H2SO4 ------>-----------------------------------------------------------------7.

CuSO4 + Zn(NO3)2 ----->

-----------------------------------------------------------------8.Na2CO3 + CuCl2 ------>

-----------------------------------------------------------------9.CuSO4 + NH4Cl ------>

-----------------------------------------------------------------10.NaOH + HNO3 ------>

-----------------------------------------------------------------11.FeCl3 + NH4OH ------>

-----------------------------------------------------------------12.Na2SO3 + HCl ------>

-----------------------------------------------------------------

Questions and Problems

1. The formation of three classes of substances caused double replacement reactions to occur in this experiment. Name them.a.b.c.

2. Write the equation for the decomposition of carboric acid.

3. Using the three criteria for double replacement reactions, together with the solubility chart of page 59, predict whether a double replacement reaction will occur in each example below. If reactin will occur, complete and balance the equation, using arrows to indicate gases and precipitates. If you believe no reaction will occur, write "no reaction" at

the right side of the equation.

a. NH4OH + HCl ---->b. Na2S + CuSO4 ---->c. NaC2H3O2 + HCl ---->d. NaOH + NH4Cl ---->e. (NH4)2 SO4 + KNO3 ---->f. K2CrO4 + Pb(NO3)2 ---->g. K2CO3 + HNO3 ---->h. BiCl3 + KOH ---->i. NaC2H3O2 + CuSO4 ---->

These two experiments should have given you plenty of practice in writing single and double replacement reactions. Can you pick out types of reactins when you see them? Go back to the twenty exercises starting on page 50. Classify each as to type of reaction. Most are single and double replacements but there are a couple of others also so review all five types.

For 1-5, balance the following equations:

1. KNO3 ----> KNO2 + O22. Fe + H2O ----> Fe3O4 + H23. CO2 + NaOH ---> NaHCO34. Zn + H2SO4 ---> ZnSO4 + H25. Fe Cl2 + Na3PO4 ---> Fe3(PO4)2 + Na Cl6. Write a balanced equation for the formation of magnesium nitride from its elements.7. Write a balanced equation for the decomposition of mercury (II) oxide to form free elements.8. Write a balanced equation for the combustion of propane

(C3H8).For 9-18, write a balanced equation and indicate the reaction type. Use for gases and for precipitates where appropriate. If no reaction is predicted, write the formulas for the reactants only.

9. Magnesium bromide + chlorine ---> magnesium chloride + bromine10. Sodium + water ---> sodium hydroxide + hydrogen gas11. Potassium nitrate ---> potassium nitrite + oxygen gas12. Zinc + hydrochloric acid --->13. Calcium oxide + hydrochloric acid --->14. Aluminum nitrate + ammonium hydroxide --->15. Potassium chlorate ---> potassium chloride + oxygen gas16. Zinc chloride + ammonium sulfide --->17. Mercury (II) sulfate + ammonium sulfide --->18. Silver nitrate + potassium chloride --->

CHAPTER 2 -- STOICHIOMETRY

SECTION 2 --- MOLES AND REACTIONS

By now you should be perfect on writing chemical equations since you have had plenty of practice. Are you wondering how all of this relates to what you studied in Chapter 1? That is precisely what we will learn in this section. This is a very useful area of quantitative chemistry. As moles provided the key for converting between grams and molecules, moles will be the key in working with relationships within balanced chemical equations. After spending all of that time on the previous section, you may feel a little rusty on moles and need to review.

Objectives:

1. Given a formula equation, translate it into an English sentence using both moles and molecules.

2. Given a chemical equation, state the chemical equivalence relations.

3. Given a chemical equation and the moles of one substance, calculate the moles of another substance within that equation.

4. Given a chemical equation and the moles of one substance, calculate the grams of another substance within that equation.

5. Given a chemical equation and the grams of one substance, calculate the grams of another substance within that equation.

The first question in Chapter 1 was about the meaning of the formula H2O. The answer at the beginnig of the chapter was that H2O represented one molecule of water which consisted of two atoms of hydrogen and one atom of oxygen. At the end of the chapter we said that H2O not only represented the above but also one mole of water molecule which consisted of two moles of hydrogen atoms and one mole of oxygen atoms.

How would you read the balanced equation 2H2 + O2 ---> 2H2O?This equation could be read in two equally correct ways. It could be two molecules of hydrogen gas mixed with one molecule of oxygen gas reacts to give two molecules of water. Pictorially, this is represented by:

2 molecules H2 + 1 molecule O2 ---> 2 molecules H2O. The equation could also be read: two moles of hydrogen gas added to one mole of oxygen gas yields two moles of water. Pictorially, this would take quite a long time to draw since each one mole, as you remember, is 6.02 x 1023 molecules. Even when we want to talk about moles, our pictorial representations will have to be of molecules. You will find it helpful to make models for these reactions. There will be models provided for your individual use.

Look at the equation 2H2 + O2 ---> 2H2O again. This balanced equation represents the mole ratio by which the reactants combine to form the product. What would happen if you combined three moles of hydrogen with one mole of oxygen? Look at the following models:

3 H2 + 1 O2 ---> ? water (H2O)

Water is still H2O. In this case you should have 2H2O formed with a hydrogen mole (or molecule in the picture) left over. Hydrogen is in excess. The reaction will still occur but the ratio of reactants to products will still be 2:1:2.

What would be produced if you combined four moles of hydrogen with one mole of oxygen?

+ ----> ?

4 H2 + 102 ----> ?

This is really the same question as before. Hydrogen is again in excess. Two moles of water should be formed with two moles of hydrogen left over.

What would be produced if four moles of hydrogen were combines with two moles of oxygen?

+ --->

4 H2 + ---> ? H2O

If you used the models, you could see that there would be enough atoms present to form four water molecules. Notice that 4H2 + 202 ---> 4H2O is really a multiple of the original equation.

What would be produced if one mole of hydrogen were combined with a half mole of oxygen? Did you answer that one mole of water formed or that it was not possible since not enough was present? You were correct if the first was your answer. One to one-half to one is still the same combining ratio as two to one to two. Remember that we are working with moles in this problem. Although we do not talk about half of a molecule, there is no reason not to talk about half a mole or 3.01 x 1023 molecules reacting.

What would be produced if three moles of hydrogen were combined with excess oxygen? Did you say three moles of water were produced or did you say two moles of water with one mole of hydrogen left over? Look at the following picture:

3 H2 + xs O2 ---> ? H2O

(xs means excess)

There is no reason that all three moles of hydrogen in our problem should not react. Three moles of water would be produced. Notice that as long as one of the reacting reagents is in excess it is only necessary to specify the amount of the other reactant.

If five moles of water were produced, how many moles of oxygen must have racted to form the water? This is the reverse of the problem above. Since the balance equation is still 2H2 + O2 ---> 2H2O, the same reacting ratios hold. You may be able to work this in your head and obtain 2.5 moles of oxygen for the answer.

Most problems in the lab do not work out as nicely. The dimensional analysis method of working problems comes in handy here. According to the equation,m it takes two moles of hydrogen and one mole of oxygen to form two moles of water. Although two moles of hydrogen are not equal to one mole of oxygen or two moles of water, in this balanced equation they can be said to

be "chemically equivalent". This could be indicated as 2 mole H2 1 mole O2 2 mole H2O with the symbol indicating chemical equivalence. These can then be used as unit factors as

2 moles H2 or 1 mole O2 or 1 mole O2 and so on. In a1 mole O2 2 moles H2 2 mole H2O

problem, the amount of oxygen needed to form five moles of water could be set up like this:

5 mole H2O x 1 mole O2 = 2.5 mole O2) . The "mole H2O" units divide 2 mole H2O

out and leave "mole O2" for the units. Do not forget to include all of the necessary units.

Try one more problem before we look at the exercises. Use dimensional analysis to help you.

Example:Oxygen gas is prepared by heating KClO3 in the presence of a catalyst.

KCl is the other product in the reaction. If 0.0244 moles of KClO3 decompose completely, how many moles of oxygen would form?

This problem has a different wrinkle. You must first write and balance the equation. The only reactant is KClO3 and the two products are KCl and O2. Were you able to deduce that from the statement? Did you remember that oxygen was diatonic? Learning to read what the question asks is an important part of working problems.

2 K Cl O3 ---> 2 K Cl + 3O2

Now that we have the equation and have balanced it, we must look at what we have and want.

2 K Cl O3 ------> 2 K Cl + 3 O2 0.0244 moles ? moles

We hav 0.0244 moles of potassium chlorate and want the number of moles of oxygen. It is often helpful to underline the known and unknown. The equation says that two moles of K Cl O3 will form three moles of O2. As was said before, it is possible to have more or fewer than two moles of K Cl O3. The coefficients do not indicate the absolute numbers that must be present.

0.0244 moles K Cl O3 x 3 mole O2 + 0.0366 mole O2 formed2 mole K Cl O3

0.0366 mole O2 would be formed.To check your understanding, work the following exercises. Work these

on your own paper and show all of your work.

Exercises:

1. If 0.0 3195 mole of copper reacts with sulfur to form copper (II) sulfide, how many moles of sulfur would be needed to completely react with the copper? How many moles of copper (II) sulfide would be formed?

