chapter 2: basic review worksheet - mr. weiss'...

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~ Chapter 2: Basic Review Worksheet

1. What is matter? Ofwhat is matter composed?

2. What is an element, and what is a compound? Give examples of each.

3. Explain the differences among. a gas, a liquid, and a solid.

4. What is meant by the term chemical property? What is meant by the term physical .property?

5. What is meant by the term chemical change? What is meant by the term physical change?

6. What are alloys? Provide an example.

7. What is a mixture? Provide an example.

8. What is a solution? Provide an example.

_ 9. What is meant by the term pure substance?

10. What is the difference between a homogeneous mixture and a heterogeneous mixture?

11. What are some of the techniques by which mixtures can be resolved into their components?

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Chapter 3: Basic Review Worksheet

1. What is an element? How many elements are presently known? How many of these occurnaturally, and how many are human-made? Which elements are most abundant on theearth?

2. What are the three fundamental particles that compose all atoms? Indicate the electricalcharge and relative mass ofeach of these particles. Where is each type ofparticle found inthe atom?

3. What is meant by the term nt.(clear atom?

4. What are isotopes?

5. Are most elements found in nature in the elemental or in the combined form? Why? Nameseveral elements that are usually found in the elemental form.

6. Give the names of some of the families ofelements in the periodic table.

7. Which general area ofthe periodic table contains the metallic elements? Which generalarea contains the nonmetallic elements? _i

8. What are ions? How are ions formed from atoms? To what do the terms cation and anionrefer?

9. What are some general physical properties of ionic compounds such as sodium chloride?How do we know that substances such as sodium chloride consist ofpositively andnegatively charged particles?

10. Write the symbol and atomic number for each ofthe following elements: magnesium, tin,lead, sodium, hydrogen, chlorine, and silver.

11. Write the name and atomic number for each of the following elements:

aHe c. Se e.Pb. B d. Ba f. Sr

12. Write the name and chemical symbol for each of the following elements:

a. 19 c. 1 e. 82b. 12 d.6 f. 2

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13. Indicate the number ofprotons, neutrons, and electrons in isolated atoms having thefollowing nuclear symbols:

a. iRe b. ;; CI c. i~Ca

14. What simple ion does each of the following elements most commonly form?

a.Mg c. Ba e. 0b. F d. Na f. CI ,

15. For each of the following simple ions, indicate the number ofprotons and electrons the ioncontains.

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c. Br-

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3. What general type ofelement is involved in type II compounds?

6. What is an oxyanion? Give two examples.

7. What is an acid? Give two examples (one that contains oxygen and one that does not).

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4. What two systems are used to show the charge ofthe cation in a type II ionic compound?Give examples of each system for two compounds.

1. When writing the name of an ionic compound, which is named first, the anion or thecation? Give an example. .

2. What ending is added to the root name ofan element to show that it is a simple anion in atype I ionic compound? Give an example,

5. Describe the system used to name type III binary compounds (compounds ofnonmetallicelements). Give four examples illustrating the method.

8. Name each ofthe following binary ionic compounds:

a. NaCI c. MgBr2 e. CaSb. K20 d. Alh f. SrO

9. Without consulting the text, name each of the following polyatomic ions:

a. NH4+ c. N03- e. CI04-b. sol- d. OIr f. pol-

1Q. Name each ofthe following compounds:

a. N02 b. ICI c. CO

11. Write the fonnula for each of the following compounds:

a. potassium sulfide c. calcium sulfateb. hydrochloric acid d. copper(II) bromide

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1. Explain the scientific meaning of uncertginty.

