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BC Science CHEMISTRY 11

SAMPLE UNITDraft Material, Not Final Form

THE MOLEA Chemical Unit of Measurement

Unit

5

1Mole Concept—Draft Version—Copyright Edvantage Interactive 2010

Unit 5 The Mole – A Chemical Unit of Measurement

By the end of this unit, you should be able to complete the following tasks:

• Explain the significance and use of the mole• Perform calculations involving the mole• Determine relationships between molar quantities of gases at STP• Perform calculations involving molecular and empirical formulas to identify a substance• Describe concentration in terms of molarity• Perform calculations involving molarity

_________________________________________________________________________________________________

You will need to refer to your Data Booklet for this unit._________________________________________________________________________________________________

By the end of this unit, you should be able to relate the concept of the mole to the quantitative properties of matter.

Go to bcscience.com for more information and links._________________________________________________________________________________________________

By the end of this unit you should know these key terms:• aqueous• atomic mass unit• Avogadro’s hypothesis• chemical species• density• diatomic molecule• dilution• empirical formula • molarity• molar mass • molar solution• molar volume • mole • molecular formula • molecular mass• percentage composition • relative atomic mass• solute• solution• solvent• standard solution• STP

The mole is at the centre of the chemical measurement.

2 Mole Concept—Draft Version—Copyright Edvantage Interactive 2010

5.1 Relative Atomic Mass and the Mole – Counting Atoms and Molecules

Warm Up1. Complete the grocery list by filling in the missing units (Figure 5.1a).

One __________ eggs, Two __________milk, Three _________flour

2. From your answer to #1, what are the three ways that we typically express amounts of matter?

(a) number of items (b) ____________________ (c) __________________

Relative Mass

When we communicate quantities of any matter, whether it is items on a grocery list or incredibly small atoms, we do so in three ways. We state number, volume, and mass. Consider a basket of identical grapefruits and a basket of identical oranges. One of each fruit placed separately on a balance might show the mass of a grapefruit to be 400 g and the mass of an orange to be 200 g, giving a mass ratio of 2-to-1. Two grapefruits having a mass of 800 g and two oranges having a mass of 400 g would still exhibit a 2-to-1 mass ratio. As long as the number of each fruit placed on the scale was the same, the mass ratio would be the same. So if we ignored units and assigned an orange a mass of 1, a grapefruit would have a mass, relative to the orange, of 2.

Suppose you wanted to measure a staple and a grain of rice, each of which is too small to register a mass on an ordinary balance (Figure 5.1b). The data above can help you. Any equal number of identical staples and identical grains of rice measured separately on a balance will give you the same mass ratio as will a single staple and a single rice grain.

Relative Masses of Atoms

Relative mass can be used to measure entities that are far smaller than paper clips and pushpins. For example, atoms are far too small for us to measure their mass directly. Yet, relative and even individual masses of atoms can be determined by measuring and comparing the masses of a known large number of each. As atoms are far too small to see, how are we to know when we have a certain number of any type of atom?

To answer this question, it helps to understand the work of British schoolteacher John Dalton, who proposed the atomic theory in 1808. Having studied the work of other scientists, Dalton proposed that all matter was composed of tiny particles he called atoms. He believed that atoms of different elements had unique masses, and that any one atom was much too small to weigh on a balance. So he attempted to find the relative masses of different atoms by studying the composition of compounds containing those atoms.

Dalton knew that water had a constant composition in which 8 g of oxygen were always present for every 1 g of hydrogen. He did not know the formula for water, so he assumed it was simply HO. Since a 1-to-1 O-to-H atom ratio always gave an 8-to-1 O-to-H mass ratio, he concluded that one atom of oxygen must be eight times heavier than one atom of hydrogen. He therefore assigned a hydrogen atom a relative mass of 1 and an oxygen atom a relative mass of 8. Dalton studied other compounds using similar assumptions to produce the first table of relative atomic masses. Many of Dalton’s formula assumptions and relative mass values were later proven incorrect. However, this method for determining relative masses works if you know the correct formulas of the compounds containing those atoms.

Figure 5.1a We express amounts of matter in several different ways.

Figure 5.1b One hundred staples and one hundred grains of rice have the same mass ratio as one staple and one grain of rice, 2:1.


A way to determine some of those formulas was discovered in the early 1800s. In 1809, French chemist Joseph-Louis Gay-Lussac measured the volumes of gases that reacted to form compounds (Figure 5.1c). He found that gases measured at the same temperature and pressure always reacted in whole number volume ratios.

• 1 volume hydrogen gas + 1 volume chlorine gas reacts to produce 2 volumes hydrogen chloride gas

• 2 volumes hydrogen gas + 1 volume oxygen gas reacts to produce 2 volumes gaseous water

Notice in the reaction above that one volume of a reacting gas produces two volumes of a product gas containing that element. This indicates that each of the reacting particles must actually be diatomic molecules, which are molecules containing two atoms of that element. Some other examples of elements that form diatomic molecules include oxygen, nitrogen, fluorine, bromine, and iodine.

Volumes of Gases

Soon after, an Italian chemist named Amadeo Avogadro (Figure 5.1d) interpreted Gay-Lussac’s work and developed an hypothesis.

Avogadro’s hypothesis may be difficult to accept at first reading. Some gas particles are much larger than others and it does not make sense to think that 100 large particles would occupy the same volume as 100 small particles. For example, imagine that the

Figure 5.1c Gay-Lussac was an avid hot-air balloonist and conducted some of his experiments aloft.

Figure 5.1d Amadeo Avogadro, 1776-1856

Avogadro’s hypothesis states that equal volumes of different gases, measured at the same temperature and pressure, contain equal numbers of particles.

large particles are grapefruits and the small particles are grapes. Suppose you distributed the100 grapefruits evenly throughout the volume of an airplane hangar and distributed 100 grapes evenly throughout the volume of an identical airplane hangar. Because the distance between particles is so great, the size of each particle is unrelated to the space that an equal number of each will fill. Avogadro’s hypothesis is only possible if the distance between the gas particles is much greater than the size of the gas particles themselves.

We can use Avogadro’s hypothesis to determine correct formulas. A 2-to-1 volume ratio of hydrogen to oxygen in water must mean a 2-to-1 atom ratio in the formula. This means that water’s formula must be H2O and not HO as Dalton had assumed. With the correct formula, the 8-to-1 mass ratio of oxygen to hydrogen in water must also mean that the relative mass of one oxygen atom compared to one hydrogen atom is 16-to-1.

Avogadro’s hypothesis also helps us determine relative masses of atoms. If we have equal volumes of different gaseous elements at the same temperature and pressure, we know that each contains an equal number of atoms. Therefore, the relative masses of each volume will also represent the relative masses of the individual atoms. The measured masses of 5.00 L samples of different gases are displayed in Table 5.1.

Table 5.1 Masses of 5.00 L Samples of Gases

Gas Accepted Formula Measured MassApproximate Mass (relative to hydrogen)

Hydrogen H2 0.45 g 1

Nitrogen N2 6.25 g 14

Oxygen O2 7.15 g 16

It took about 50 years for Avogadro’s hypothesis to gain acceptance in the scientific community. However, as accurate formulas for compounds were eventually determined, a set of reliable relative atomic masses was established.


Atomic Mass Unit

Modern tools, such as the mass spectrometer, analyze vaporized samples of different elements to determine relative masses of atoms more precisely than Dalton and earlier scientists were able to do. We now use a different standard than hydrogen to compare masses. The agreed-upon standard is the atomic mass unit (abbreviated amu or simply u), which is defined as exactly 1

12 the mass of a single carbon-12 atom. In other words, one carbon-12 atom is considered to have a mass of exactly 12 atomic mass units. You can find mass numbers in the periodic table that represent the masses of an atom of each element relative to a carbon-12 atom (Figure 5.1e).

The Mole

You may recall our earlier question: As atoms are far too small to see, how are we to know when we have a certain number of any type of atom? Suppose that you want to determine the amount of a chemical species, such as carbon-12 atoms. A chemical species, sometimes called a particle, or chemical entity, is any atom, molecule, or ion. Chemical species are extremely tiny and it would take a huge number of them to register a mass in grams on a balance. How many carbon-12 atoms would you need to have a mass of exactly 12 g? (Figure 5.1f )• Calculated as precisely as possible, the number of carbon-12 atoms in 12 g is

602 213 670 000 000 000 000 000.• Expressed in scientific notation, the value is 6.0221367 × 1023. This number is called the

“mole” or “Avogadro’s number,” named in his honour.