2. If 0.0178 moles of magnesium are burned in air to form magnesium oxide, how many moles of oxygen would be needed? How many moles of magnesium oxide would be formed?

3. In the reaction 2 Fe (s) + 2 H2O(1) + O2(g) ---> 2 Fe(OH)2(s) how many moles of Fe(OH)2 will form when 0.25 mole O2 reacts with iron?

*4. According to the equation in question 3, how many moles of Fe(OH)2 would be formed if 2.5 mole Fe, 5.2 mole H2O, and 1.5 mole O2 were combined? Why?

*5. In the equation A l (OH)3 + H2SO4 ---> A l2(SO4)3 + H2O, how many moles of sulfuric acid (H2SO4) are needed to ract with 0.250 mole A l(OH)3?

The next pages contain an experiment on mole ratios.

Experiment: Mole Ratios and Chemical Reactions

In this experiment you will investigate the relationships between moles of reacting chemical species.

1. Weigh a clean, dry 250 ml Erlenmeyer flask to thenearest 0.0lg. Keep the flask on the balance pan. ________

2. Add between 15.0 and 16.0g CuSO4 5H2O to the flask.(Adjust the balance so that it weighs between 15.0gand 16.0g more than the flask. Add the salt untilthe pan begins to drop. Stop adding the salt andadjust the balance to the nearest 0.0lg)

Weight of CuSO4 5H20 and flask ________

3. Add 200 ml of distilled water to the flask todissolve the salt. Swirl the flask to help indissolving. It may be necessary to gently heatthe flask.

4. Weigh out a 3.0g tuft of steel wool to the nearest0.0lg. The steel wool may be placed directly onthe balance pan but be sure to blow off any smallloose steel particles before recording the finalweight.

Weight of steel wool (Fe) ________ moles of Fe ________

5. Add the steel wool to the flask. Swirl the contentsof the flask intermittently for 5-10 minutes or untilthe steel wool has disappeared.

6. Let the solid copper particles which have formed settleto the bottom. Carefully decant, or pour off the liquidas is shown in the picture to the right. Wash the residuewith 10 ml of distilledwater, and decant again.Repeat the washing procedureat least three times.

Prepare a filtering setup by foldinga piece of filter paper and fittingit into a funnel. Moisten the paperwith distilled water.

Add 10 ml of distilled water to the flask

containing the washed residue. Swirl the

contents of the flask and filter the

contents. The tip of the funnel should touch the collection

beaker so that a steady filtrate stream

can run down the side.

Wash any residue in the flask intothe filter paper. Rotate the flaskto remove any residue from it.

Carefully remove the filter paper andcontents. Place it on a watch glassand place in a drying oven at no morethan 100 C over night. If any of thepaper accidently tore off, include thison the watch glass. Weigh the driedpaper and contents to the nearest 0.0lg.

Weight of filter paper and copper ________

7. Weigh ten sheets of filter paper. To find the averageweight of one sheet, divide the weight of the ten byten. _________

8. Weight of copper formed during the reaction _________Moles of copper formed _________

9. Calculate the ratio of moles of iron (in the steel wool) to the moles of copper involved in the reaction.

Questions and Problems

1. On the basis of this experiment, write the equation for the reaction between iron in steel wool and on aqueous solution of CuSO4 5H2O.

2. Was all of the copper ion in CuSO4 5H2O converted to metalliccopper? Explain your answer.

3. How many atoms of iron and how many atoms of copper were involved in your experiment?

4. Though it is predominantly iron, steel wool does contain a small percentage of other elements. How does this reflectupon your experimental results?

5. If you had evaporated the filtrate (liquid) of step 6 to dryness, what would you expect the composition of the residue

to have been?

Since you have had practice on the mole type problems, look at the following problem. Try this on your own before you look at the solution.

Example:

How many grams of calcium acetate can be produced from 1.50 moles of acetic acid (H C2H3O2) when it reacts with limestone according to the following equation?

CaCO3 + 2HC2H3O2 ----> Ca(C2H3O2)2 + CO2 + H2O

You know how to do this problem if the answer requests moles. Work it out in moles.

1.50 moles HC2H3O2 x 1 mole Ca(C2H3O2)2 = 0.750 mole Ca(C2H3O2) 2 moles HC2H3O2

The answer, however, should be in grams. In Chapter 1, you learned how to change moles to grams by using the weight of one mole.

0.750 mole Ca(C2H3O2)2 x 158.17gCa(C2H3O2)2 = 118.63g 1 mole Ca(C2H3O2)2 or

119g Ca(C2H3O2)2

This problem could be set up as a continuous dimensional analysis problem.

1.50 moles HC2H3O2 x 1 mole Ca(C2H3O2)2 x 158.17gCa(C2H3O2) 2 moles H(2H3O2 1 mole Ca(C2H3O2)2

= 119gCa(C2H3O2)2

Example:

How many grams of copper are needed to replace silver in 0.0235 moles of silver nitrate dissolved in water? One of the products is copper (II) nitrate.

What answer did you get? This problem is similar to the last experiment. The reactin is of a type that you did earlier. What is the name of this type of reaction? The first step is to write to the balanced chemical equation for the single replacement reaction.

Cu + 2 AgNO3 ---> Cu(NO3)2 + 2Ag? 0.0235 moles

It is set up just like the previous problem.

0.0235 moles of AgNO3 x 1 mole Cu x 63.54gCu = 0.747g Cu 2 mole AgNO3 1 mole Cu

Did you have trouble reading the problem? The first part asks, "How many grams of copper. . ." so the unknown must be copper. The next part is, ". . .are needed to replace silver in 0.0235 moles of silver nitrate dissolved in water?" Reading this correctly requires that you recognize what happens in a single displacement reaction. The silver must be one of the products.

Try these problems on your own.

Exercises:

1. Calculate the number of grams of oxygen required to burn 2.40 mole of C2H6.

2. How many grams of K Cl O3, would be produced from the reaction of 3.55 mole of Cl2 with KOH, according to the following

balanced equation?

3 Cl2 + 6 KOH ---> 5 K Cl + K Cl O3 + 3 H2O

Look at the following problem.

How many grams of silver would a silver nitrate solution? Theequation is

Cu + 2 AgNO3 ---> Cu(NO3)2 + 2 Ag 175g ? g

Do you recognize that this problem is the reverse of an earlier example? Attempt to solve it on your own and then read the following discussion. At this point you should recall the information found from the balanced chemical equation. The coefficients in front of the formulas represent molecules or moles. Either way, they represent the number of particles that combine in a chemical reaction, not the weights of particles. Using pictorial representations we have

Cu 2 Ag NO3 ---> Cu(NO3)2 + 2Ag

To work a problem of the type above, a good method is to change the weight of what we know to moles (number of particles). Since the equation tells us how the particles will numerically combine, we can use this information.

In a continuous dimensional analysis calculation:

175g of Cu x 1 mole Cu x 2 moles Ag x 196.97g Ag = 1085g Ag 63.54gCu 1 mole Cu1 mole Ag = 1090g Ag

or by sig. liq.

This corresponds to 2.75 moles This is the final con-This is the mole conversion factor version of moles of and gives the moles of silver or silver to grams of

5.51 moles silver

The word "stoichiometry" was used in this section. It means that mathematics of chemistry and entails the type of problems that we have been doing. These are very useful for real laboratory situations as you will see in the next experiment. Before you perform the experiment, look at the following example.

Example:

In the decomposition of 3.00 grams of potassium chlorate, how many

grams of oxygen will be formed? How many molecules of oxygen?

Work out the first part yourself. If you are not sure of the equation, look back in this section since we have already looked at an example using this reaction. I get 1.17 grams of oxygen for my answer.

The answer to the second part of the question can found in two ways. You could begin with the 3.00 grams of potassium chlorate or begin with 1.17 grams of oxygen. The two set-ups follow:

3.00g K ClO3 x 1 mole K ClO3 x 3 mole O2 x 6.02x10 23 molecules 122.55g K ClO3 2 mole K ClO3 1 mole O2

2.20 x 10 22 molecules O 2

1.17g O2 x 1 mole O2 x 6.02 x 10 23 molecules O 2 = 2.20x1022moleculesO2 32.0g O2 1 mole O2

Experiment: Synthesis of Acetylsalicylic Acid

Purpose: To use stoichiometry problems in an actual laboratory situation.

Theory: By knowing the formulas for the reactants and products in a chemical reaction, it is possible to compute the theoretical

yield of acetylaslicylic acid or aspirin. this involves the use of chemical equations and mole relations. The theoretical yield is what would ideally be expected for the product. In many reactions a lesser yield than expected is obtained. Some of the reasons

involve the purification procedures. Other reasons, inherent within the system itself, will be converted next semester.

A percentage yield may be computed by

Actual Yield in grams x 100 = % yieldTheoretical Yield in grams

Chemicals needed: Salicylic acidAcetic anhydrideConc. surfuric acid0.5 M acetic acid (cold)

Caution: Acetic anhydride fumes irritate the eyes and nasal membranes. Both it and the conc. sulfuric

acid will cause skin burns. Perform the experiment under a hood or in an area with plenty of ventilation.

Equipment needed:

Test tubeWater bath (400 ml beaker will work)Ice bath (Ice and water in a beaker)Bunsen BurnerSuction flaskVaccum tubingMeans of suction (your teacher will instruct

you on this.)