2. Explain how a unit is related to a measurement.

3. Explain the terms conversion factor and equivalence statement.

4. For each of the following, make the indicated conversion:

a. 122.4 x 105 to standard scientific ~otationb. 5.993 x 10-4 to ordinary decimal notation

5. For each ofthe following, make the indicated conversion:

a. 6.0 pt to litersb: 6.0 pt to gallonsc. 5.91 yd to metersd. 62.5 mi to kilometerse. 88.5 cm to millimeters

6. Evaluate each of the following mathematical expressions, being sure to express the answer fI'I'i;to the correct number of significant figures:

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a. 10.20 + 4.1 + 26.0001 + 2.4b. (1.091 - 0.991) + 1.2c. (4.06 + 5.1)(2.032 - 1.02)d. (67.21)(1.003)(2.4)

7. Make the indicated temperature conversions:...a. 541 K to Celsius degreesb. 221°C to kelvins

8. Given the following mass, volume, and density information, calculate the missing quantity: .

a. Mass = 121.4 g; volume = 42.4 cm3; density = ? g/cm3

b. Mass = 0.721 lb; volume = 241 cm3; density = ? g/cm3

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2. Defme molar mass. Using K20 as an example, calculate the molar mass from the atomicmasses ofthe elements.

1. What does the average atomic mass ofan element represent? What unit is used for averageatomic mass?

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4. For 5.00~g samples ofeach ofthe following substances, calculate the number ofmoles ofthe substance present, as well as the number ofatonis of each type present in the sample:

a. Cu(s)b. NH3(g)c. KCI03(s)d. Ca(OH)2(S)

3. What is meant by the percent composition by mass for a compound? Determine the percentcomposition by mass for water.

5. For the compounds in question 4, calculate the percent by mass ofeach element present inthe compounds.

6. Defme, compare, and contrast what are meant by the empirical and molecular formulas fora substance. Give an example ofeach.

7. A 1O.00-g sample ofa compound that consists ofcarbon and hydrogen is found to consistof7.99 g ofcarbon and 2.01 g of hydrogen. What is the empirical formula for thecompound?

8. The molar mass ofthe compound in question 7 is 30.07 g/mol. What is the molecularformula of the compound?

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4. What does it mean to balance an equation?

1. What kind ofvisual evidence indicates that a Qhemical reaction has occurred? Give areaction that illustrates each type ofevidence you have mentioned.

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3. How are the physical states ofreactants and products indicated when writing chemicalequations?

2. What are the substances indicated to the left of the arrow called in a chemical equation? Tothe right of the arrow?

5. What do the coefficients in a balanced chemical equation represent? What do the subscriptsin a balanced chemical equation represent? Which can be changed when balancing achemical equation?

6. Balance the following chemical equations:

a. FeCb(aq) + KOH(aq) ~ Fe(OH)3(S) + KCI(aq)b. AgC2H30 2(aq) + HCI(aq) ~ AgCI(s) + HC2H302(aq)c. SnO(s) + C(s) ~ Sn(s) + C02(g)d. K20(s) + H20(l) ~ KOH(aq)

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1. Considering the reaction represented by the (unbalanced) equation

determine the number of moles ofNH3(g) that can be produced from the following:

a. 0.20 mol N2(g) reacts completely with H2(g).b. 0.30 mol H2(g) reacts completely with N2(g).


Mg(s) + HCI(aq) ~ MgClz(aq) + H2(g)

determine the mass ofH2(g) that can be produced from the following:

a. 10.0 g Mg(s) reacts completely with HCI(aq).b. 20.0 g HCI(aq) reacts completely with Mg(s).

3. What is meant by a limiting reactant in a particular reaction? What does it mean to say thatone or more ofthe reactants are present in excess?


determine the limiting reactant in each ofthe following cases:

a. 4.0 mol H2(g) reacts with 3.0 mol 02(g).b. 10.0 g H2(g) reacts with 10.0 g 02(g).c. 10.0 mol H2(g) reacts with 10.0 mol 02(g).d. 5.0 g H2(g) reacts with 30.0 g 02(g).

5. What do we mean by the theoretical yield for a reaction? What is meant by the actualyield?

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Answer Key ~.

Chapter 2: Basic Review Worksheet1. Defining what scientists mean by "matter" often seems circular to students. Scientists say

that matter is something that "has mass and occupies space" without ever really explainingwhat it means to "have mass" or to "occupy space"! The concept ofmatter is so basic andfundamental that it becomes difficult to give a good textbook definition other than to saythat matter is the "stuff' of which everything is made.