Quick Check1. Dalton attempted to determine the relative masses of atoms by ___________________________________2. Avogadro’s hypothesis is only possible if _______________________________________________________3. A 5.00 L container of helium gas contains “X” atoms of helium. A 15.0 L container of argon gas at the same

temperature and pressure will contain ___________ atoms of argon.

Figure 5.1e The average mass of hydrogen is 1.01 u, which means that one hydrogen atom is approximately 1

12 the

mass of a single carbon atom.

1.01

The mole is the SI base unit amount of pure substance that contains the same number of chemical species as there are atoms in exactly 12 g of carbon-12. This number is usually expressed to three significant figures or 6.02 × 1023. “Mole” is abbreviated as mol and given the symbol n (Figure 5.1g).

Figure 5.1f 6.0221367 × 1023 of carbon atoms

Figure 5.1g The number of entities in 6.02 × 1023 is often referred to as Avogadro’s number or Avogadro’s constant.

n


One mole of any entity equals 6.02 × 1023 of that entity. In other words, one mole of carrots equals 6.02 × 1023 carrots. One mole of atoms equals 6.02 × 1023 atoms. You can work with fractions of moles and multiples of moles. • 0.500 mol of ions = 3.01 × 1023 ions• 10.0 mol of molecules = 6.02 × 1024 molecules • 1.00 mol = 1000 mmol (millimoles)• 1.00 mol = 0.001 kmol (kilomole)

You can use an analogy to help you picture how big 1 mol is. Suppose that an imaginary meteorological phenomenon called “El Pea-ño” suddenly causes garden peas to fall over the entire province of British Columbia and cover the ground with peas 1 m deep. Is that amount equal to 1 mol of peas? NO, the layer contains only 1018 peas. Imagine that more peas then fall over every continent on Earth and cover all the continents with a layer 1 m deep. Now there are 1021 peas on the ground. The weather then gets much worse and the oceans freeze. Soon the entire land and sea area of Earth is buried under 1 m of peas. You still do not have 1 mol of peas. Suppose you search the galaxy for 250 planets the same size as Earth and cover each of them with peas 1 m deep. Finally, you have 1 mol of peas.

Quick Check1. What is the definition of an atomic mass unit?

________________________________________________________________________________________2. What is a chemical species?

________________________________________________________________________________________3. How many entities does Avogadro’s number represent?

________________________________________________________________________________________4. Complete the following chart.

Number of Moles Number of Entities

3.01 × 1023

0.75

1.5 × 1024

10.0

Counting Chemical Species with Avogadro’s Number

As you continue in this unit, you will discover how the mole allows us to not only communicate quantities as number, mass, or volume, but also to convert between each of these quantities.

The easiest way to calculate numbers of particles is to use the equivalence statement of 1 mol = 6.02 × 1023 as a conversion factor. Conversion factors may be written in reciprocal form, depending on the calculation being done (Figure 5.1h).

Figure 5.1h Converting moles and number of particles


As with any problem involving the application of conversion factors, it is important to begin with the end in mind. • To determine the number of particles, use 6.02 × 1023 species

1 mol• If given the number of a chemical species and asked to calculate how many moles it

contains, you would need to invert the factor and use 1 mol6.02 × 1023 species

The following are steps in the solution process for questions involving moles. Some questions that you will encounter will require only one conversion, while others may require more than one.

Step 1: Read the question carefully to determine the units required in the answer and the units given in the question. In other words, begin with the end in mind.

Step 2: Look for equivalence statements containing the given and required units.

Step 3: Use those equivalence statements to construct the conversion factors that allow you to change the given units into the desired units.

Step 4: Arrange the conversion factors into a calculation sequence such that, when the calculation is performed, the given units will cancel by multiplication and the required units will remain.

6.02 × 1023 species1 mol

or 1 mol6.02 × 1023 species

Sample ProblemHow many atoms of neon are contained in 5.0 mol of neon?

What to Think About

Begin with the end in mind to help you plan your use of conversion factors. The unit in our answer must be atoms of neon and we are given moles of neon.

How to Do It

We know from the definition of a mole that: 1 mol Ne = 6.02 x 1023 atoms Ne This equivalence statement allows us to write two conversion factors that are simply reciprocals of each other:

6.02 x 1023 atoms Ne1 mol Ne

and 1 mol Ne

6.02 x 1023 atoms NeChoose the first conversion factor to multiply by the given quantity because this ratio cancels the given units and converts them to the desired units. Note that the correct conversion factor for this problem lists the final desired units in the numerator position and the units that must cancel in the denominator position.

5 mol Ne x 6.02 x 1023 atoms Nemol Ne

= 3.01 × 1024 atoms Ne


Practice Problems—Converting Moles to Entities

1. 50.0 mol of C contain how many atoms of C?

2. How many molecules of H2O are contained in 30.0 mol of H2O?

3. 13.5 mol of argon contain how many atoms of argon?

Sample ProblemGold is the most malleable metal. It can be hammered into sheets called “gold leaf” that are only several hundred atoms thick. A bar of gold half the size of a deck of cards could form a sheet large enough to cover a basketball court. How many moles of gold would 5.00 × 102 atoms of gold represent?

What to Think About

Look for an equivalence statement that equates atoms of gold to moles of gold.

How to Do It 1.00 mol Au = 6.02 × 1023 atoms AuConstruct a conversion factor that cancels atoms and leaves moles of gold. Multiply the given quantity by this conversion factor.

5.00 × 102 atoms Au x 1 mol AU6.02 x 1023 atoms Au

= 8.30 × 10-22 mol Au

Practice Problems—Converting Entities to Moles

1. 2.01× 1023 atoms of He equal how many moles of He?

2. 1.81 × 1024 atoms of sodium represent how many moles of sodium?

3. 6.02 × 1024 molecules of sugar, C12H22O11 represent how many moles of sugar?


Working with Subscripts

The subscripts in a chemical formula tell you how many atoms or ions of each element there are. For example, in 1 mol of water (H2O) there are 6.02 × 1023 molecules of water. Each molecule of water has three atoms (1H and 2 O), so in 1 mol of water there would be 3 × 6.02 × 1023 atoms.

Sample ProblemHow many fluoride ions are contained in 2.5 moles of CaF2?

What to Think About

Beginning with the end in mind, a careful read of the problem shows you need to calculate a number of particles (in this case, fluoride ions, F-). The first form of the conversion factor will allow cancelling of the given unit moles and the conversion to the unit representing the particles, fluoride ions.

There is one other issue here and it is an important one. What kind of particle is CaF2? This is an ionic compound and consists of formula units. Each formula unit contains two fluoride ions. A second conversion factor must be added to account for this fact.

How to Do It

2.5 moles CaF2 x 6.02 x 1023 F – CaF2

1 mol CaF2 x 2 F – ions

1 F – CaF2

= 3.0 × 1024 F – ions

Notice that the answer to this problem makes sense. As one mole contains 6.02 × 1023, it is logical that 2.5 moles would contain two and one-half times Avogadro’s number of CaF2 units and twice that again number of fluoride ions. Always check to see that your answer is logical.

Practice Problems—Converting Compounds to Particles

1. (a) How many moles of hydrogen atoms are in 1 mol of ammonia molecules (NH3)?

(b) How many atoms of hydrogen are there in 1 mol of ammonia molecules?

2. How many atoms in total are in 6 mol of water?

3. The nitrate ion, NO3− can impair the immune system and cause stress to some aquatic species if

levels rise too high. When it combines with hydrogen to form nitric acid (HNO3) it becomes a highly toxic and corrosive acid. How many nitrate ions are contained in 15 mol of HNO3(aq)?


A Mole of Paper Clips in Space and Time Activity

Question

How big is one mole of paper clips?

Concepts and Theories

The mole is a convenient and useful term for counting large quantities of entities. You know that one mole of paper clips equals 6.02 × 1023 paper clips, but can you picture just how big that number really is? Suppose you linked one mole of paper clips together end-to-end. Would that be enough paper clips to reach to Pluto and back? How long would it take to count one mole of paper clips?