Procedure:Come to class prepared. Although the experiment is not difficult, it may be necessary to wait for equipment. The early bird gets the Buchner funnel. As with any experiment, keep your goggles on the entire time.

Weigh to the nearest 0.0lg a 3.01g or 3.0g sample of salicylic acid. Place the solid in your largest test tube. Working under the hood, carefully add 4.5 ml of acetic anhydride, followed by 15 drops of conc H2SO4.

Place the test tube in a water bath maintained at a temperature of 60 C over a low Bunsen burner flame. The temperature is maintained at 60 C + 5 by alternately placing and removing the burner as needed. Heat the system, with intermittent shaking, for 20 minutes.

After the heating cycle has ended, cool the contents of the test tube in an ice bath. Add 5 ml of distilled water to the tube and stir the mixture for 5 minutes while it is kept cold. Be careful not to punch a hole in the bottom of the test tube while stirring. Cut a piece of filter paper to fit the bottom of the Buchner funnel and filter the crystals with the funnel and a suction flask.

Buchner funnel

to vacuum pump or asirator

filtratim flask

Wash the residue with three 5-ml portions of ice cold 0.5 M acetic acid. Wash with one portion, filter; wash with a second, filter; wash a third time, and filter the third portion. Remove the filter paper and crystals. Place on a watch glass and allow to dry overnight in the drying oven.

Fill out the data chart, calculate the theoretical yield from the equation, and calculate the percentage yield of the synthesis.

Acetic anhydride

H SO4H3CCO- O - OCCH3 + C6H4 COOH OH -------> C6H4COOHOOCCH3 + H3 C-COOH

Added in xs

Data: 1. Weight of salicylic acid = g

2. Weight of dried aspirin and filter paper = g

3. Weight of filter paper = g

4. Weight of aspirin = g

5. Theoretical yield (from equation) = gshow calculations:

6. Percentage yield of aspirin = g

To finish up this section, you need to try some exercises.

1. How many grams of diphosphorus dioxide (P4O10) must be reacted with water to produce 4.00g of phosphoric acid (H3PO4)?

P4 O10 + H2O ----> H3PO4

2. How many moles of nitrogen dioxide may theoretically be produced when copper reacts completely with 21.0 grams of concentrated nitric acid?

Cu + HNO3 -----> NO2 + CuO + H2O

3. How many moles of barium hydroxide can be neutralized by the reaction with 145 grams of sulfuric acid?

Ba(OH)2 + H2SO4 ---> BaSO4 + H2O

4. How many grams of zinc chloride can be prepared by reacting 10.0g of zinc withexcess hydrochloric acid?

5. If excess sulfuric acid reacts with 30.0g of sodium chloride, how many grams of hydrogen chloride are produced?

6. Calculate the number of grams of lead (II) chloride produced by reacting 0.400 mole of chloride ions with excess lead (II) ions.

Pb2+ + 2Cl- ---> Pb Cl2The following is a balanced equation for the combustion of the

organic substance butadiene in oxygen:2 C4H6(g) + 11 O2(g) ----> 8 CO2(g) + 6 H2O(g)

7. What weight of CO2 will be produced by the combustion of 40.0 grams of C4H6?

8. How many molecules of CO2 will be produced by the combustion of 40.0 grams of C4H6?

*9. How many atoms of hydrogen will be produced by the combustion of 40.0 grams of C4H6?

10. How many moles of O2 must be consumed to produce 54.0 grams of H2O by this reaction?

11. How many molecules of O2 must be consumed to produce 54.0 grams of H2O by this reaction?

12. Write the following equation as an English sentence in two ways.2H2O + 2 NaCl ----> 2H2 + Cl2 + Cl2 + 2Na + O2

13. What is the Law of Conservation of Matter:What is conserved in the equation in 12?

A compound hs the following composition by weight.C, 57.1% H, 4.8% O, 38.1%

14. Calculate the simplest formula for this compound.

15. If the molecular weight is shown to be 126, what then must be the true molecular formula?

DO NOT WRITE ON THIS SHEET

Activity Series

K Most ActiveCaNaMgAlZnFeNiSnPbH2

CuAgAu Least Active

Solubility Rules

1. Compounds of the Group IA metals and NH4+ are soluble.

2. Binary compounds of Cl-, Br-, and I- are soluble, except for AgCl, Hg2Cl2, and PbCl2. (PbCl2 is soluble in hot water.)

3. Compounds containing SO42- are soluble, except BaSO4, SrSO4 , and PbSO4.

CaSO4 and Ag2SO4 are only slightly soluble.

4. Compounds containing NO3- group are all soluble.

5. Binary compounds of S2- are insoluble except for the sulfides of the metals of Group IA, IIA, and NH4+.

6. Compounds containing the OH- group are insoluble except those of Group IA and Sr, Ba, and Ra.

7. Compounds of metals containing CO32-, PO43-, and SO32- are insoluble

except for those of Groups IA and NH +.

CHAPTER 2 - STOICHIOMETRY

SECTION 3 - LIMITING REAGENTS

In many of the exercises you were given the weight of only one substance. Often you had to assume that you had an excess of any other needed reagents. How do you determine which is the limiting reagent (the one that makes the difference) and which is in excess of you are given the mass of both? This section will address that problem.

Objectives

1. Define limiting reagent.

2. Given the amounts of the reactants in a reaction, determine the limiting reagent.

3. Given the amounts of the reactants in a reaction, determine the amount of excess reagent left after the reaction is completed.

You will perform a short experiment to begin this section.

EXPERIMENT:

You will use exactly 10.0 ml of 6 M HCl. Place two pieces of zinc (each approximately 2 cm x 1 cm, strip form) in evaporating with the 10.0 ml of acid. In a second dish place 10 pieces of zinc. Wait until the reaction has stopped and the questions below.

Questions:

1. When two strips of zinc are used, which reagent, the acid or the zinc, is in excess? Why?

Test the solutions in each dish with litmus paper. (In acid blue litmus turns red. In base red litmus turns blue.)

Does this observation agree with your answer to the first question?

2. When 10 strips of zinc are used, which reagent is the limiting reagent? Why?

Again test the solution with litmus and compare to your answer to 2.

3. If more acid were used with the two strips, would you expect more product?

If more acid were used with the ten strips, would you expect more product?

4. Write a balanced chemical equation for this reaction.

In the first part of Section 2 in this chapter, we looked at the formation of water reaction:

2H2 + O2 = 2H2O

One of the accompanying questions asked how many moles of water could be formed from four moles of hydrogen and one mole of oxygen. By using pictorial representations we can again see what is occuring.

We begin with four water mole-cules (since I did not feel likedrawing out 4 x 6.02 x 1023 molecules) and one oxygen

molecule.

Each water molecule consists oftwo hydrogen atoms and one oxygen atom.

The two oxygen atoms can "grab"two hydrogen atoms each. Twohydrogen molecules remain un-reacted.

Four moles of hydrogen and one mole of oxygen can form only two moles of water with two moles of hydrogen unreacted.

In the above example oxygen is called the limiting reagent. All of the oxygen is completely consumed during the reaction and the amount of oxygen originally present will determine how much product is formed. Hydrogens was present is excess of the amount needed for complete reaction of the oxygen.

Look at the following exercises:

1. A mother has four children and three lollipops. Would the children or the lollipops be the limiting "reagent" in this case? 2. There are seven lab benches which can have four students apiece in the chemistry room. There are also twenty four student desks in the room. Which of these is the limiting factor on possible student enrollment per section in that classroom. 3. How much H2SO4 can be prepared from 5.0 moles of SO2, 1.0 mole of O2, and an unlimited quantity of H2O according to the equation

2SO2 + O2 + H2O = 2H2SO4?

4. What would be the limiting reagent, according to the equation 2Fe + 2H2O + O2 = 2Fe(OH)2 ,

if 2.5 mole Fe, 5.2 mole H2O, and 1.5 mole O2 were put together in a way that the reaction above could take place?

How did you do on the last two exercises? Essentially they were the

same as the water example. If you had trouble, reread the material on water and then ask your teacher for help. We have used the idea of limiting reagent in some earlier laboratory experiments. In the experiment where hydrated copper (II) sulfate was mixed with steel wool, which reagent, the steel wool or the copper (II) sulfate, was used in the calculation? Why was that used instead of the other reagent? Would it have made any difference in your answer? Which was the limiting reagent in the aspirin experiment? Look at the following example:

If 32.0 g of NaOH reacts with 75.0 g of H2SO4, which one of the reagents will be in excess?

There are many ways in which this problem can be solved. Each way, however, requires that a balanced equation be written. Why?

2NaOH + H2SO4 = Na2SO4 + 2H2O

Once again, what does this equation say? (Besides - Nothing, an equation can't talk!) The equation shows that two moles (or particles or molecules) of sodium hydroxide react with one mole of sulfuric acid to form one mole of sodium sulfate and two moles of water. The example is given in grams and not moles.

32.0 g NaOH x 1 mole NaOH = 0.800 mole NaOH 40.0 g NaOH

75.0 g H2SO4 x 1 mole H2SO4 = 0.765 mole H2SO498.1 g H2SO4

There are fewer moles of H2SO4 so does this mean that NaOH is in excess with H2SO4 limiting? If you answered yes, you forgot the stoichiometry or combining mole ratio of the equation. If you answered nothing, you really are not doing a very good job of reading this. According to the mole ratio, two moles of sodium hydroxide must have one mole of sulfuric acid.