2. Chemists tend to give a functional definition of what they mean by an "element": Anelement is a fundamental substance that cannot be broken down into any simplersubstances by chemical methods. Compounds, on the other hand, can be broken down intosimpler substances (the elements of which the compound is composed). Examples includewater as a compound and gold as an element.

3. A gas is a substance that has no fixed volume or shape; in addition, the particles making upa gas are spread relatively far apart. A liquid is a substance that has a definite volume buttakes the shape of its container; also, the particles making up a liquid are relatively closetogether. A solid is a substance that has a fixed shape and volume and is rigid; the particlesmaking up a solid are generally packed together and orderly.

4. The chemical properties of a given substance indicate how that substance reacts with othersubstances. The physical properties of a substance are the inherent characteristics of the ~.

substance, which result in no change in the composition of the substance when we measureor study these properties.

5. A chemical change for a substance results in the substance being converted into anothersubstance or substances. A physical change for a substance is a change in the substance thatdoes not alter the identity or composition of the substance; physical changes typicallyrepresent changes in only the physical state (solid, liquid, vapor) of the substance.

6. An alloy is a substance that contains a mixture of elements and has metallic properties.Brass and steel are examples of alloys.

7. A mixture is a combination of two or more substances that may be varied in itscomposition. Most commonly in chemistry a mixture is a combination of two or more puresubstances (either elements or compounds). Dirt is an example ofa mixture.

8. A solution is a particular type ofmixture that appears completely homogeneousthroughout. A solution can be made by dissolving sugar in 'Yater, for example.

9. Pure substances are either elements or compounds. A pure substance always has the samecomposition. A pure substance cannot be a mixture.

10. A homogeneous mixture is a mixture that is the same throughout; a homogeneous mixtureis also called a solution. A heterogeneous mixture is a mixture that contains regions withdifferent properties. ..

11. Mixtures can be resolved into their components by distillation or filtration. If'$..


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4. No. Some elements can be composed of molecules. For example, the oxygen that we tfIIII!'breathe is made of diatomic oxygen molecules (02).

,5. When we analyze sulfur dioxide, for example, we notice that each and every molecule

consists of one sulfur atom and two oxygen atoms, and on a mass basis, sulfur dioxideconsists of 50% each of sulfur and oxygen. Thus sulfur dioxide has a constant composition.The reason the mass percent of all sulfur dioxide is constant is because of a constantnumber of atoms of each type present in the compound's molecules.

6. Yes. For example, if a scientist anywhere in the universe analyzed sulfur dioxide, he or shewould find the same composition. If a scientist finds something that does not have the samecomposition, then the substance cannot be sulfur dioxide.

7. Answers will vary. Examples may include dirt, sand, paper, and chunky peanut butter.

8. All solutions are mixtures, but not all mixtures are solutions. Only homogeneous mixturesare solutions.

9. Answers will vary. Examples include: a. a sugar-water solution, homogeneous; b. air,homogeneous (although air consists of more than two gases, it is mainly oxygen andnitrogen); c. rubbing alcohol, homogeneous; d. brass, homogeneous.

10. Filtration and distillation are both physical methods. They do not involve a change in thechemical makeup ofthe substances that are separated.

Chapter 3: Basic Review Worksheet1. An element is a pure substance that cannot be broken down into simpler substances by

chemical means. There are presently more than 110 elements recognized, of which 88occur in nature (the remaining have been synthesized by nuclear processes). The mostabundant elements (by mass) on the earth are oxygen (49.2%), silicon (25.7%), andaluminum (7.50%), with less than 5% of each of the other elements present.

2. The three fundamental particles from which atoms are/composed are electrons, protons,and neutrons. The properties of these particles are summarized below:

Particle Relative Mass Relative Charge Location

proton 1836 1+ nucleusneutron 1839 none nucleuselectron 1 1- outside the nucleus

3. The expression nuclear atom indicates that we view the atom as having a dense center ofpositive charge (called the nucleus) around which the electrons move through primarilyempty space.