Average distance from the Earth to Pluto 5.85 billion km

Number of seconds in a year 31 536 000

Events and Objects1. Use a centimetre ruler to find the length of a standard paper clip. 2. Determine the length of a mole of standard paper clips linked end-to-end (ignore loss of length at each link). 3. Use your favourite search engine to determine the present population of Earth. Then calculate how long it would

take the population of Earth to count one mole of paper clips, if each person counts one paper clip per second, 24 hours a day, 365 days a year.

Records and Transformations4. (a) Could one mole of paperclips reach to Pluto and back from Earth? Explain.

(b) Could you make the journey to and from Pluto more than once? If so, how many times?

5. How many years would it take the population of Earth to count one mole of paper clips, if each person counts one paper clip per second, 24 hours a day, 365 days a year?


Review Questions

1. Hemoglobin is the oxygen transport molecule in red blood cells. Myoglobin is the oxygen binding molecule in muscle tissues. A 1.00 × 10-4 mol sample of hemoglobin molecules is found to have a mass of 6.80 g. A sample containing an equal number of myoglobin molecules is measured and found to have a mass of 1.70 g.

(a) Calculate the mass of one molecule of hemoglobin relative to one molecule of myoglobin.

(b) Determine the mass of 1 mol of hemoglobin molecules and the mass of 1 mol of myoglobin molecules.

2. What does Avogadro’s hypothesis state?

3. What is the definition of an atomic mass unit?

4. What is the definition of a mole?

5. One of the statements (a), (b), (c), or (d) is incorrect. Circle the incorrect statement and explain why it is incorrect.

(a) The number of atoms in 1 mol of aluminum is the same as the number of atoms in 1 mol of magnesium.

(b) In 1 mol of water, H2O there is the same number of atoms as in 1 mol of methane, CH4.

(c) There is the same number of ions in 1 mol of calcium sulphate, CaSO4, as there is in 1 mol of sodium chloride, NaCl.

(d) The number of molecules in 1 mol of ammonia, NH3, is the same as the number of molecules in 1 mol of chlorine, Cl2.

6. How many bromine molecules does 0.50 mol of Br2 contain?

7. A sample of CO gas contained 6.00 × 1022 molecules. How many moles of carbon monoxide does this represent?


8. How many moles of iodine molecules (I 2 ) are contained in 9 × 1023 molecules of iodine?

9. How many oxygen atoms (O2) are in 5 mol of oxygen molecules?

10. Lithium is the lightest metal and least dense solid element. How many lithium moles does 4.2154 × 1023 lithium atoms equal?

11. How many moles do 9.58 × 1023 flouride ions represent?

12. Battery acid is a concentrated aqueous solution of sulphuric acid, H2SO4. Convert 2.86 × 1024 molecules of H2SO4 into moles.

13. (a) Carbon dioxide, produced by respiration in plants and animals, causes the slightly acidic nature of unpolluted rain. How many molecules of CO2 are in a 0.725-mol sample?

(b) How many atoms are in the sample?

14. The male luna moth can detect specialized chemicals known as pheromones in order to find a mate. If a moth can detect at least 1.70 × 109 molecules of the pheromone C12H22O, how many moles of the substance must be present?

15. Glycerol, C3H5(OH)3, is a colourless, viscous chemical found in cough syrup, toothpaste, and soaps. Calculate the number of hydrogen atoms in 4.50 mol of glycerol.

16. Bluestone is an attractive mineral with the chemical name copper(II) sulphate pentahydrate. How many molecules of water are in a 1.75-mol sample of bluestone?

luna moth


5.2 Expressing and Calculating Molar Quantities – The Chemist’s Favourite Unit

Warm UpThe cells in your body use glucose, C6H12O6, as an important source of energy.

1. How many molecules are in 1 mol of glucose? _________________

2. How many carbon atoms are in 1 mol of glucose? _________________

3. How many atoms in total are in 1 mol of glucose? __________________

Measuring Moles

The mole is more than just a counting unit of convenience. The definition of the mole as the “number of atoms in exactly 12 grams of carbon-12” specifies two quantities: a number of species in a fixed mass of matter. You can think of the two parts of this definition as a question and an answer. The question is “How many atoms of carbon-12 are contained in 12 grams of carbon-12?” and the answer is “Avogadro’s number of 6.02 × 1023.”

If 1 mol of carbon-12 atoms has a mass of 12 grams, then 1 mol of any other atom will have a molar mass equal to that element’s relative mass expressed in grams (Figure 5.2b). Stated another way, if you have a mole of any element, you can place the unit “grams” after that element’s mass number in the periodic table (Table 5.2a).

Table 5.2a Mass of Some Elements

Mass of 1 Atom Mass of 1 Mol

Oxygen 16.0 u 16.0 g

Sodium 23.0 u 23.0 g

Iron 55.8 u 55.8 g

Figure 5.2a Glucose (C6H12O6)]

Figure 5.2b Examples of molar masses

A mole of a pure substance has a mass in grams exactly equal to the substance’s atomic or molecular weight.

Quick Check1. (a) Complete the following table by adding the missing quantities for the mystery substance.

Number of Moles Number of Atoms Number of Grams

63.01 × 1023 12

6.02 × 1023

210

(b) What element is the mystery substance? ____________________________________________

2. (a) What does one atom of sodium weigh? _____________________________________________

(b) What do 6.02 × 1023 atoms of sodium weigh? ______________________________________


The Molar Mass

Unlike the equivalency between the mole and the number of species it represents, the mass of one mole of a chemical species is not a constant value. Rather it depends upon the identity of the species and its atomic makeup (Figure 5.2c).

Figure 5.2c The mass of 1 mol of a chemical species depends on the atoms that make it up.

The molar mass of a pure substance is the mass of one mole of the substance expressed in g/mol.

Simply stated, the molar mass of any element or compound is the atomic or molecular mass expressed in units of grams. The conversion factors may be written in reciprocal form, depending on the calculation being done (Figure 5.2d).

Figure 5.2d The molar mass of an element is its atomic or molecular mass with units of grams.

(molar mass) g1 mol

or

1 mol(molar mass) g

Figure 5.2e Molar mass is the mass of 1 mol of an element or compound.


Sample ProblemPhosphoric acid, H3PO4, gives cola its tart taste. What is the mass of 0.0154 mol of phosphoric acid?

What to Think About

First, consider the units for your answer. As you are calculating a mass, you will need an answer in grams. This requires use of the first form of the conversion factor, (molar mass) g/mol.

Next, calculate the molar mass of the substance. Be sure to determine the formula correctly, or the molar mass will be incorrect and so will the final answer.

How to Do It

0.0154 mol H3PO4 × __________ g H3PO4

mol H3PO4 = ___________ g H3PO4

For phosphoric acid:

(3 x 1.0 gmol

H) + (1 x 31.0 g1 mol

P) + (4 × 16.0 gmol

O) =

98.0 gmol

H3PO4

Insert the molar mass and complete the calculation, ensuring that your answer has the appropriate number of significant figures.

0.0154 mol H3PO4 x 98.0 g H3PO41 mol H3PO4

= 1.51 g H3PO4

Practice Problems—Converting Moles to Grams

1. Chromium ions are not only responsible for the beautiful colors of rubies and emeralds, but also for the colour change that allows a breathalyzer to work. How many grams of chromium are contained in 3.5 mol of chromium?

2. Sulphuric acid is the second most commonly produced acid in the world. How many grams of sulphuric acid are produced by a chemical plant, making 55 kmol of this acid for sale? (Hint: Be sure you have the correct formula for sulphuric acid.)

3. Potassium nitrate is useful not only as a fertilizer, but also in the manufacture of explosives. How many kilograms of potassium nitrate are contained in a 250 mol sample?


Sample ProblemHow many moles of copper are contained in a 1984 Canadian penny? In 1984, a penny was pure copper and it weighed 3.08 grams.

What to Think About

Your answer must be in units of moles. As the mass of the penny is given in grams, the second form of the conversion factor, 1 mol/(molar mass)g allows you to cancel the grams and convert to moles.

Look up the molar mass to one decimal place. Apply the appropriate rules for determining the answer with the correct number of significant figures.

Notice that according to SI convention, whenever an answer contains a decimal and is smaller than 1, the decimal should be preceded by a 0.