0.800 mole NaOH x 1 mole H2SO4 = 0.400 mole H2SO4 2 mole NaOH

0.800 mole of NaOH needs only 0.400 mole of H2SO4 so we have plenty of H2SO4 present for all of the NaOH to react. The sodium hydroxide is the limiting reagent and the acid is present in excess.

If you had chose to work with sulfuric acid first, you could have deduced that it was present in excess by a similiar mole ratio.

0.765 mole H2SO4 x 2 mole NaOH = 1.53 mole NaOH 1 mole H2SO4

Here, there is not enough NaOH present to react with all of the sulfuric

acid so the acid must be in excess with the base (NaOH) limiting.

A third method involves using both reagents as limiting to calculate the amount of product.

32.0 g NaOH x 1 mole NaOH x 2 mole H2O x 18.0 g H2O = 14.4 g H2O 40.0 g NaOH 2 mole NaOH 1 mole H2O

75.0 g H2SO4 x 1 mole H2SO4 x 2 mole H2O x 18.0 g H2O = 27.6 g H2O 98.1 g H2SO4 1 mole H2SO4 1 mole H2O

In this case the reagent that produced the least amount of product is limiting and the other reagent is in excess.

Try the following exercises:

1. When solutions of lead (II) nitrate, Pb(NO3)2, and sodium chloride, NaCl, are mixed, the following reaction occurs:

Pb(NO3)2 + 2NaCl = PbCl2 + 2NaNO3

How many grams of PbCl2 can be prepared from 1.00 g of Pb(NO3)2 and 1.00 g of NaCl?

2. Which is th limiting reagent if 5.00 g og copper metal reacts with a solution containing 20.0 g of AgNO3? 3. If 20.0 g of KOH reacts with 15.0 g of (NH4)2 SO4, calculate the following: a) the moles of K2SO4 produced and b) the grams of NH3 produced. 4. It is necessary to prepare the maximum possible amount of magnesium acetate by a reaction involving 15.0 g of iron(III) acetate with either 10.0 g of MgC2O4 or 15.0 g of MgSO4. Both reactions produce the desired magnesium acetate and a precipitate. Which reaction will give the most Mg(C2H3O2)2 and how many grams will be produced? 5. A 36.0 g sample of calcium hydroxide, Ca(OH)2, is allowed to react with a 54.0 g sample of phosphoric acid, H3PO4. How many grams of calcium phosphate could be produced? If 45.2 g of calcium phosphate could be produced in an actual run of this reaction, what is the percent yield? (Refer back to the aspirin experiment for percent yield.) These exercises really require that you understand moles and chemical reactions. They are also meant to be somewhat challenging. The exercises that follow are review exercises for the entire chapter.

Chapter Review Exercise:

1. At elevated temperatures the compound NF3 decomposes to N2 and F2 . Find the amounts of N2 and F2 formed by the decomposition of: a) exactly 3 mole of NF3, b) 0.268 mole of NF3.

2. Direct reaction of P with Cl2 can form PCl5 under suitable conditions. Find the mass of Cl2 required to form 2.70 mole of PCl5 3. Find the mass of AsCl3 formed by the reaction of 0.133 mole of Cl2 with arsenic.

4. Carbon disulfide, CS2, is a very flammable substance that reacts with O2 to form carbon dioxide and sulfur dioxide. Find the mass of O2 that is required to react with 9.34 g of CS2. Find the mass of carbon dioxide and of sulfur dioxide formed in this reaction.

5. A certain oxide of lead is converted by H2 to lead metal and water. One mole of the oxide reacts with 8.1 g of H2 and forms 622 g of lead. Find the formula of the oxide.

6. Quantities of 11.1 g of H2 and 33.3 g of Cl2 are mixed. Find the mass of hydrogen chloride that forms.

7. The reaction of carbon with calcium oxide produces carbon monoxide and calcium carbide, CaC2. A total of 2.45 g of CaC2 is isolated from the reaction of 5.00 g of calcium oxide with 2.50 g of carbon. Find the percent yield of the CaC2.

8. The oxide of an unknown element is believed to have the formula XO2. Heating a 25.2 g sample of the compound decomposes it completely to form 3.2 g of O2. Find the atomic weight of X, the unknown element.

Alternate Problems

1. An important step in the manufacture of nitric acid is the reaction of ammonia with O2 to form NO and water. Find the amount of O2 required to react completely with 76.0 g of ammonia, NH3. Find the mass of water and of NO produced.

2. The Cl2 formed by the decomposition of 1.30 mole of PCl3 is used to convert carbon to CCl4. Find the amount of CCl4 formed. (2PCl3 = 2P + 3Cl2, C + 2Cl2 = CCl4)

3. An important reaction that takes place in a blast furnance during the production of iron is the formation of iron metal and

carbon dioxide from Fe2O3 and carbon momoxide. Find the mass of Fe2O3 required to form 910 kg of iron. Find the amount of carbon dioxide that forms in this process.

4. Incomplete combustion of hydrocarbons is an important cause of air pollution because it produces carbon monoxide. Find the difference between the mass of O2 required to convert one mole of octane, C6H18, completely to CO and the mass of O2 required to convert one mole of octane completely to CO2.

5. Reaction of tungsten with Cl2 forms WCl6. Find the mass of the unreacted starting material when 12.6 g of tungsten is treated with 13.6 g of Cl2 and the reaction takes place. Find the mass of WCl2 that forms.

6. When HgO is heated, it decomposes to mercury and O2. Careful technique makes it possible to isolate 62.7 g of mercury from the decomposition of 75.8 g of the oxide. Find the percent yield of the reaction.

7. Find the mass of sodium chloride and the mass of water needed to prepare 125 g of an 18.0% sodium chloride solution by mass.

8. A mixture of 50.0 g of S and 100 g of Cl2 reacts completely to form S2Cl2 and SCl2 and no other products. Find the mass of S2Cl2 formed.

DO NOT WRITE ON THIS SHEET

Activity Series

K Most ActiveCaNaMgAlZnFeNiSnPbHCuAgAu Least Active

Solubility Rules

1. Compounds of the Group IA metals and NH4+ are soluble.

2. Binary compounds of Cl-, Br-, and I- are soluble, except for AgCl, Hg2Cl2, and PbCl2. (PbCl2 is soluble in hot water.) 3. Compounds containing SO4

2- are soluble, except BaSO4, SrSO4 and PbSO4. CaSO4 and Ag2SO4 are only slightly soluble. 4. Compounds containing NO3- group are all soluble.

5. Binary compounds of S2- are insoluble except for the sulfides of the metals of Group IA, IIA, and NH . 6. Compounds containing the OH- group are insoluble except those of Group IA and Sr, Ba, and Ra. 7. Compounds of metals containing CO3

2-, PO42-, and SO3

2- are insoluble except for those of Groups IA and NH4+.

CHAPTER 2

Section 1

Page 42

+ ------->

These molecules of hydrogen react with one molecule of nitrogen to form two molecules of ammonia.

3H2 + N2 ---> 2NH3

Page 44

3Ca(OH)2 + 2H3PO4 ---> Ca3(PO4)2 + 6H2O

Page 48

1. BaCl2 + (NH4)2CO3 ----> BaCO3 + 2NH4Cl2. 2KClO3 ---> 2KCl + 3O23. Al(OH)3 + NaOH ---> NaAlO2 + 2H2O4. 2Fe(OH)3 + 3H2SO4 ---> Fe2(SO4)3 + 6H2O5. 2Na + 2H2O ---> 2NaOH + H26. 3Mg + N2 ---> Mg3N27. 2Mg + O2 ---> 2MgO8. 2AgNO3 + CuCl2 ---> 2 AgCl + Cu(NO3)29. C2H6O + 3 O2 ---> 2CO2 + 3 H2O10. 3FeCl2 + 2Na3PO4 ---> Fe3(PO4)2 + 6NaCl11. 4HNO3 + 1Sn ---> 1SnO2 + 4NO2 + 4H2O12. 2NH4OH + 2AgNO3 ---> 2NH4NO3 + Ag2O + H2O