4. Isotopes represent atoms ofthe same element that have different atomic masses. Isotopesare atoms of a given element that have different numbers of neutrons in their nuclei.

5. Most elements are too reactive to be found in nature in other than the combined form.Aside from the noble metals gold, silver, and platinum, the only other elements commonlyfound in nature in the uncombined state are some of the gaseous elements. (such as 02,. N2,He, Ar, etc.), and the solid nonmetals carbon and sulfur.


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6. Group Family Name1 Alkali metals2 Alkaline earth elements6 . Chalcogens (not used commonly)7 Halogens8 Noble gase~

7. Based on arrangement by electronic structure, the metallic elements tend to be toward theleft-hand side of the chart, whereas the nonmetallic elements are found toward the righthand, upper side. Since metallic nature increases going downward within any verticalcolumn (as the outermost shell gets farther from ,the nucleus), there are also some metallicelements among the lower members of groups at the right-hand side of the table (manyperiodic tables indicate the dividing line between metallic and nonmetallic elements with acolored "stairstep").

8. Ions are electrically charged particles formed from atoms or molecules that have gained orlost one or more electrons. Positively charged ions are called cations, whereas negativeions are termed anions.

9. Ionic compounds typically are hard, 'crystalline solids with high melting and boiling points.Ionic substances such as sodium chloride, when dissolved in water or when melted,conduct electric currents. Chemists have taken this evidence to mean that ionic substancesconsist of positively and negatively charged particles (ions).

10.' Name Symbol Atomic Number

magneSiUm Mg 12tin Sn 50lead Pb 82sodium Na 11hydrogen H 1chlorine Cl 17silver Ag 47

II. a. helium, 2; b. boron,S; c. selenium, 34; d. barium, 56; e. phosphorus 15; f. strontium 38.

12. a. potassium, K; b. magnesium, Mg; c. hydrogen, H; d. carbon, C; e. lead, Pb; f. helium,He.

13.a.2p,2n,2e;b. 17p,20n, 17e;c.20p,20n,20e

14. a. Mg2+; b.F-; c. Ba2+; d. Na+; e. 0 2-; f. cr

IS. a. 19p, 18e; b. 35p, 36e; c. IIp, We

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identical; these atoms are different fr~m the atoms of all other elements; (4) atoms of one ~

element can combine with atoms of another element to form a compound, and such acompound will always contain the same relative numbers and types of atoms; and (5)atoms are rearranged into new groupings during an ordinary chemical reaction, and noatom is ever destroyed and no new atom is ever created during such a reaction. Studentsshould explain these in their own words.

3. A given compound always contains exactly the same relative masses of its constituentelements. This statement is termed the law ofconstant composition. The law of constantcomposition is a result of the fact that a given compound always contains the same typesand numbers of each constituent atom. For example, water's composition by mass (88.8%oxygen, 11.2% hydrogen) is a result of the fact that each water molecule contains oneoxygen atom (relative mass 16.0) and two hydrogen atoms (relative mass 1.008 each). Thelaw of constant composition is important to our study of chemistry because it means thatwe can always assume that any sample of a given pure substance, from whatever source,will be identical to any other sample.

4. The various isotopes of an element have virtually identical chemical properties because thechemical properties of an atom are a function of the electrons in the atom (not the nucleus).The physical properties of the isotopes of an element (and compounds containing thoseisotopes) may differ because of the difference in mass of the isotopes.

5. Isolated atoms do not form ions on their own but are induced to gain or lose electrons bysome other species (which loses or gains the electrons).

6. An ionic compound could not possibly exist ofjust cations or just anions. There must be a ".~balance of charge, or the compound would be very unstable (like charges repel each other).