How to Do It

3.08 g Cu x 1 mol Cu(molar mass) g

= _____________ mol Cu

3.08 g Cu x 1 mol Cu63.5 g Cu

= 0.0485 mol Cu

Practice Problems—Converting Grams to Moles

1. A 27.0 g sample of beryllium contains how many moles of beryllium?

2. The most common form of oxygen is the diatomic gas, O2, found in our atmosphere. Ozone is produced in the upper atmosphere and has the formula O3. How many moles are contained in a 96.0 g sample of ozone?

3. Smelling salts are used to revive an unconscious boxer. A capsule of smelling salts contains 500.0 mg of (NH4)2CO3. How many moles is this?


Multi-Step Calculations with Molar Quantities

You can identify multi-step mole conversions by the noticeable lack of the unit “mole” in the question. Consider the mole to be the “base unit” for amount in mole conversion problems. In any multiple step problem, usually your first step is to convert to moles.

Sample ProblemHydrogen monofluoride, HF(g), is a particularly interesting gas because it can be used to etch glass. The white lines on the beakers and graduated cylinders in your lab may well have been made with this acidic gas. Determine the mass in kilograms of 3.9 × 1027 molecules of HF.

What to Think About

Since you have been given molecules to find kg, first use Avogadro’s number in the denominator to cancel molecules and convert to moles.

Next, convert to mass and calculate the molar mass for HF.

Convert grams to kilograms. Recall that 1 kg is exactly 103 or 1000 g.

How to Do It

3.9 x 1027 molecules x 1 mol HF

6 .02 x 1023 molecules x

x = _________ kg

Molar mass HF = (1 × 1.0 g/mol H) + (1 × 19.0 g/mol F) = 20.0 g/mol HF



20.0 g HF1 mol HF

x

_______ = _________ kg



20.0 g HF1 mol HF

x

1 kg HF1o3 g HF

= 1.3 × 102 kg

Practice Problems—Calculating Molar Quantities

1. What is the mass of a sample containing 4.154 × 1032 atoms of mercury?

4.154 x 1032 atoms Hg x 1 mol Hg6.02 × 1023 atoms Hg x g Hg

mol Hg = _______ g Hg

2. Up to 1.44 × 105 kg of various oxides of nitrogen are emitted by a gas-burning electrical plant in one year. Assuming this entire mass to be nitrogen dioxide, how many oxygen atoms would be present in this gas sample? (Hint: Remember to finish the problem by applying the two atoms O/one molecule NO2 factor.)

3. Up to 5.0 × 1030 molecules of ammonia gas are also released into the airshed above an electrical plant like the one described in question 2. What is the mass (in tonnes) of this gas? (Given: 1 tonne = 1000 kg)


How Long Is Your Name? Activity

Question

How long is a line of calcium carbonate formula units laid end to end?


Calcium carbonate is commonly referred to as chalk. The line in question is composed of the calcium carbonate required to write your name. Assume the length of one CaCO3 formula unit is 450 pm. Note that 1 pm (picometre) = 10-12 m.

Events and Objects1. Write a brief procedure outlining how you might collect the data needed.2. If your teacher instructs you, perform the activity yourself. Otherwise, use the following data to perform your

calculation.

Initial mass of chalk 15.075 g

Mass of chalk once name was written 14.992 g

Mass of chalk required to write your name

Records and Transformations3. How long is the line of calcium carbonate formula units that make up your name?

4. The distance from Earth to the Sun is 1.5 × 108 km. How does the length of your name compare?

How Big Is Your Mouth? Activity

QuestionHow many molecules of water can your lab partner hold in his or her mouth?


Events and Objects1. Write a brief procedure outlining how you might collect the data required.2. If your teacher instructs you, perform the activity with a partner. Otherwise, use the following sample

data to perform the calculation.

Initial mass of paper cup filled with water 184.194 g

Mass of paper cup once a mouthful of water has been removed 125.048 g

Mass of waster held in the mouth all at once without swallowing

Records and Transformation

3. How many molecules of water can your lab partner hold in his or her mouth?


Review Questions

1. 1 mol Al = __________ g Al

2. 2.99 g Na = __________ mol Na

3. What is the molecular mass of each of the following? (a) arsine gas (AsH3)

(b) table salt (NaCl)

(c) washing soda (sodium carbonate monohydrate)

4. What is the molar mass of each of the following? (a) methane gas (CH4)

(b) photographic hypo (Na2S2O3 • 5H2O)

(c) asbestos (magnesium silicate)

5. Cycling enthusiasts often prefer bicycles made with titanium frames. Titanium is resistant to corrosion and fatigue, has a significantly lower density than steel, and seems to have a natural shock-absorbing ability. Suppose a high-quality titanium bicycle frame contains 1300 g of titanium.

(a) How many moles of titanium does this frame contain?

(b) How many atoms of titanium does this number of moles represent?

6. The black centre of a pencil, often called pencil “lead,” is actually a form of carbon called graphite, which consists of layers of carbon atoms. When you write with a pencil, you rub these layers of carbon atoms onto the writing surface. How many moles of carbon are deposited onto the sheet of paper if the mass of the paper increases by 0.030 mg?

7. Ethane, C2H6, is a component of natural gas. How many moles of ethane are contained in 15.0 kg of ethane?

8. Calcium oxalate is a poisonous compound found in rhubarb leaves. A forensic sample contains 52.4 mg of the compound. How many moles of carbon are present in the sample?

9. How many molecules are there in 1.000 mg of the organic solvent, carbon tetrachloride?


10. How many conversion factors would be required to convert kilograms of hydrogen monofluoride gas to molecules of HF? Show the unit analysis necessary to perform the problem.

11. What is the mass of 100 atoms of platinum? Answer to three significant figures.

12. (a) A readily available cleaner is incorrectly called TSP or trisodium phosphate. What is the mass of 4.27 × 1024 formula units of Na3PO4?

(b) Why is TSP an incorrect name for this compound?

13. How many atoms are in a 14.56 g sample of sodium hydrogen sulphate, the active ingredient in toilet cleaner?

14. What mass of sodium ions is contained in 6.80 × 1024 formula units of sodium chloride?

15. How many carbon atoms are present in a 500.0 mg tablet of acetaminophen, C8H9NO2?

16. How many moles of carbon are in a 7.50 g spoonful of sucrose, C12H22O11?

17. Potassium permanganate is a fungicide that is effective at curing athlete’s foot. What mass of KMnO4 contains 4005 atoms of oxygen?

18. Sulphuric acid is the electrolyte in car batteries. How many hydrogen ions are present in 98.1 g of sulphuric acid?

19. Given 2.00 million ammonium ions, what is the mass in kilograms?


5.3 Avogadro’s Hypothesis and Standard Molar Volume – STP

Warm Up

1. How many molecules of carbon dioxide are contained in the Figure 5.3a balloon? _________2. How many moles of carbon dioxide is this? _____________________3. How many atoms of oxygen are in the balloon? _________________4. If the density of carbon dioxide gas is 1.96 g/L, what is the balloon’s volume? _______________________________________________________________

Comparing Gas Volumes

In section 5.1, you were introduced to the work of French chemist, Joseph-Louis Gay-Lussac and Italian chemist Amadeo Avogadro. Although Avogadro belonged to a large family of lawyers, he was a skilled scientist. He earned his bachelor’s degree by the age of 16 and had a PhD at only 20 years old. Avogadro and Gay-Lussac spent many hours studying gases. Since elastic containers like the balloon were not yet invented, they frequently used animal bladders to contain the gases that they studied. Avogadro’s work eventually led him to publish his hypothesis in 1811: “Equal volumes of different gases, measured at the same temperature and pressure, contain equal numbers of particles.”

Figure 5.3a This inflated balloon has a volume of 22.4 L.

Quick Check Helium gas has such a low density that it is frequently used in party balloons because it causes them to float. Another noble gas, xenon, has such a high density that a balloon filled with it would drop to the floor.

1. What is the molar mass of: He __________ Xe __________

2. Sketch two balloons, one filled with He, the other filled with Xe. Each balloon should contain the same number of gas particles. How do the volumes compare?

3. What can you conclude about the arrangement of particles in the gas state? (Hint: How big are the particles compared to the overall volume of the balloon? What do you notice about the space between the particles?)