13. NaOH + NH4Cl ---> NaCl + NH3 + H2O14. 2NaOH + 2AgNO3 ---> 2NaNO3 + Ag2O + H2O15. CuSO4 5H2O ---> CuSO4 + 5H2O

Page 50

1. NaCl + AgNO3 ---> AgCl + NaNO32. Zn + CuSO4 ---> ZnSO4 + Cu3. 2AgNO3 + Cu ---> Cu(NO3)2 + 2Ag4. BaCl2 + Na2SO4 ---> 2 NaCl + BaSO45. ZnCl2 + (NH4)2S ---> ZnS + 2 NH4Cl6. Ni + 2HCl ---> NiCl2 + H27. CaCO3 ---> CaO + CO28. 2FeCl3 + 3Na2CO3 ---> 6NaCl + Fe2(CO3)39. CaCO3 + 2HCl ---> CaCl2 + H2O + CO210. Cu + 2H2SO4 ---> CuSO4 + 2H2O + 2SO211. Magnesium bromide + chlorine ---> magnesium chloride + bromine12. Calcium hydroxide + carbon dioxide ---> calcium carbonate + water 13. Nickel + hydrochloric acid ---> nickel (II) chloride + hydrogen14. Zinc + lead(II) acetate ---> lead + zinc acetate15. Ammonium nitrite ---> nitrogen + water16. Silver nitrate + copper ---> copper(II) nitrate + silver17. Iron(II) sulfide + hydrochloric acid ---> hydrogen sulfide +

iron (II) chloride18. Calcium carbonate + hydrochloric acid ---> calcium chloride +

water + carbon dioxide19. Aluminum hydroxide + sulfuric acid ---> aluminum sulfate +

water20. Calcium hydroxide + ammonium sulfate ---> calcium sulfate +

ammonia + water

Page 65

1. 2KNO3 ---> 2KNO2 + O22. 3Fe + 4H2O ---> Fe3O4 + 4H23. CO2 + NaOH ---> NaHCO34. Zn + H2SO4 ---> ZnSO4 + H25. 3FeCl2 + 2Na3PO4 ---> 6Fe3(PO4)2 + NaCl6. 3Mg + N2 ---> Mg3N27. 2MgO ---> 2Hg + O28. C3H8 + 5O2 ---> 3CO2 + 4H2O9. MgBr2 + Cl2 ---> HCl2 + Br2(e)10. 2Na + 2H2O ---> 2NaOH + H211. 2KNO3 ---> 2KNO2 + O212. Zn + 2HCl --->ZnCl2 + H213. CaO + 2HCl ---> No reaction14. Al(NO3)3 + 3NH4OH ---> Al(OH)3 + 3 NH4NO3

15. 2KClO3 ---> 2KCl + 3O216. ZnCl2 + (NH4)2S ---> ZnS + 2NH4Cl17. HgSO4 + 2NH4NO3 ---> No reaction18. AgNO3 + KCl ---> AgCl + KNO3

-Section 2-

Page 72 - Exercises

1. Cu + S ---> Cu S0.03195 mole? ?

0.03195 mole Cu x 1 mole S = 0.03195 mole S 1 mole Cu

0.03195 mole Cu X 1 mole CuS = 0.03195 mole CuS 1 mole Cu

2. 2Mg + O2 ---> 2MgO 0.0178 moles ? moles ? moles

0.0178 moles Mg x 1 mole O2 = .00890 mole O2 2 mole Mg

0.0178 mole Mg x 2 mole MgO = 0.0178 mole MgO 2 mole Mg

3. 2 Fe(s) + 2H2O + O2(g) ---> 2 Fe(OH)2(s) 0.25 mole ? moles

0.25 mole O2 x 2 mole Fe(OH)2 = 0.50 mole Fe(OH)2 1 mole O2

4. 2 Fe(s) + 2 H2O(l) + O2(g) ---> 2 Fe(OH)2(s)2.5 mole Fe 5.2 mole 1.5 mole ? moles2.5 mole Fe would form 2.5 mole Fe(OH)2 if plenty of the othertwo were present.5.2 mole H2O would form 5.2 mole Fe(OH)2 if plenty of the other two

were present. 1.5 mole O 2 would form 3.0 mole Fe(OH)2 if plenty of the other two were present. 2.5 mole Fe will produce the least amount of product or 2.5 mole Fe(OH)2. There would be plenty of H2O and O2 present for all of the Fe to react but enough Fe for either of the other two to be used up completely. The amount of Fe present is what limits this reaction. 5. 2Al (OH)3 + 3H2SO4 ---> Al2(SO4)3 = 6H2O

0.250 moles ? moles

0.250 mole Al(OH)3 x 3 mole H2SO4 = 0.375 moles H2SO4 2 mole Al(OH)3

Page 77

1. 2C2H6 + 7O2 ---> 4CO2 + 6H2O2.40 mole ?g

2.40 mole C2H6 X 7 mole O2 x 32.00g O2 = 269g O2 2 mole C2H6 1 mole O2

2. 3 Cl2 + 6 KOH ---> 5 KCl = KClO3 + 3H2O3.55 mole ? g

3.55 mole Cl2 x 1 mole KClO3 x 122.56g KClO3 = 145g KClO3 3 mole Cl2 1 mole KClO3

Page 83

1. P4O10 + 6H2O ---> 4H3PO4 ?g 4.00g

4.00g H PO x 1 mole P4O10 x 283.88g P4O10 = 284g P4O10 4 mole H3PO4 1 mole P4O10

2. Cu + 2HNO3 ---> 2NO2 + CuO + H2O 21.0g ? g

21.0g HNO3 x 1 mole HNO3 x 2moles NO2 x 46.01gNo2 = 14.2gNO2 68.01gHNO3 2 moles HNO3 1 mole NO2

3. Ba(OH)2 + H2SO4 ---> BaSO4 + 2H2O?moles 145g

145g H2SO4 x 1 mole H2SO4 x 1 mole Ba(OH)2 x 171.35gBa(OH)2 = 98.08g H2SO4 1 mole H2SO4 1 mole Ba(OH)2

253 Ba(OH)2

4. Zn + 2 HCl ---> ZnCl2 + H210.0g ?g

10.0g Zn x 1 mole Zn x 1 mole ZnCl2 x 136.28gZnCl2 = 65.37gZn 1 mole Zn 1 mole ZnCl2

20.8gZnCl2

5. 2NaCl + H2SO4 ---> HCl + Na2SO430.0g ?g

30.0gNaCl x 1 mole NaCl x 1 mole HCl x 36.469HCl = 9.36gHCl 58.44g 2 moleNaCl 1 mole HCl

6. Pb2+ + 2 Cl- ---> PbCl2 0.400 mole ? g

0.400 mole Cl- x 1 mole PbCl2 x 278.109PbCl2 = 55.6g PbCl2 2 mole Cl- 1 mole PbCl2

7. 2 C4H6 (g) + 11 O2(g) ---> 8 CO2(g) + 6 H2O(9) 40.0g

40.0g x 1 mole C4H6 x 8 mole CO2 x 44.019CO2 = 130.189CO2 54.099C4H6 2 mole C4H6 1 mole CO2

8. 40.0g C4H6 x 1 mole C4H6 x 1 mole C4H6 x 8mole CO2 x 6.02x10 23 molecules CO2 = 1.78 x 1024 molecules CO2

9. 40.09C4H6 x 1 mole C4H6 x 6 moles H2O x 6.02 x 10 23 molecules H 2O 54.09C4H6 2moles C4H6 1 mole H2O

x 2 atoms H = 2.68 x 10 24 H atoms 10. 54.0g H2O x 1 mole H2O x 11 moles O2 x 32.00g O2 = 5.50 molesO2 18.019H2O 6 molles H2O 1 mole O2

11. Use # 10 to make this simpler5.50 moles O2 x 6.02 x 10 23 molecules O 2 = 3.31 x 10 24

moleculesO2 12. Two molecules of water react with two molecules of sodium

chloride to form two molecules of hydrogen gas and one molecule of chlorine gas (oops, another mistake. My typist was not a chemist.) and two molecules of sodium metal and one molecule of oxygen.

Two moles of water react with two moles of sodium chloride to form two moles of hydrogen, one mole of chlorine gas, two moles of sodium metal atoms, and one mole of oxygen. 13. In a chemical reaction, matter can be neither created nor destoyed. These are the same number of atoms on both sides of the equation. The mass is the same on both sides of the equation. 14. Assume 100 grams

57.19 C x 1mole C = 4.75 moles C C 4.75 H 4.75 O2.3812.01gC

4.8g H x 1 mole H = 4.75 moles H C 4.75 H 4.75 O 2.38 1.01 g H 2.38 2.38 2.38

38.1 g O x 1 mole O = 2.38 mole O 16.00g O C2H2O

15. C2H2O adds up to 42

126 = 3 There are 3 -C2H2O units so the true molecular

42 formula is C6H6O3. Does this problem look like one you already? It should. I thought that you needed a little review here. Answers to Exercises

page 87

1. The lollipops are the limiting reagent for, hopefully, obvious reasons.

2. 7 lab benches x 4 students = 28 students benches 1 bench

24 desks desks

The limit for enrollment would be the number of desks.

3. 2 SO2 + O2 + H2O ---> 2H2SO4Given: 5.0 moles 1.0 mole x S ? moles5.0 moles SO2 x 2 moles H2SO4 = 5.0 moles H2SO4

2 moles SO2

1.0 mole O2 x 2 moles H2SO4 = 2.0 mole H2SO4 1 mole O2

There is only enough O2 present to form 2.0 mole H2SO4 so O2 must be the limiting reagent. 4. 2Fe + 2H2O + O2 ---> 2Fe(OH)2

Given: 2-5 mole5.2 mole 1.5 mole

In words:

2.5 mole of Fe would produce 2.5 moles Fe(OH)2 and require 2.5 mole H2O and 1.25 mole of O2 to do so. There is excess of H2O and O2. Fe is the limiting reagent. 7.87 x 10-2 mole Cu x 2 mole AgNO3 = 0.157 mole AgNO3

1 mole Cu

Thre are only 0.118 moles of AgNO present, however, so Cu must be present in excess with AgNO3 limiting. 3. 2KOH + (NH4)2SO4 ---> K2SO4 + 2NH3 + 2H2O

This was probably a difficult equation for you. This type was covered in the double replacement type of reaction earlier.