7. a.9p, IOn, 10e;b. 12p, 12n, 10e; c. 26p, 30n, 23e

8. a. 37p, 36e; b. 26p, 24e; c. Ip, 2e; d. 13p, lOe; e. 17p, 18e;f. 8p, lOe

9. RbR, RbCI, Rb20; FeH2, FeCh, FeD; AIR3, AICh, Ah03

Chapter 4: Basic Review Worksheet,I. When naming ionic compounds, we name the positive ion (cation) first. Sodium chloride is

an example.

2. For simple binary Type I ionic compounds, the ending -ide is added to the root name of theelement that is the negative ion (anion). For example, the Type I ionic compound formedbetween potassium and sulfur, K2S, is named potassium sulfide. Potassium is the cation,and sulfur is the anion (with the suffix -ide added).

3. Type II compounds involve elements tha~ form more than one stable ion, and so it is'necessary'to specify which ion is present in a given compound.

4. Type II compounds are named by either oqwo systems, the "ous-ic" system (which isfalling out of use), and the "Roman numeral" system, which is preferred by most chemists.For example, iron forms two types of staple ions: Fe2+ -and Fe3

+. Iron can react with oxygento form either oftwo stable oxides, FeO or Fe203, depending on which cation is involved.Under the Roman numeral naming system, FeD would be named iron(lI) oxide to show ~,that it contains Fe2

+ ions; Fe203 would be named iron(lIl) oxide to indicate that it contains

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6. Several families of polyatomic anions contain an atom of a given element combined withdiffering numbers of oxygen atoms. Such anions are called oxyanions. For example, sulfurforms two common oxyanions, S032- and sol-.

7. Acids in general are substances that produce protons (H+ ions) when dissolved in water.HCI and H2S04 are two examples.

8. a. sodium chloride; b. potassium oxide; c. magnesium bromide; d. aluminum iodide; e.calcium sulfide; f. strontium oxide.

9. a. ammonium ion; b. sulfite ion; c. nitrate ion; d. hydroxide ion; e. perchlorate ion; f.phosphate ion.

10. a. nitrogen dioxide; b. iodine monochloride; c. carbon monoxide

Fe3+ ions. The Roman numeral used in a name corresponds to the charge of the specific ionpresent in the compound. Under the less-favored "ous-ic" system, for an element that formstwo stable ions, the ending -ous is used to intlicate the lower charged ion, so FeO andFe203 would be namedferrous oxide andferric oxide, respectively. The "ous-ic" systemhas fallen out of favor because it does not indicate the actual charge on the ion but only thatit is the lower or higher charged of the two. This can lead to confusion. For example, Fe2+is calledferrous ion in this system, but Cu2+is called cupric ion (because there is also aCu+ ion).

In writing the names for such compounds, the element listed first in the formula is namedfirst (using the full name of the element), and then the second element in the formula isnamed as though it were an anion (with the -ide ending). Since there often may be morethan one compound possible involving the same two nonmetallic elements, the namingsystem for Type III compounds goes one step further than the system for ionic compoundsby explicitly stating (by means of a numerical prefix) the number of atoms of each of thenonmetallic elements present in the molecules of the compound. For example, carbon andoxygen (both nonmetals) form two common compounds, CO and CO2. To indicate clearlywhich compound is being discussed, the names of these compounds indicate explicitly thenumber of oxygen atoms present by using a numerical prefix.

CO carbon monoxide (mon- or mono- is the prefix meaning "one")C02 carbon dioxide (di- is the prefix meaning "two")

The prefix mono- is not normally used for the first element named in a compound whenthere is only one atom of the element present, but numerical prefixes are used for the firstelement if there is more than one atom ofthat element present. For example, nitrogen andoxygen form many binary compounds:

NO nitrogen monoxideN02 nitrogen dioxideN20 dinitrogen monoxideN20 4 dinitrogen tetroxide (tetra- or tetr- means "four")


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11. a. K2S; b. HCI; c. CaS04; d. CuBr2

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· 5.