4. What special properties do gases have that liquids and solids do not have? _____________________________________________________________

5. What happens to the volume of a gas if the temperature is increased? _____________________________________________________________

6. How would the volume of a gas sample be affected if the pressure on the sample increases? ________________________________________


Volumes of Chemical Species

Consider the balloons you have drawn above. Avogadro’s hypothesis tells us that the two balloons should have exactly the same volume with the He atoms being farther apart than the Xe atoms, because the Xe atoms are so much larger. Gas particles, no matter what their identity, occupy less than 0.01 percent of the entire volume, so both balloons would be mainly empty space. Thanks to Avogadro’s discovery, there is a universal volume that can be applied to a mole of any gas. This volume depends on the conditions of temperature and pressure the gas is measured in.

An increase in temperature results in an increase in volume. An increase in pressure decreases the volume. Because gas volume changes with variations in pressure and temperature, it has been necessary to establish a set of standard conditions for measuring the volume of a mole of gas. These conditions have varied over the years depending upon the organization and the country involved in making the measurements. Chemists in British Columbia and many parts of the world use the current NIST (National Institute of Standards and Technology) values. Standard Conditions of Temperature and Pressure (STP) imply a temperature of 0ºC and a pressure of 1 atmosphere or 101.3 kPa (kilopascals). A pressure of 1 atmosphere is the pressure exerted on objects by the atmosphere at sea level. Measurement of 1 mol of gas under these conditions will always result in a volume of 22.4 L (Figure 5.3b).

at STP

Figure 5.3b 1 mol of gas

Molar volume is the volume occupied by 1 mol of a gas at a given temperature and pressure. Under STP conditions, 1mol of any gas occupies 22.4 L of volume (Figure 5.3c).

Molar volume conversion factors may be written in reciprocal form, depending on the calculation being done (Figure 5.3d).

Figure 5.3c Molar volume

22.4 L or 1 mol at STP

1 mol 22.4 L

Figure 5.3d Molar volume conversion factors


Practice Problems—Converting Moles to Volume at STP

1. Diphosphine, P2H6, has an unpleasant garlic-like odour. What is the volume of 55 kmol of diphosphine at STP?2. Air is a mixture of gases, primarily nitrogen and oxygen. How many mL of air are equivalent to 2.5 × 10-3 moles

at STP?3. H2S gas is released from rotten eggs. What volume of H2S gas is equivalent to 17.05 g of the chemical at STP?

17.05 g H2S x mol x = L

Sample ProblemHow many moles of neon gas are present in a 75.0 mL neon sign at STP?

What to Think AboutWhat conversion factors are required to do the problem?

Recall that 1 mol = 22.4 L at STP.

Recall that 1 L = 1000 mL.

Considering the answer is in units of moles, what number and unit should start the problem?

Insert the conversion factors with the appropriate units and convert. Check that the answer has the correct number of significant figures and that the answer makes sense.

75.0 mL is a very small volume. Considering that a mole is 22.4 L, this very small answer does seem reasonable.

How to Do It

75.0 mL Ne x ___________________________ = ________ mol

75.0 mL Ne x 1 L

1000 ml x 1 mol22.4 L = 3.35 x 10-3 mol

Sample ProblemNitrogen dioxide is an air pollutant. What is the volume of 0.500 mol of NO2 at STP?

What to Think About

As we are trying to find a vol-ume, the answer will be in litres. This means we need to use the

conversion factor 22.4 L1 mol .

Note that the answer to the problem makes sense. A mole of any gas has a volume of 22.4 L at STP, so half a mole will have a volume of half of 22.4 L.

How to Do It

0.500 mol of NO2 × 22.4 L1 mol = 11.2 L


Density Relates the Mass of Matter to Its Volume

The two quantities mass and volume are often used to measure samples of matter in the laboratory. Both of these quantities are extensive. That is, they depend on the amount of matter present. Neither quantity can be used alone to identify a sample of matter. There is, however, a quantity that relates the mass of a particular sample of matter to its volume. This is an intensive property and can be used to identify a sample. The property is density. Density is the mass of a particular volume of matter, commonly expressed in units of g/mL or g/L. The density of a sample depends upon its temperature.

Practice Problems—Converting Volumes to Moles at STP

1. A cartesian diver is an enclosed plastic container of air that rises and falls in water depending on the pressure exerted on it. How many moles of air are in a 2.5 L cartesian diver at STP?

2. Experienced scuba divers like to breathe nitrox, which is air that has a higher than usual oxygen-to-nitrogen ratio. How many moles of nitrox are in a 35 L tank at STP?

3. Ammonia gas is evolved when a capsule of smelling salts is broken. The harsh odour serves to revive someone who is losing consciousness. How many molecules of ammonia are contained in 450 mL of ammonia gas at STP?

450 mL NH3(g) x _______ L

mL x _______ mol

mL x _______ molecules

mol =

density (D) = mass (m)

volume (V)

The density of a solid is commonly used to help identify it. Density may also be used in a wide variety of calculations for solids. However, when it comes to the mole, density is particularly important for liquids and gases. This is because we rarely weigh liquids and gases to determine their quantity. Rather, we measure their volume. Consequently, if we wish to determine how many moles of a liquid or gas we have, given its volume, it is very helpful to know its density.

Density is really just an equivalency between the mass and volume of a particular substance. It may be used in any of the following forms:

g1 mL or 1 mL

g and g

1 L or

1 Lg


Sample ProblemThe density of an unknown noble gas is 1.79 g/L at STP. What is the molar mass of the gas? What noble gas is this?

What to Think About

What are the units of molar mass? g

molWhat do you know about any gas that might be used to start the problem?

Use density to convert to appropriate units for molar mass. Check your periodic table and identify the unknown noble gas.

How to Do It

22.4 L1 mol × 1.79 g

1 L = 40.1 g

mol

The noble gas is Argon.

Practice Problems—Using Density to Find Volume

1. Most disposable lighters contain a mixture of natural gases under pressure. A typical gas mixture has a

density of 2.33 gL

at STP. What is the molar mass of this gas mixture?

2. What is the density of krypton gas at STP?

1 mol22.4 L

X _____ = _____ gL

Sample ProblemSilicon dioxide is better known as the common mineral, quartz. It has a density of 2.64 g/mL. What is the volume of a 0.769 mol piece of quartz?

What to Think About

In this problem, the answer will be in units of mL, hence you must begin the problem with the number of moles, rather than the density. This allows you to convert a value with a single unit into a single unit answer.

The conversion factors required are the molar mass and the density. In this problem, the density will have to be inverted as the answer is in mL.

How to Do It

0.769 mol x ________ gmol

x _________ mLg

= __________ mL

0.769 mol x 60.1 g1 mol

x 1 mL2.64 g = 17.5 mL


Practice Problems—Determining Volume

1. Carbon disulphide liquid is spewed out during volcanic eruptions. How many liters of CS2 are formed during an eruption that produces 85 mol of CS2? The density of CS2 is 1.26 g/mL.

2. Silicon tetrafluoride is another volcanic eruption product and is produced in huge plumes released by active volcanoes. The gas has a density of 4.69 g/L. What is the volume of 3.846-tonne sample of SiF4 gas? (1 tonne = 1000 kg)

3. H2S gas is released from rotten eggs. It is also produced by stink bombs sold in fireworks stands around Halloween. How many millilitres of H2S gas are contained in a 0.328 g sample of the chemical with a density of 1.365 g/L?

Mystery Gas Activity

Question

What is the identity of a mystery gas?


A student was assigned the task of determining the molar mass of an unknown elemental gas (Figure 5.3e). The student measured the mass of a sealed 525 mL round bottom flask that contained dry air. The student then flushed the flask with the unknown gas, resealed it and measured the mass again. Both the air and the unknown gas were measured in a chilly laboratory at sea level such that the temperature and pressure were STP conditions.

Events and Objects

The data for the experiment is shown in the table below.

Volume of sealed flask 525 mL

Mass of sealed flask and dry air 124.700 g

Mass of sealed flask and unknown gas 124.768 g

Figure 5.3e Mystery gas at STP

Records and Transformations1. Calculate the mass of the dry air that was in the sealed flask. The density of

dry air is 1.12 g/L at STP conditions.

2. Calculate the mass of the sealed flask with no air in it.

3. Calculate the mass of the unknown gas that was added to the sealed flask.

4. Using the information above, calculate the value of the molar mass of the unknown gas. (Hint: You may wish to determine the density of the unknown gas first.)