20.0g KOH x 1 mole KOH = 0.356 mole KOH 56.11g KOH

15.0g (NH4)2SO4 x 1 mole (NH4)2SO4 = 0.114 mole (NH4)2SO4 132.14

0.356 mole KOH x 1 mole (NH4)2SO4 = 0.178 mole (NH4)2SO4 2 mole KOH

We do not have enough ammonium sulfate for all the KOH to react so ammonium sulfate is limiting. a) 0.114 mole (NH4)2SO4 x 1 mole K2SO4 = 0.114 mole (NH4)2SO4

1 mole (NH4)2SO4

b) 0.114 mole (NH4)2SO4 x 2 mole NH3 x 17.03g NH3 = 3.88g NH3 mole (NH4)2SO4 1 mole NH3

If all of MgSO4 is to be used up, 0.0831 mole of Fe(C2H3O2)3 is needed. How much do we have? only 0.644 mole. Here , Fe (C2H3O2)3 is the limiting reagent. .0644 mole Fe (C2H3O2)3 x 6 mole Mg(C2H3O2)2x 142.20g Mg(C2H3O2)2 = 4 mole Fe(C2H3O2)3 1 mole

13.7g Mg(C2H3O2)2

Reaction 2 will produce the greatest yield.

5. 3Ca(OH)2 + 2H3PO4 ---> Ca3(PO4)2 + 6H2O36.0g 54.0g ?

Part 1 of this problem is to find the limiting reagent.Mw's - Ca(OH)2 74.10g/mole

H3PO4 98.00g/mole Ca3(PO4)2 310.19 g/mole

36.0g Cu(OH)2 x 1 mole Ca(OH)2 = 0.486 mole Ca(OH)2 74.10g

54.0g H3PO4 x 1 mole H3PO4 = 0.551 mole H3PO4 98.00g

.486 mole Ca(OH)2 x 2 mole H3PO4 = 0.324 mole of H3PO4 3 mole Ca(OH)2

Do we have enough H3PO4 for all of the Ca(OH)2 to react? Yes, we have plenty so Ca(OH)2 is the limiting reagent.

Chapter Review Exercises

Page 91

1. 2NF3 ---> N2 + 3F2

a) (exactly) 3 mole ?g ?g

3 mole NF3 x 1 mole N2 x 28.01g N2 = 42.02g N2 2 mole NF3 1 mole N2

3 mole NF3 x 3 moles F2 x 38.00g F=2 = 171.0g F2 2 mole NF3 1 mole F2

b) 2 NF3 ---> N2 + 3F2 0.268 mole NF3 x 1 mole N2 x 28.01gN2 = 3.75g N2

2 mole NF3 1 mole N2

.268 mole NF3 x 3 mole Fe x 38.00gFe = 15.3g F2 2 mole NF3 1 mole F2

2. 2p + 5Cl2 ---> 2 PCl5 (or you can use P4) ?g 2.70 mole

2.70 mole PCl5 x 5 mole Cl2 x 70.91g Cl2 = 478.62g =479gCl2 2 mole PCl3 1 mole Cl2

3. 2 A s + 3Cl2 ---> 2AsCl3 0.113 mole ?

0.113 mole x Cl2 2 mole AsCl3 x 181.28gAsCl3 13.66g = 13.7AgCl3 3 mole Cl2 1 mole AsCl3

6. H2 + Cl2 ---> 2HCl11.1g 33.3g ---> ?g

Cl2 is so much heavier than H2 that I would guess 33.3g Cl2 is the limiting reagent, but we should check it out anyway.

11.1g H2 x 1 mole H2 = 5.51 mole H2; 33.3gCl2 x 1 mole Cl2 2.02 70.91g Cl2

=0.470 mole Cl2

Since H2 and Cl2 react in a 1:1 mole ratio, Cl2 is definitely the limiting reagent

0.470 mole Cl2 x 2 mole HCl x 36.45g HCl = 34.3g HCl 1 mole Cl2 1 mole HCl

7. 3C + CaO ---> CO + CaC22.50g 5.00g 2.45g <--- actual find theoretical

* Find the moles of C and CaO2.50gC x 1mole = .208 mole C ; 5.00gCaO x 1 moleCaO =

12.01g 56.08

0.0892 mole CaO

* Let's assume CaO is limiting and see if there is enough C present0.0892 mole CaO x 3 mole C = 0.2675 mole C

1 mole CuO * We do not have 0.2675 mole of C. C must be limiting. We do have 0.208 mole of C.

0.208 mole C x 1 mole CaC2 x 64.10gCaC2 = 4.44g CaC2 3 mole C 1 mole CaC2

* Find the percent yield% yield = Actual yield x 100 = 2.45g x 100 = 55.2%

Theoretical yield 4.44g

Chapter 2 - Alternate Problems

1. 4NH3 + 5O2 ---> 4NO + 6H2O76.0g ?g ?g ?g

76.0gNH3 x 1 mole NH3 X 5 MOLE O2 X 32.00gO2 = 178.51G = 179g O2 17.03g NH3 4NH3 1 mole O2

76.0gNH3 x 1 mole NH3 x 4 mole NO x 30.01gNO = 134g NO 17.03gNH3 4 mole NH3 1 mole NO

We could just subtrac 134g NO from (76.0 + 179 O2) to find the mass of H2O (Law of Conservation of Mass) but I'll work it out anyway.

76.0gNH3 x 1 mole NH3 x 6 mole H2O x 18.02gH2O = 121gH2O 17.03gNH3 4 mole NH3 1 mole H2O

2. 2 PCl3 ---> 2 P + 3Cl2 ; C + 2Cl2 ---> CCl41.30 mole 1.30 mole PCl3 x 3 mole Cl2 = 1.95 mole Cl2 formed

2 mole PCl3

This 1.95 mole Cl2 can then be used in the second equation as is. Don't try to adjust the moles.

C + 2Cl2 ---> CCl4 1.95 mole ?g

1.95 mole Cl2 x 1 mole CCl4 x 153.82g CCl4 = 150gCCl4 2 mole Cl2 1 mole CCl4

3. Fe2O3 + 3CO ---> 2Fe + 3CO2?g 910kg ?g

910kgFe x 1 mole Fe x 2 mole Fe2O3 x 159.70gFe2O3= 5200KgFe2O355.85gFe 1 mole Fe 1 mole Fe2O3

It looks as if either might be limiting. Try one and compare with the moles of the other.

12.6gW x 1 mole W x 3 mole Cl2 = 0.206 mole Cl2 183.85gW 1 mole W

We do not have quite 0.206 moles of Cl so W is in excess.

* To answer the first part of the problem we must find the grams of W that reacted:

0.192 mole Cl2 x 1 mole W x 183.85gW = 11.8g W 3 mole Cl2 1 mole W

* To find the mass of unreacted W, subtract:12.6g W - 11.8gW = 0.8gW left over

* Now we need to find the mass of WCl6 that forms. Either work the stoichiometry or add 11.8gW to 13.6gCl2=

25.4g

0.192 mole Cl2 x 1 mole WCl6 x 396.57gWCl6 = 25.4gWCl6 3 mole Cl2 1 mole WCl6

6. 2HgO ---> 2Hg + O275.8g 62.7g

75.8g HgO x 1 mole HgO x 2 mole Hg x 200.59gHg = 70.2gHg <-- 216.59gHgO 2 mole HgO 1 mole Hg

% yield = actual x 100 62.7gHg x 100 = 89.3% theoretical 70.2gHg

Answers to ExercisesPage 87

1. The lollipops are the limiting reagent for, hopefully, obvious reasons.

2. 7 lab benches x 4 students = 28 students benches 1 bench

24 desks desks

The limit for enrollment would be the number of desks.

3. 2 SO2 + O2 + H2O --> 2H2SO4Given: 5.0 moles 1.0 mole xs ? moles5.0 moles SO2 x 2 moles H2SO4 = 5.0 moles H2SO4

2 moles SO2

1.0 mole O2 x 2 mole H2SO4 = 2.0 mole H2SO4 1 mole O2

There is only enough O2 present to form 2.0 mole H2SO4 so O2 must be the limiting reagent.

4. 2Fe + 2H2O + O2 ---> 2Fe(OH)2Given: 2-5 mole 5.2 mole 1.5 mole

In words:2.5 mole of Fe would produce 2.5 mole Fe(OH)2 and require 2.5 mole

H2O and 1.25 mole of O2 to do so. There is excess of H2O and O2. Fe is the limiting reagent.

Answers to Exercises, page 90:

1. Pb(NO3)2 + 2NaCl ---> PbCl2 + 2NaNO31.00g1.00g ?g

Which is the limiting reagent?

1.00g x 1 mole Pb(NO3)2 = 3.02 x 10-3 mole Pb(NO3)2 331.2Og

1.00g x 1 mole NaCl = 1.71 x 10-2 mole NaCl 58.44

3.02 x 10-3 Pb(NO3)2 x 2 mole NaCl = 6.04 x 10 -3 mole NaCl 1 mole Pb(NO3)2

Since 6.04 x 10-3 mole of NaCl is needed to react with 3.02 x 10-3 mole Pb(NO3)2 and there is 1.71 x 10-2 mole of NaCl present, the NaCl must be present in excess with Pb(NO3)2 the limiting reagent.