2. A unit tells us what scale or standard is being used to represent the results of themeasurement. .

3. Dimensional analysis is a method of problem solving that pays particular attention to theunits of measurements and uses these units as if they were algebraic symbols that multiply,divide, and cancel. Consider the following example. A dozen of eggs costs $1.25. Supposewe want to know how much one egg costs and also how much three dozen eggs will cost.To solve these problems, we need to make use of two equivalence statements:

1 dozen eggs = 12 eggs1 dozen eggs = $1.25

T~e first of these equivalence statements is obvious: Everyone knows that 12 eggs are"equivalent" to one dozen. The second statement also expresses an equivalence: If you givethe grocer $1.25, he or she will give you a dozen eggs. From these equivalence statementswe can construct the conversion factors we need to answer the two questions. For the firstquestion (what does one egg cost), we can set up the calculation as follows:

$1.25 = $0.104 = $0.1012 eggs

as the cost of one egg. Similarly, for the second question (the cost of3 dozen eggs), we canset up the conversion as follows:

3 dozen x $1.25 = $3.751 dozen

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as the cost of three dozoo eggs. See Section 5.6 ofthe text for how we construct conversion IfI!'Afactors from equivalence statements.

4. a. 122.4 x 105 = (1.224 x 102) x 105 = 1.224 x 107

b. 5.993 x 10-4 = 0.0005993

5. a. 6.0 pt x 1 qt x 1 L = 2.8 L2 pt 1.0567 qt

b. 6.0 pt x .!....9!. x 1 gal = 0.75 gal2 pt 4 qt ..

c. 5.91 yd x 1 m = 5.40 m1.0936 yd

d. 62.5 mi x 1 km = 101 km0.62137 mi

e. 88.5 em x 10 mm = 885 mmlcm

6. a. 10.20 + 4.1 + 26.001 + 2.4 = 42.701 = 42.7 (one decimal place)b. [1.091 - 0.991] + 1.2 = 1.3 (one decimal place)c. (4.06 + 5.1)(2.032 - 1.02) = (9.16)(1.012) = (9.2)(1.01) = 9.3d. (67.21)(1.003)(2.4) = 161.8 = 1.6 x 102 (only 2 significant figures)

7. a. 541 K - 273 = 268°Cb. 221°C + 273 = 494 K ~

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Atoms N = 1.77 x 1023 molecules x 1 N atom = 1.77 x 1023 atoms N1 molecule

Mass of the element present in 1 mol of the compound x 100Mass of 1 mol of the compound

Atoms H = 1.77 x '1023 molecules x 3 H atoms = 5.31 x 1023 atoms H1 molecule


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C. KCI03 (molar mass = 122.6 g)

Mol KCI03= 5.00 g x 1 mol = 0.0408 mol KCI03122.6 g

0.0408 mol KCI03 x 6.022 x 1023

formula units = 2.46 x 1022 formula units KCI031 mol ~

Atoms K = 2.46 x 1022 formula units x 1 Katom = 2.46 x 1022 atoms K1 formula unit •

b. NH3 (molar mass = 17.03 g)

Mol NH3 = 5.00 g x 1 mol = 0.294 mol17.03 g

Molecl:1les NH3 = 0.294 mol x 6.022 x 1023

molecules = 1.77 x 1023 molecules1 mol

128.1 g/24.3 mL = 5.27 g/rnL. We can get the extremes ofthe densities by dividing themaximum mass by the minimum volume and the minimum mass by the maximum volume.Doing so, we get 128.2 g/24.2 mL = 5.30 g/mL and 128.0 g/24.4 mL = 5.25 g/mL.

Chapter 6: Basic Review Worksheet1. The average atomic mass of an element represents the weighted average mass, on the

relative atomic scale, of all the isotopes of an element. Average atomic masses are usuallygiven in terms of atomic mass units

2. The molar mass of a compound is the mass in grams of one mole of the compound (6.022 x

1023 molecules of the compound) and is ca1cuhited by summing the average atomic massesof all the atoms present in a molecule (or empirical formula unit for an ionic substance) ofthe compound. For example, a unit of the compound K20 contains two potassium atomsand one oxygen atom. The molar mass is obtained'by adding up the average atomic masses

. ofthese atoms: Molar mass K20 = 2(39.10 g) + 1(16.00 g) = 94.20 g.