5. Determine the identity of the unknown gas. (Hint: Remember, not all of the gases on the periodic table are monatomic.)


Review Questions

1. How many litres of carbon dioxide are in a15 mol- sample at STP?

2. How many moles of air are contained in a set of human lungs with a volume of 4.8 L measured at STP?

3. Acetylene gas, C2H2 is used as a fuel in welding torches. What is the mass of the C2H2 contained in a canister that delivers 1400 L of acetylene at STP?

4. Dentists sometimes use N2O gas as a partial anesthetic while extracting wisdom teeth. What is the volume of a 25.0 g sample of N2O gas at STP?

5. P2O5 gas is produced each time you strike a safety match. What is the density of this gas at STP?

6. Nitrogen gas is released in an explosive reaction when an airbag expands. How many nitrogen atoms are contained in a 28.75 L sample of nitrogen at STP?

7. (a) A 2.40 L glass bulb contains 8.98 g of a noble gas at STP. What is the molar mass of the gas?

(b) Identify the gas.

8. Bromine liquid, Br2 has a density of 3.53 g/mL. How many bromine atoms are contained in a 15.0 mL sample of Br2?

9. A power plant emits 2.41 × 107 L of sulphur dioxide gas each year. What is the mass (in kilograms) of this gas sample?

10. What is the density of liquid octane if 250 mmol of C8H18 has a volume of 40.5 mL?


11. What is the mass of the hydrogen atoms contained within 8390 L of ammonia gas at STP?

12. Mercury (Hg) is a liquid metal at room temperature. What is the volume of 1 mol of mercury? Its density is 13.6 g/mL.

13. Benzene, C6H6, is a common organic solvent with a density of 0.878 g/mL. How many atoms are in a 0.250 L-sample of benzene?

14. How many molecules of arsine gas, AsH3, are contained in a 7.65 × 10-6 L-sample at STP?

15. Dry air is approximately 80 percent nitrogen and 20 percent oxygen by amount. Calculate the average density of dry air.

16. Uranium hexafluoride is an extremely dense gas used in nuclear chemistry. What is its density at STP?

17. What is the density of gold if 1.47 mol are contained in a 15.0 cm3 piece?

18. How many moles of hydrogen atoms are contained in 44.8 L of methane gas, CH4?


Section 5.4 Formulas of Compounds – Analyzing Composition

Warm UpConsider a cookie with a mass of 5.00 g that contains 2.00 g of chocolate chips (Figure 5.4a). You can express the amount of chocolate chips in the cookie as a percentage of the cookie’s mass.

% chocolate chips = mass chocolate chipsmass cookie

= 2.00 g5.00 g × 100%

= 40%

1. Complete the chart below.

Mass cookie Mass chocolate chips % chocolate chips by mass

Double chocolate fudge cookie 3.80 g 0.80 g

Decadent chocolate cookie 5.50 g 1.00 g

2. Cookie with the most chocolate: __________________________________3. Cookie with the largest percentage of chocolate: _____________________

Percentage Composition

Forensic investigators collect samples from a crime scene to analyze. How do they identify the unknown samples? They use the principles of chemistry in a variety of ways. One method of analyzing an unknown sample is to determine its percentage composition.

The percentage composition of a compound is the percentage by mass of each type of atom in the compound.

Figure 5.4a Just like a cookie, chemical compounds also have a composition. Chemical compounds are made up of atoms of elements.

Sample ProblemWhat is the percentage composition of sugar with a formula of C12H22O11 (Figure 5.4B)?

What to Think About

First, calculate the molar mass of sugar.

Then, find the percentage of each element by dividing the total mass of each element by the molar mass of the compound.

How to Do It

molar mass = 12(12.0 g) + 22(1.00 g) + 11(16.0 g) = 342.0 g

% C = 12 (12.0 g)

342.0 g x 100% = 42.1%

% H = 22 (1.00 g)

342.0 g x 100% = 6.4%

% O = 11 (16.0 g)

342.0 g x 100% = 51.5%

Figure 5.4b A sugar molecule with 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms


Practice Problems—Calculating the Percentage Composition

1. Adipic acid is a white crystalline powder used to produce nylon. Its formula is H2C6H8O4. Calculate its percentage composition.

2. Ammonium sulphate is a common fertilizer used to reduce the pH of soil. Its formula is (NH4)2SO4. Calculate its

percentage composition.

3. Ibuprofen is a common pain reliever and anti-inflammatory used to treat headaches and relieve symptoms of arthritis. Its formula is C13H18O2. What is the percentage composition?

Sample ProblemEpsom salts, MgSO4 • 7H2O, are used to soothe sore feet in a warm bath. Calculate the percentage of water in Epsom salts.

How to Do Itmolar mass MgSO4 • 7H2O = 24.3 g + 32.0 g + 4(16.0 g) + 7(18.0 g) = 246.3 g

% H2O = 7(18.0 g)246.3 g × 100% = 51.2%

Practice Problems—Calculating the Percentage Composition of Molecules

1. Calculate the percentage of NH3 in Ag(NH3)2Cl.

2. Calculate the percentage of water in Ba(OH)2 • 8H2O.

3. Calculate the percentage of water in CuSO4 • 5H2O.

Another application of percentage composition is to describe the amount of water or other molecules in complex compounds.


Empirical and Molecular Formulas

The composition of a pure substance can be shown in two ways. • The molecular formula states the actual number of atoms of each element in a

compound.• The empirical formula states the smallest whole-number ratio of atoms of each

element.

Organic chemistry is the study of compounds and reactions involving carbon. There are millions of organic compounds, and they have very different properties and reactivities. Many have the same empirical formula, but each has a different molecular formula and therefore different properties. For example, a group of compounds called alkenes have an empirical formula of CH2. Some examples of alkenes are C2H4 (ethene), C3H6 (propene), C4H8 (butene) and C8H16 (octene).

The compound octane is a hydrocarbon and a component of gasoline (Figure 5.4c). The molecular formula for octane is C8H18. As shown in Figure 5.4c, one molecule of octane is made of 8 carbon atoms

and 18 hydrogen atoms The empirical formula for octane is C4H9 because the lowest whole-number ratio of atoms of C to atoms of H is 4 to 9.

Another example of an organic compound is glucose, which has a molecular formula of C6H12O6. The subscripts 6, 12, and 6 have the greatest common factor of 6, so we divide each by 6 to find the lowest integer subscripts. That gives us an empirical formula of C1H2O1. We do not show the number 1 as a subscript in a formula, so the empirical formula becomes CH2O.

Figure 5.4c Octane

Quick Check1. Complete the following table. Molecular Formula Empirical Formula

H2O2

H2O

C4H8Cl2

C6H12O3

Figure 5.4d Each molecule of H2O has 2 H atoms and 1 O atom.

Determining an Empirical Formula

The molecular formula for water is H2O (Figure 5.4d). Each water molecule is made of two hydrogen atoms and one oxygen atom. Therefore, in a dozen water molecules, there are two-dozen hydrogen atoms and one-dozen oxygen atoms. In a mole of water molecules, there are 2 mol of hydrogen atoms and 1 mol of oxygen atoms.

The subscripts in a formula are the mole ratios of atoms of each element in a chemical compound. You can use the mole ratios to calculate the empirical formula of an unknown compound.


Practice Problems—Determining Empirical Formulas

1. A compound is 18.7% Li, 16.3% C and 65.0% O. Determine its empirical formula.

2. A compound is 9.93% C, 58.6% Cl and 31.4% F. Determine its empirical formula.

3. A compound contains 69.2 g Ag, 10.3 g S and 20.5 g O. Determine its empirical formula.

Sample ProblemDetermine the empirical formula of a compound that is 65.5% carbon, 5.5 hydrogen, and 29.0% oxygen.

What to Think About

First, determine the mass of each element. Assume that there are 100 g of the compound and convert the percentages to mass: 65.5 g carbon, 5.50 g hydrogen, and 29.0 g oxygen.

Next, convert mass to moles for each element. Be sure to carry at least 4 significant figures.

You could write the formula as C5.458H5.500O1.813, but we typically use whole numbers as subscripts in a formula. To get whole numbers, simply divide all of the mole numbers by the smallest mole number calculated; in this case, 1.813.