3.02 x 10-3 mole Pb(NO3)2 x 1 mole PbCl2 x 2.78.10gPbCl2 = 1 mole Pb(NO3)2 1 mole PbCl2

0.840g ofPb(NO3)2

There are other ways (even easier) of working this problem.

2. Cu + 2AgNO3 ---> 2Ag + Cu(NO3)25.00g 20.0g

5.00g x 1 mole Cu = 7.87 x 10-2 mole Cu 63.54gCu

20.0g x 1 mole AgNO3 = 0.118 mole AgNO3 169.88g AgNO3

7.87 x 10-2 mole Cu x 2 mole AgNO3 = 0.157 mole AgNO3 1 mole Cu

There are only 0.118 mole of AgNO3 present, however, so Cu must be present in excess with AgNO3 limiting.

3. 2KOH + (NH4)2SO4 ---> K2SO4 + 2NH3 + 2H2O

This was probably a difficult equation for you. This type was covered in the double replacement type of reaction earlier.

20.0gKOH x 1 mole KOH = 0.356 mole KOH 56.11gKOH

15.0g(NH4)2SO4 x 1 mole (NH4)2SO4 = 0.114 mole (NH4)2SO4 132.14

0.356 mole KOH x 1 mole (NH4)2SO4 = 0.178 mole (NH4)2SO4 2 mole KOH

We do not have enough ammonium sulfate for all the KOH to react so ammonium sulfate is limiting.

a) 0.114 mole (NH4)2SO4 x 1 mole K2SO4 = 0.114 mole (NH4)2SO4 1 mole (NH4)2SO4

b) 0.114 mole (NH4)2SO4 x 2 mole NH3 x 17.03gNH3 = 3.88gNH3 mole (NH4)2SO4 1 mole NH3

4. Reaction 1: 4Fe(C2H3O2)3 + 6MgC2O4 ---> 6Mg(C2H3O2)2 + 2Fe2(C2O4)3

15.0g 10.0g ?

MW of Fe(C2H3O2)3 = 232.99g/mole MgC2O4 = 112.33g/mole

15.0gFe(C2H3O2)3 x 1 mole = 6.44 x 10= -2 mole Fe(C 2H3O2)3 232.99g

10.0g MgC2O4 x 1 mole = 8.90 x 10 -2 mole MgC 2O4 112.33g

By stoichiometry8.90 x 10-2 mole MgC2O4 x 4 mole Fe(C2H3O2)3 = 5.93 x 10-2 mole

6 mole MgC2O4of Fe(C2H3O2)3

We have enough Fe(C2H3O2)3 for all of MgC2O4 to react soMgC2O4 is limiting.

8.90 x 10-2 mole MgC2O4 x 6 mole Mg(C2H3O2)2 x 14240g mg(C2H3O2)2 6 mole MgC2O4 1 mole Mg (C2H3O2)2

12.7g Mg(C2H3O2)2

Reaction 2:4Fe(C2H3O2)3 + 6MgSO4 ---> 6Mg(C2H3O2)3 + 2Fe2(SO4)3 15.0g 10.0g

MW of MgSO4 = 120.38

15.0g MgSO4 x 1 mole MgSO4 = .125 mole MgSO4 120.38gMgSO4

.125 mole MgSO4 x 4 mole Fe(C2H3O2)3 .0831 mole of Fe(C2H3O2)3 6 mole MgSO4

4. cont'd

If all of MgSO4 is to be used up, 0.0831 mole of Fe(C2H3O2)3 is needed. How much do we have? - only .0644 mole. Here , Fe(C2H3O2)3 is the limiting reagent..0644mole Fe(C2H3O2)3 x 6 mole Mg(C2H3O2)2 x 142.20g Mg(C2H3O2)2 =

4 mole Fe(C2H3O2)3 1 mole

13.7g MG(C2H3O2)2Reaction 2 will produce the greatest yield.

5. 3Ca(OH)2 + 2H3PO4 ---> Ca3(PO4)2 + 6H2O36.0g54.0g ?

Part 1 of this problem is to find the limiting reagent. MW's - Ca(OH)2 74.10g/mole

H3PO4 98.00g/mole Ca3(PO4)2 310.19g/mole

36.0gCu(OH)2 x 1 mole Ca(OH)2 = 0.486 mole Ca(OH)2 74.10g

54.0g H3PO4 x 1 mole H3PO4 = 0.551 mole H3PO4

98.00g

.486 mole Ca(OH)2 x 2 mole H3PO4 = 0.324 mole of H3PO4 3 mole Ca(OH)2

Do we have enough H3PO4 for all of the Ca(OH)2 to react?Yes, we have plenty so Ca(OH)2 is the limiting reagent.

5. Part 2 of this problem is to calculate the theoretical yield. Start with the limiting reagent.

0.486 mole Ca(OH)2 x 1 mole Ca3(PO4)2 x 310.19g = 50.3g 3 mole Ca(OH)2 1 mole Ca3(PO4)2

The theortical yield (maximum possible) is 50.3g of Ca3(PO4)2

Part 3 of this problem is to calculate the percent yield.

Percent yield = Actual yield x 100 Theoretical yield

= 45.2g x 100 50.3g

= 89.9%

Chapter Review Exercises

Page 91

1. 2NF3 ---> N2 + 3F2a) (exactly) 3 mole ?g ?g

3 mole NF3 x 1 mole N2 x 28.01gN2 = 42.02gN2 2 mole NF3 1 mole N2

3 mole NF3 x 3 mole F2 x 38.00gF2 = 171.0gF2 2 mole NF3 1 mole F2

b) 2NF3 ---> N2 + 3F20.268 mole ?g ?g.268 mole NF3 x 1 mole N2 x 28.01gN2 = 3.75gN2

2 mole NF3 1 mole N2

.268 mole NF3 x 3 mole Fe x 38.00gFe = 15.3gF2 2 mole NF3 1 mole F2

2. 2P + 5Cl2 ---> 2PCl5 (or you can use P4)?g 2.70 mole

2.70 mole PCl5 x 5 mole Cl2 x 70.91gCl2 = 478.62g = 479gCl22 mole PCl3 1 mole Cl2

3. 2As + 3Cl2 ---> 2AsCl3 0.113 mole ?

0.113 mole x Cl2 2 mole AsCl3 x 181.28gAsCl3=13.66g =13.7gAgCl3 3 mole Cl2 1 mole AsCl3

4. CS2 + 3O2 ---> CO2 + 2SO29.34g ?g9.34gCS2 x 1 mole CS2 x 3mole O2 x 32.00gO2 = 11.78 = 11.8gO2

76.14gCS2 1 mole CS2 1 mole O2

5. PbyOx + H2 ---> Pb + H2 OWe are trying to find x and y.1 mole 8.1g 622g One mole of PbyOx reacts with 8.1g H2 to form 622gPb. There are

many ways to solve this problem. This is only one way.Change all grams to moles:622gPb x 1 mole Pb = 3.00 mole Pb

207.19gPb8.1gH2 x 1 mole H2 = 4.00 mole H2

2.02gH2

The Pb caould only have come form PbOx so we must have three mole of Pb present originally.

The 4.00 mole of H2 must have formed 4.00 mole of H2O (to balance the hydrogens). This means that there had to be 4.00 mole of O atoms present originally.

This gives a formula of Pb3O4.Check this:Pb3O4 + 4H2 ---> 3Pb + 4H2O685.57 8.1g = 622g + 72.06

693 694694

6. H2 + Cl2 ---> 2HCl11.1g 33.3g---> ?g

Cl2 is so much heavier than H2 that I would guess 33.3gCl2 is the limiting reagent, but we should check it out anyway.

11.1gH2 x 1 mole H2 = 5.51 mole H2; 33.3gCl2 x 1 mole Cl2 =2.02 70.91gCl2

= 0.470 mole Cl

Since H2 and Cl2 react in a 1:1 mole ratio, Cl2 is definitely the limiting reagent.

0.470 mole Cl2 x 2 mole HCl x 36.45g HCl = 34.3gHCl 1 mole Cl2 1 mole HCl

7. 3C + CaO ---> CO + CaC22.50g 5.00g 2.45g <--- actual, find theoretical

* Find the moles of C and CaO2.50gC x 1 mole = .208 mole C; 5.00gCaO x 1 mole CaO = 0.0892

12.01g 56.08 moleCaO

* Let's assume CaO is limiting and see if there is enough C present.

0.0892 mole CaO x 3 mole C = 0.2675 mole C 1 mole CaO

* We do not have 0.2675 mole of C. C must be limiting. We do have 0.208 mole of C. 0.208 mle C x 1 mole CaC2 x 64.10gCaC2 = 4.44gCaC2

3 mole C 1 mole CaC2

* Find the percent yield % yield = Actual yield x 100 = 2.45g x 100 = 55.2%

Theoretical yield 4.44g

8. XO2 ---> x + O225.2g 3.2g

* How much X do we have? Subtract 25.2g - 3.2g = 22.0g* How many moles of x do we have? According to this we have 3.2gO2 x 1 mole O2 = .10 mole O2.