3. The percent composition (by mass) of a compound shows the relative amount of eachelement present in the compound on a mass basis. For compounds whose formulas areknown (and whose molar masses therefore are known), the percentage of a given elementpresent in the compound is given by

The percent composition of water, therefore, is 11.2% hydrogen and 88.8% oxygen.

4. a. Cu (molar mass = 63.55 g)

Mol Cu =5.00 g x I mol = 0.0787 mol63.55 g

Atoms Cu = 0.0787 mol x 6.022 x 1023

atoms = 4.74 x 1022 atoms1 mol

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Atoms Cl = 2.46 x 1022 formula units T 1 Cl atoms = 2.46 x 1022 atoms Cl1 fonnula unit

Atoms 0 = 2.46 x 1022 formula units x 3 o atoms = 7.38 x 1022 atoms 01 fonnula unit

d. Ca(OHh (molar mass = 74.096 g)

Mol Ca(OH)2 = 5.00 g x 1 mol = 0.0675 mol Ca(OHh74.096 g .

0.0675 mol Ca(OH)2 x 6.022 x 1023

fonnula units = 4.06 x 1022 formula units Ca(OH)21 mol

Atoms Ca = 4.06 x 1022 formula units x 1 Caatom = 4.06 x 1022 atoms Ca1 fonnula unit

Atoms 0=4.06 x 1022 formula units x 20 atoms = 8.12 x 1022 atoms 01 fonnula unit

Atoms H = 4.06 x 1022 formula units x 2 H atoms = 8.12 x 1022 atoms H1 fonnula unit


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5. a. 100% Cu

b. NH3: %N = 14.01 gN x 100% = 82.27% N17.03 g

%H= 3(1.008gH) x 100%= 17.76%H17.03 g

c. KC103: %K = 39.10 g K x 100% = 31.89% K122.6 g

%Cl = 35.45 g Cl x 100% = 28.92% Cl122.6 g

%0= 3(16.00g0) x 100%=39.15%0122.6 g

d. Ca(OH)2: %Ca = 40.08 g Ca x 100% = 54.09% Ca74.096 g

%0 = 2(16.00 g 0) x 100% = 43.19% 074.096 g

%H = 2(1.008 g H) x 100% = 2.721 % H74.096 g

6. The empirical formula of a compound represents the smallest ratio of the relative numberof atoms of each type present in a molecule of the compound, whereas the molecularformula represents the actual number of atoms of each type present in a real molecule ofthe compound. For example, both acetylene (molecular form\.!la C2H2) and benzene(molecular formula C6H6) have the same relative number ofcarbon and hydrogen atoms(one hydrogen for each carbon atom) and so have the same empirical formula (CH).

7. 7.99 g C x 1 mol = 0.665 mol C12.01 g

2.01 g H x I mol = 1.99 mol H1.008 g

1.99/0.665 = 2.99. Thus there are three hydrogen atoms for every one carbon atom. Theempirical formula is CH3•

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may be changed. For example, when a piece ofzinc is added to an aqueous copper(I1} ion ~

solution (which is bright blue), the Cu2+ ions·are reduced to copper metal, and the blue \ IF"}'"

color of the solution fades as the reacti<;m takes place.

Zn(s) + Cu2\aq) --+ Zn2(aq) + Cu(s)

blue solution redlblack solid

In many reactinns of ionic solutes, a.solid precipitate forms when the ions are combined.For example, when a clear, colorless aqueous solution of sodium chloride is added to a

. clear, colorless solution of silver nitrate, a white solid of silver chloride forms and settlesout of the mixture.

AgN03(aq) + NaCI(aq) -4 NaN03(aq) + AgCI(s)

In some reactions, particularly in the combustion of organic chemical substances withoxygen, gas, heat, and light (a flame) may be produced. For example, when methane(natural gas) is burned in oxygen, a luminous flame is produced, and heat energy isreleased:

CH4(g) + 202(g) --+ CO2(g) + 2H20(g) + energy

2. The substances to the left of the arrow in a chemical equation are called the reactants;those to the right of the arrow are referred to as the products.