How to Do It

mol C = 65.5 g × 1 mol12.0 g

= 5.458 mol C

mol H = 5.50 g × 1 mol1.00 g

= 5.500 mol H

mol O = 29.0 g × 1 mol16.0 g

= 1.813 mol O

C5.458 H5.500 O1.813

1.813

becomes C3.01H3.03O1.

The empirical formula is C3H3O.

Sometimes, when you divide by the smallest mole number, the mole ratios do not result in whole numbers.• If the ratio ends in 0.50 (half ), multiply everything through by 2.• If the ratio ends in 0.33 or 0.67 (thirds), multiply everything through by 3.• If the ratio ends in 0.25 (quarter), multiply everything through by 4.


Finding the Molecular Formula of a Compound

Recall that the molecular formula states the actual number of atoms of each element in a compound, whereas the empirical formula states the smallest whole-number ratio of atoms of each element.

Practice Problems—More Practice Determining Empirical Formulas

1. A compound is found to contain 50.80% Zn, 16.05% P, and 33.14% O. Determine its empirical formula.

2. A sample of Aspirin® was analyzed and found to contain 60.0 g carbon, 4.40 g hydrogen, and 35.6 g of oxygen. Determine the empirical formula for Aspirin®.

3. A sample of a compound was found to contain 69.9 g Fe and 30.1 g O. Determine its empirical formula.

The molecular formula is a multiple of the empirical formula. The molar mass of the molecular formula is a multiple of the molar mass of the empirical formula.

Sample ProblemA compound is 72.3 g iron and 27.6 g oxygen. Determine its empirical formula.

What to Think About

We multiply everything by 3 to get rid of the thirds: 3 (Fe1O1.330 )

How to Do Itmol Fe = 72.3 g × 1 mol

55.8 g = 1.296 mol Fe

mol O = 27.6 g × 1 mol16.0 g = 1.725 mol O

Fe1.296/1.296O1.725/1.296 becomes Fe1O1.330.

The empirical formula is Fe3O4.

Sample ProblemA substance has an empirical formula of CH2 and a molar mass of 42.0 g

mol . Determine its molecular formula.

What to Think About

First, calculate the molar mass of the empirical formula.

Then compare the molecular molar mass to the empirical molar mass.

How to Do It

12.0 g + 2(1.00) g = 14.0 gmol

42.0 g14.0 g

= 3.00

The molecular formula is 3(CH2), which is C3H6.


Practice Problems—Molecular Formulas

1. Vinegar has an empirical formula of CH2O and a molar mass of 60.0 g

mol. Determine its molecular formula.

2. A compound has an empirical formula of C3H4. Circle the molar masses that could be consistent with possible molecular formulas.

120.0 gmol

55.0 gmol

20.0 gmol

80.0 gmol

3. A sample of antifreeze was analyzed and contained 38.7 g carbon, 9.70 g hydrogen, and 51.6 g oxygen.

It has a molar mass of 62.1 gmol

. Determine its molecular formula.

Molecular and Empirical Formula Activity

Question

How does the molar mass of the molecular formula compare to that of the empirical formula?


Consider a substance with an empirical formula of CH. To calculate its molar mass we add the molar mass of one C to the molar mass of one H. If the same substance has a molecular formula of C6H6, we have increased both the number of each type of atom by six, and the molar mass of the substance by 6:

CH = 13.0 g

mol × 6

6(CH) = 6(13.0 g

mol )

C6H6 = 78.0 g

mol

Events and Objects1. Using a molecular model kit, build the molecule C2H4 and complete its line in the following table.2. Build C4H8 and complete its line in the table.

Molecular Formula

Molar Mass Diagram of Molecule

Empirical Formula

Molar Mass Diagram of Empirical Structure

C2H4

C4H8

Records and Transformations3. By what factor can you multiply the number of atoms in C2H4 to get C4H8? __________________________

4. By what factor can you multiply the molar mass of C2H4 to get the molar mass of C4H8? _______________


Review Questions

1. Menthol is a strong-smelling compound obtained from mint oil and used in cough drops. It has a formula of C10H20O. Calculate its percentage composition.

2. Sodium acetate trihydrate (NaCH3COO • 3H2O) is an inexpensive salt, commonly used in pickling foods. Calculate the percentage of water in this substance.

3. Complete the following chart.

CompoundMolecular Formula

Empirical Formula

ribose

(a naturally occurring

sugar in the body found

in vitamin B-12)

C5H10O5

lactic acid

(formed from glucose

and used by working

muscles for energy)

C3H6O3

acetic acid

(vinegar)C2H4O2

4. Trinitrotoluene (TNT) is an explosive and has the formula C7H5O6N3. Calculate the percentage of nitrogen in this compound.

5. (a) Explain why knowing only the empirical formula of a compound is not enough to identify an unknown sample.

(b) What other important piece of information is needed to identify an unknown sample?

6. A compound used in paints, enamels, and ceramics was analyzed and contained 0.5068 mol barium, 0.5075 mol carbon, and 1.520 mol oxygen. Determine its empirical formula.

7. Benzoic acid is a compound used as a food preservative. The compound contains 68.8% carbon, 4.95% hydrogen, and 26.2% oxygen. Determine its empirical formula.


8. A substance has an empirical formula of NH2 and a molar mass of 32.1 g/mol. What is its molecular formula?

9. A sample of ascorbic acid, also known as vitamin C, was analyzed and found to contain 40.9 g of carbon, 4.58 g of hydrogen, and 54.5 g of oxygen. Ascorbic acid has a molar mass of 176.1 g/mol. Determine the molecular formula of ascorbic acid.

10. Caffeine is a stimulant found in coffee and chocolate. A sample was analyzed and found to contain 49.48% carbon, 5.19% nitrogen, and 16.48% oxygen. Determine its empirical formula.

11. (a) Propane (C3H8) is a gas commonly used as a fuel. Calculate the percentage composition of propane.

(b) Using the percentages you obtained in (a), perform calculations to prove that the empirical formula is C3H8.

12. A 100.0 g sample of a component of gasoline was analyzed. It contained 92.29 g carbon. The rest of the sample was hydrogen. If the molar mass of the compound was 78.0 g/mol, determine its molecular formula.

36 Mole Concept—Draft Version—Copyright Edvantage Interactive 2010r

5.5 A Concentration Measure — Molarity

Warm UpYou can make nectar for a hummingbird feeder by dissolving 820 g of sucrose (C12H22O11) in enough water to make 2.0 L of solution (Figure 5.5a).

1. Convert 820 g of sucrose to moles. _____________________________________________________________________

2. Determine how many moles of sugar are present in each litre of the solution. _____________________________________________________________________

3. The amount of sucrose dissolved a typical soft drink is about 0.25 mol per litre. How does the sugar content of hummingbird nectar compare with soft drinks? _____________________________________________________________________

What is a Solution?

Nickel(II) chloride (NiCl2) is a beautiful lime-green powder. When the powder is dissolved in water, the result is a green solution. Nickel(II) chloride solutions are used in the electroplating industry to deposit a layer of nickel metal on the surface of iron objects. This makes the iron more resistant to corrosion.• A solution is a mixture that is homogeneous – that is, it is uniform throughout, and any

one part of the solution always looks just the same as any other part.• The main component of a solution, such as water in the above example, is called the

solvent.• The substance that dissolves into solvent, such as nickel(II) chloride in the above

example, is called the solute.

Solutions in which water is the solvent are extremely common and are given a special name: aqueous (“aqua” means “water”). When nickel(II) chloride is dissolved in water, it is called an “aqueous solution of nicke(II) chloride.” This solution can be symbolized as NiCl2(aq). The (aq) indicates that the solvent is water. (Figure 5.5b)

Concentration

Suppose a solution of NiCl2(aq) contains 4.0 mol of NiCl2 in every litre of the solution. This quantity is a measure of the solution’s concentration. A solution’s concentration is the number of moles of solute in each litre. For example, a solution containing 4.0 mol per litre is twice as concentrated as

another solution containing only 2.0 mol per litre.

Figure 5.5c shows one way to represent the composition of a solution. Diagrams like this help to show the relationship between the total number of moles of solute present in a solution, the volume of the solution, and the concentration of the solution. A solution of know concentration is called a standard solution.

Figure 5.5a Hummingbirds at a feeder

Figure 5.5b Water is the solvent in an aqueous solution

Figure 5.5c This solution has a concentration of 4 moles per litre


Quick Check1. Circle the solution in each pair that is more concentrated.