32.00gO

This .10 mole O2 must have come from XO2. We must also have .10 mole XO2 present and then .10 mole x.

* If .10 mole of X weighs 22.0g, we can easily figure the weight of one mole.

22.0g of X = 22Og/1 mole.10 mole

The atomic weight of X must be 220.

Chapter 2 - Alternate Problems

1. 4NH3 + 5O2 ---> 4NO + 6H2O76.0g ?g ?g ?g

76.0gNH3 x 1 mole NH3 x 5 mole O2 x 32.00gO2 = 178.51g = 179gO2 17.03gNH3 4NH3 1 mole O2

76.0gNH3 x 1 mole NH x 4 mole NO x 30.01gNO = 134gNO_______ 17.03gNH3 4 mole NH3 1 mole NO ______ ______

We could just subtract 134gNO from (76.0 + 179gO2) to find the mass of H2O (Law of Conservation of Mass)

76.0gNH3 x 1 mole NH3 x 6 mole H2O x 18.02gH2O = 121gH2O17.03gNH3 4 mole NH3 1 mole H2 O

2. 2PCl3 ---> 2P + 3Cl2 ; C + 2Cl2 ---> CCl41.30 mole 1.30 mole PCl3 x 3 mole Cl2 = 1.95 mole Cl2 formed

2 mole PCl3

This 1.95 mole Cl2 can then be used in the second equation as is. Don't try to adjust the moles.

C + 2Cl2 ---> CCl4 1.95 mole ?g

1.95 mole Cl2 x 1 mole CCl=4 x 153.82gCCl4 = 150gCCl4 2 mole Cl2 1 mole CCl4

3. Fe2O3 + 3CO ---> 2Fe + 3CO2?g 910kg ?g910kgFe x 1 mole Fe x 2 mole Fe2O3 x 159.70gFe2O3 =

55.85gFe 1 mole Fe 1 mole Fe2O3

5200 kg Fe2O3

I did not change to grams because I wanted to find answer is kg. You could have converted or could have used 1k mole Fe 55.85kg Fe

910kgFe x 1 mole Fe x 3 mole CO2 x 44.01gCO2 = 2150kgCO2 55.85gFe 1 mole Fe 1 mole CO2

4. 2C6H18 + 21O2 ---> 12CO2 + 18H2O1 mole ?g

1 mole C6H18 x 21 mole O2 x 32.00g = 336gO2 2 mole C6H18 1 mole O2

2C6H18 + 15O2 ---> 12CO + 18H2O

1 mole

1 mole C6H18 x 15 mole O2 x 32.00gO2 = 240.gO2 2 mole C6H18 1 mole O2

Difference = 336g - 240g = 96g

5. W + 3Cl2 ---> WCl612.6g 13.6g

This is a limiting reagent - type problem.12.6gW x 1 mole W = .0685 mole W

183.85gW

13.6g Cl2 x 1 mole Cl2 = 0.192 mole Cl2 70.91

It looks as if either might be limiting. Try one and compare with the moles of the other.

12.6gW x 1 mole W x 3 mole Cl2 = 0.206 mole Cl2 183.85gW 1 mole W

We do not have quite 0.206 moles of Cl2 so W is in excess.

* To answer the first part of the problem we must find the grams of W that reacted:

0.192 mole Cl2 x 1 mole W x 183.85gW = 11.8gW 3 mole Cl2 1 mole W

* To find the mass of unreacted W, subtract:12.6gW - 11.8gW = 0.8g W left over

* Now we need to find the mass of WCl6 that forms. Either work the stoichiometry or add 11.8gW to 13.6gCl2 = 25.4g

0.192 mole Cl2 x 1 mole WCl6 x 396.57gWCl6 = 25.4gWCl6 3 mole Cl2 1 mole WCl6

6. 2HgO ---> 2Hg + O275.8g 62.7g

75.8gHgO x 1 mole HgO x 2 mole Hg x 200.59gHg = 70.2gHg 216.59gHgO 2 mole HgO 1 mole Hg

% yield = Actual x 100 = 62.7gHg x 100 = 89.3%Theoretical 70.2gHg

7. This is a difficult kind of problem for you.An 18.0% NaCl solution by mass is (assumeing 100g of solution)

18.0gNaCl What is the 100g of solution? The solution100g solution

is made up of 18.0gNaCl and 82g of water. We have 18.0gNaCl 82.0gH2O

125g of solution x 18.0gNaCl = 22.5gNaCl 100g solution

125g solution - 22.5gNaCl = 102.5gH2O

8. 3S + 2Cl2 ---> S2Cl2 + SCl250.0g 100g ?g

All of the S and Cl2 have reacted. We need to find the moles of either present and use this.

50.0gS x 1 mole S x 1 mole S2Cl2 x 135.03gS2Cl2 = 70.2gS2Cl2 32.06gS 3 mole S 1 mole S2Cl2

Check this by using Cl2100gCl2 x 1 mole Cl2 x 1 mole S2Cl2 x 135.03gS2Cl2=95.2gS2Cl2

70.91gCl2 2 mole Cl2 1 mole S2Cl2

-Oops- Despite what the problem says, Cl2 must be present in excess so theoretical mass of S2Cl2 formed must me 70.2gS2Cl2

This laboratory experiment should preceed Chapter 1, Section 3. An entire class period may be needed to perform this experiment.

Partial Weight Analysis of a Compound

The blue copper (II) sulfate found in the lab is an example of hydrate. Hydrates have definite amounts of water incorporated periodically within the crystalline structure. The five moles of water molecules per mole of copper sulfate can be removed by heating. This enables you to determine the percentage of water of hydration present in a sample.

Weigh a clean, dry crucible to the nearest 0.0lg. ________g

Grind up crystalline hydrated copper (II) sulfate

in a mortar. Weigh out between 2.50 and 3.50 gramsof the salt in the crucible. Be sure to clean upany spilled copper sulfate from the balance panbefore you record the weight.Weight of sample and crucible. _________g

Weight of sample, calculated _________g

4. Place the crucible and its contents on a pipestem triange set on a ring clamped to a ring stand.

Heating the system gently for 3 to 5 minutes to avoid splattering. Continue to heat but more strongly for 10 more minutes. Record any observations during this time.

Observations:

Obtain a desiccator from the lab or make one from a large jar. The

bottom of the jar should have 1/2 inch solid anhydrous calcium chloride. A

commercial laboratory desiccator may have the lid coated with petroleum jelly so do not set

the lid down flat on the table. The top should be placed upon the jar as

soon as it is filled and not removed until necessary. The crucible will

rest on a wire screen or plate placed above the calcium chloride.

At the end of the 10-minute heating, allow the crucible to cool on the stand for 2 minutes and then use tongs to transfer it to the desiccator.

5. After the crucible and contents have cooled to room temperature in the closed desiccator, weigh the crucible and solid.

Weight of crucible and residue. _________g

6. Calculate the weight of the residue _________g

7. Calculate the weight of the water loss by the original sample_________g

8. Calculate the percentage of water of hydration in the sample_________g

9. Place a few drops of water on the cooled, weighed residue.Feel the bottom of the crucible before and after doing this. Record any observations.

Observations:

An entire clas period may be needed to perform this experiment.

Experiment: The formula of a compound

Each compound has its own distinct formula, as was just seen in the previous section, has its

own constant composition. In this experiment you are going to make a compound and then calculate

its simplest formula.

1. Weigh a clean, dry crucible and record its weight to 0.0lg _________

2. Obtain a clean piece of magnesium ribbon about 35 cm in length. Use a fine piece of

steel wool to clean the ribbon. Cut the ribbon into pieces of approximately 1 cm and add to the crucible. Weigh the crucible and contents. _________

3. Weight of magnesium ribbon by difference. _________

4. Calculate the moles of magnesium (M.W. of Mg = 24.31)________

5. Set up the ring, triangle, and ringstand as in the previous experiment.Heat the covered crucible (picture 1)gently at first and then graduallyincrease the heat intensity for 2 minutes.

With your tongs, carefully tilt the lid (picture 2) and continue strongheating of the partially uncovered crucible for an additional 10 minutes.

At the end of this time, allow thecovered crucible and contents to cool.Check to see that all of the magnesiumhas reacted before covering. If thereis still unreacted magnesium, heat strongly again.

Since the magnesium reacted with both the oxygen and nitrogen in the air, it is necessary to convert the mixture of compounds to only the oxide compound.

Use a medicine dropper to add only enough distilled water to cover the contents of the crucible Carefully heat the system. Observe the odor of any vapor that has been produced. When the water has evaporated, add a little more and carefully heat again. Continue to do this until there is no longer a distinct odor produced upon heating.

When this has happened, continue careful heating until all of the water has evaporated. Strongly heat the uncovered crucible for five more minutes. Allow the crucible and its product to cool. Weigh the crucible and its contents.

Weight of crucible and contents ________

6. Determine the weight of the magnesium oxide bydifference______

7. Determine the weight of oxygen that combined with the original weight of magnesium(You know how much Mg is present in the compound.) _______

8. Calculate the number of moles of combined oxygen _______

9. Determine the simplest formula for magnesium oxide.(If you are not sure what to do here, read ahead in section 3.) _______