3. The physical states are indicated by using italic letters in parentheses after the formula: (s),(I), (g), or (aq).

4. When we "balance" a chemical equation, we adjust the coefficients of the reactants and ~

products in the equation so that the same total numbers of atoms of each element arepresent both before and after the reaction has taken place.

5. Coefficients in a balanced chemical equation represent the relative numbers of each type ofmolecule involved in the reaction. Subscripts represent the numbers of each type of atom ina particular molecule. When we balance a chemical equation, it is permitted only to adjustthe coefficients of a formula because changing a coefficient merely changes the number ofmolecules of a substance being used in the reaction without changing the identity of thesubstance.


RW-94

6. a. FeCh(aq) + 3KOH(s) -4 Fe(OH)3(s) + 3KCI(aq)

b. AgC2H30 2(aq) + HCI(aq) --+ AgCI(s) + HC2H30 2(aq)

c. 2SnO(s) + C(s) -4 2Sn(s) + CO2(g)

d. K20(s) + H20(1) --+ 2KOH(aq)

Chapter 7: Standard Review Worksheet1. All chemical reactions do produce some evidence that the reaction has occurred, but

sometimes this evidence may not be visual and may not be very obvious. For example,when very dilute aqueous solutions of acids and bases are mixed, the neutralization

~

reaction

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b. 0.30 mol H2 x 2 mol NH] = 0.20 mol NH33 mol Hz

c. 10.0 molE2 x 1 molOz = 5.00 mol O2; H2is limiting.2 molH2

b. lO.Og H2 x I mol Hz X 1 molOz x 32.00g °2 - 79Ag 02; 02 is limiting.2.016g H2 2 mol Hz 1 molOz ..

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" II


RW-101

Synthesis (combination):Ca(s) + Ch(g) ~ CaCh(s)

Decomposition:2HgO(s) ~ 2Hg(l) + 02(g)

Single displacement:Mg(s) + 2AgN03(aq) ~ Mg(N03)2(aq) + 2Ag(s)

Double displacement:Na2S04(aq) + BaCh(aq) ~ 2NaCI(aq) + BaS04(S)

Chapter 9: Basic Review Worksheet1. The balanced equation is N2+ 3H2 ~ 2NH3

a. 0.20 mol N2 x 2 mol NH j = 0040 mol NH31 mol N z

b. 20.0g HCI x 1 mol HCI x 1 mol H2 x 2.016 H2 = 0.553g H236.458g HCI 2 mol HCI 1 mol Hz

3. Although we can calculate specifically the exact amounts of each reactant needed for achemical reaction, often reaction mixtures are prepared using more or less arbitraryamounts of the reagents. However, regardless of how much of each reagent may be usedfor a reaction, the substances still react stoichiometrically, according to the mole ratiosderived from the balanced chemical equation for the reaction. When arbitrary amounts ofreactants are used, there will be one reactant that, stoichiometrically, is present in the leastamount. This substance is called the limitingreactant for the experiment. We say that theother reactants in the experiment are present in the excess, which means that a portion ofthese reactants will still be present unchanged after the reaction has ended and the limitingreactant has been used up completely.

4. The balanced equation is 2H2 + 02 ~ 2H20.

a. 4.0 mol H2 x 1 mol °2 = 2.0 mol 02; H2is limiting.2 mol H2

'.d. 5.0g H2 x 1 mol Hz x I mol °2 x 32.00g °z - 40.g O2; O2is limiting.

. 2.016g Hz 2 mol H2 1 molOz .•

2. The balanced equation is Mg + 2HCI ~ MgCh + H2

a. 10.Og Mg x 1 mol Mg x I mol Hz x 2.016 Hz = 0.829g H224.31g Mg 1 mol Mg 1 mol H2

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chapter 2: basic review worksheet - mr. weiss'...

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