(a) 5 mol of NaCl in 10 L or 9 mol of NaCl in 16 L

(b) 0.6 mol of Cu(NO3)2 in 4 L or 1.4 mol of Cu(NO3)2 in 12 L

Molarity

The molarity of a solution is the number of moles of solute dissolved in each litre of a solution. The greater the molarity, the more concentrated a solution is. For example, a solution in which the solute concentration is 6.0 mol/L is three times as concentrated as a 2.0 mol/L solution.• The concentration unit for molarity can be written in two ways:

molL

(read “moles per litre”)

M (read “molar”)

A solution in which each litre of solution contains 3.0 mol can be described either as 3.0 mol per litre or as 3.0 molar.• Square brackets are used with formulas to indicate concentration. For example, [NaCl] is

read “the concentration of NaCl” or “the molarity of NaCl.”[CaSO4] = 1.3 M” can be read “The molarity of the calcium sulphate solution is

1.3 molar.”

Quick Check1. What does the term “molarity” mean? __________________________________________________________2. What are two ways to write units for moles per litre? _________________________________________________________________________________________3. How are the following symbols read:

(a) [MgF2] = 2.5 M __________________________________________________________________________

(b) [Ca(NO3)2] = 4.31 molL

___________________________________________________________________

4. Write the following statement using chemical symbols and units: The concentration of a solution of iron(III) chloride is 3.2 molar. __________________________________________________________________

The following equation relates the volume of a solution V (in litres), the number of moles of solute dissolved in the solution n (in moles), and the concentration of a solution C (in moles per litre):

C = nVconcentration

( molL

or M)

number of moles (mol)

Volume (L)

n = CV

V = nc

C = nV


Sample ProblemWhat is the concentration of a solution in which 150.0 g of NaOH are dissolved in enough water to form 2.0 L of solution?

What to Think About

Find the total number of moles of NaOH in the solution. Then calculate the number of moles of it in each litre.

How to Do It

1. nNaOH = 150.0 g x 1 mol40.0 g

= 4.00 mol

2. C = nV

= 4.000 mol2.0 L

= 2.0 molL

n = 4 mol x x x xV = 2 L

2 mol for every litre

x xx x

Practice Problems—Finding the Number of Moles of Solute.1. How many moles of KCl are present in 35.0 mL of a 4.21 M solution?

What is the volume of the solution in litres?

What is the solution concentration?

How are concentration and volume used to find moles of solute in a solution?

2. What is the molarity of a solution prepared by dissolving 117 g of table salt (NaCl) in a total of 4.00 L of solution?

3. What is the volume of a sample of 0.100 M CuCl2 solution if the sample contains a total of 3.00 mol of CuCl2?


Dilution

Adding water to a solution decreases the concentration of the solute in the solution, because adding water increases the total volume. This gives the solute more room to spread out. Decreasing solution concentration by adding water is called dilution.

Dilution factor method of calculating dilution

Cf = Ci x Vi

Vf i = initial f = final

Sample ProblemA 3.00 L solution of 4.51 M KBr is diluted by adding 2.00 L of water to it. What is the concentration of the solution after mixing?

What to think about

Find the total volume of the solution after mixing, and then write the dilution calculation involving the dilution factor.

How to Do It

1. Vf = 3.00 L + 2.00 L = 5.00 L

2. C = 4.51 M x 300 L5.00 L = 2.71 M

Figure 5.5d When you add water, you decrease the solution concentration.

Practice Problems—Dilution

1. A 5.00 L solution of 4.00 M table sugar (sucrose, C12H22O11) is diluted by adding sufficient water until the total volume becomes 15.0 L. What is the concentration of the solution after mixing?

2. A 0.300 M solution of KNO3 with a volume of 75.0 mL is diluted by adding 125.0 mL of water. What is the molarity of the solution after mixing?

3. What is the concentration after mixing if 300.0 mL of 6.00 M H2SO4 (sulphuric acid) is combined with enough water to produce 4.00 L of solution?

Suppose we add water to a solution. What is the new [ ] after mixing?

20 mL of 16 M H2SO4

C = 16M x 20 ml80 ml

= 4M

The dilution factor is Vold

Vnew


The Mole Concept

All calculations between number of particles n, mass m, and volume of a gas V use the concept of the mole to relate these quantities to each other. For solutions, the concentration of the solute and the volume of the solution are also related to the mole through the equation n = CV.

Quick Check1. What is the volume in litres of 10.00 mL? ________________________

2. Why should volume be represented in litres when using the equation n = CV?

_________________________________________________________________________________________

_________________________________________________________________________________________

_________________________________________________________________________________________

3. What remains unchanged about a solution before and after diluting it with water?

_________________________________________________________________________________________

4. If the volume of a solution is doubled by diluting it with water, what happens to the concentration?

_________________________________________________________________________________________

Figure 5.5e The mole is at the centre of chemical measurement.


Molarity Activity

Question

How can you prepare 1.00 L of a 0.200 M solution of copper(II) chloride and then make a second solution that is 1/10th the concentration of the first one?

Events and Objects

To make a solution of copper(II) chloride with a concentration that is precisely known requires three things: • a high-quality balance to weigh out a sample of powdered

nickel(II) chloride • high-purity water, such as deionized (DI) water, in which to

dissolved the CuCl2 • precision glassware with which to precisely measure

volumes (volumetric flasks and pipettes).


1. Begin by weighing 0.200 mol (26.9 g) of CuCl2 into a 400 mL beaker.

2. Add DI water to the beaker and stir with a glass stirring rod. Be careful not to spill any solution. Wash off the glass stirring rod, with the washings staying in the beaker. If you spill any solution, you should start again.

3, Using a funnel, transfer the contents of the beaker into a 1000 mL volumetric flask, washing the beaker out into the flask so as not to lose any solution. Add DI water to the volumetric flask, carefully filling it to the 1.000 L mark. You have made a 0.200 M solution.

4. To prepare a second solution that is 1/10th the concentration of the 0.200 M solution, pour about 30 mL of 0.200 M solution into a small beaker.

5. Using a 10.00 mL pipette, transfer 10.00 mL of solution into a 100.0 mL volumetric flask. Add DI water to the flask until it fills to the 100.0 mL mark.

Records and Transformations

6. Outline a procedure to make 1000.0 mL of 0.200 M NaCl(aq), and then use some of this solution to make 100.0 mL of 0.050 M NaCl. _______________________________________________________________________________________

_______________________________________________________________________________________

_______________________________________________________________________________________

_______________________________________________________________________________________

_______________________________________________________________________________________

_______________________________________________________________________________________

_______________________________________________________________________________________

_______________________________________________________________________________________

_______________________________________________________________________________________

_______________________________________________________________________________________

Figure 5.5f A high-quality balance and volumetric glassware.


Review Questions

1. What is the concentration of Na2SO4 in a solution made by dissolving 15.0 mol in 3.00 L?

2. What is the concentration of a solution made by dissolving 3.920 mol of FeBr3 in 100.0 mL?

3. What is [NaCl] if 52.0g of it is dissolved in enough water to produce 8.00 L of solution?

4. Find the molarity of calcium nitrate prepared by dissolving 182 g of it in a sufficient amount of water to produce 600.0 mL of solution.

5. How many moles of NH3 are present in 4.00 L of 0.300 mol/L NH3(aq)?

6. What mass of NaCl is dissolved in 92.1 mL of 3.68 M solution?

7. What mass of sucrose (C12H22O11) is needed to produce 500.0 mL of 0.100 M sucrose solution?

8. What is the volume of a 0.700 M solution of potassium iodide that contains a total of 2.10 mol of the solute?


9. What is the volume of a 0.400 M solution that contains 0.480 mol of HCl?

10. What is [NaCl] in a solution prepared by diluting 25.0L of 6.0 M NaCl to a total of 100.0L?

11. Suppose 30.0 mL of 2.50 M LiBr is diluted by adding 130.0 mL of water. What is the new molarity after the dilution?

12. If 50.0 mL of a 0.187 M solution of Al(NO3)3 is diluted to a total of 1000.0 mL, what is the concentration after adding the water?

13. Suppose 9.75 L of 0.200 M Cs2S is diluted by adding 10.25 L of water. What is [Cs2S] after dilution?

14. 1.98 g of ammonium sulphate is dissolved in 250.0 mL of solution. What is the molarity?

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