Name Section Date. Chapter 6: Basic Review Worksheet 1. What does the average atomic mass of an element represent? What unit is used for average atomic mass? 2. Define molar mass. Using KO as an example, calculate the molar mass from the atomic masses of the elements. 3. What is meant by the percent composition by mass for a compound? Determine the percent composition by mass for water. 4. For 5.00-g samples of each of the following substances, calculate the number of moles of the substance present, as well as the number of atoms of each type present in the sample. a. Cus) b. NH(g) ©. KCIOs(s) d. Ca(OH)2(s) 5. For the compounds in Question 4, calculate the percent by mass of each element present in the compounds. > 6. Define, compare, and contrast what are meant by the empirical and molecular formulas for a substance. Give an example of each. 7. 410.00 g sample of a compound that consists of carbon and hydrogen is found to consist of 7.99 of carbon and 2.01 g of hydrogen. What is the empirical formula for the compound? 8. The molar mass of the compound in question 7 is 30.07 g/mol. What is the molecular formula of the compound? World of Chemistry 25 Copyright © Houghton Miffin Company. All rights reserved.‘Name Section Date Chapter 6: Review Worksheet 1. Express the atomic mass unit in grams. Why is the average atomic mass for an element typically not a whole number? 2. What does one mole of a substance represent on a microscopic, atomic basis? What does one mole of a substance represent on a macroscopic, mass basis? 3. Define molar mass. Calculate the molar mass of H3PO, from the atomic masses of the elements. 4, Describe in general terms how percent composition by mass is obtained by experiment for new compounds. How can this information be calculated for known compounds? 5. For 5.00-g samples of each of the following substances, calculate the number of moles of the ‘substance present, as well as the number of atoms of each type present in the sample. a. K2CrO,(s) b. AuCh(s) c. SiHL(g) 4. Ca(PO.)x5) 6. For the compounds in Question 5, calculate the percent by mass of each element present in the compounds. 7. What does an empirical formula tell us about a compound? A molecular formula? What information must be known for a compound to calculate its molecular formula? 8. An oxide of iron is found to be 70.0% iron by mass. Determine the empirical formula for this compound and name it. 9. A compound that consists of nitrogen and oxygen is found to be 30.4% nitrogen by mass. The molar mass of this compound is between 90g/mol and 100 g/mol. Determine the empirical and molecular formulas for this compound, World of Chemistry 27 Copright @ Houghton Mifin Company. All rights reserved.‘Name, Section Date Chapter 6: Challenge Review Worksheet 1. How does the text define a mole, and why have chemists defined the mole in this manner? 2. How do we know that 16.00 g of oxygen contains the same number of atoms as 12.01 g of carbon, and 22.99 g of sodium? How do we know that 106.0 g of NaxCOs contains the same ‘number of carbon atoms as does 12.01 g of carbon, but three times as many oxygen atoms as in 16.00 g of oxygen (0), and twice as many sodium atoms as in 22.99 g of sodium. 3. Define molar mass. Calculate the molar mass of Al;(SOs)s from the atomic masses of the elements. 4. Why must the molecular formula be an integer multiple of the empirical formula? 5. When chemistry teachers prepare an exam question on determining the empirical formula ofa ‘compound, they usually take a known compound and calculate the percent composition of the ‘compound from the formula. They then give the students this percent composition data and have the students calculate the original formula. Using a compound of your choice, first use the ‘molecular formula of the compound to calculate the percent composition of the compound. Then use this perceat composition data to calculate the empirical formula of the compound. 6. How does the percent by mass of each element present in a compound depend on the mass of the sample? 7. A151.9-mg sample of a new compound has been analyzed and found to contain the following ‘masses of elements: carbon, 82.80 mg; hydrogen, 13.90 mg; oxygen, 55.15 mg. Calculate the empirical formula of this compound. 8. One way of determining the empirical formula of a hydrocarbon (a compound that consists of hydrogen and carbon) is to bum it in the air and measure the mass of carbon dioxide and water vapor that is produced. To do this, we must assume that all of the carbon from the hydrocarbon ends up in the carbon dioxide and all of the hydrogen from the hydrocarbon ends up in the water. Suppose you burn 2.500 g of a hydrocarbon and you collect 3.66 g carbon dioxide and 1.50 g water. Determine the empirical formula for this hydrocarbon. 9. The molar mass of the compound in question 8 is found to be around 60 g/mol. Determine the molecular formula of the compound. World of Chemistry 29 Copyright © Houghton Mifflin Company. All rights reserved.significant (if a decimal point is indicated); (3) exact numbers (e.g., definitions) have an infinite ‘number of significant figures. ‘When we have to round off an answer to the correct number of significant figures (as limited my whatever measurement was least precise), we do this in a particular manner. If the digit to be removed is equal to or greater than 5, the preceding digit is increased by 1. If the digit to be removed is less than 5, the preceding digit is not changed. To perform a series of calculations involving a set of data, retain all of the digits in the intermediate calculations until arriving at the final answer, and then round off to the appropriate number of significant figures. ‘When doing arithmetic with experimentally determined numbers, the final answer is limited by the least precise measurement. In doing multiplication or division calculations, the number of significant figures in the result should be the same as the measurement with the fewest significant figures. In performing addition or subtraction, the number of significant figures in the result is limited by the measurement with the fewest decimal places. 3. a, 0,0000009814 = 9.814 x 107 b. 14.2 x 10°= 14.2 4, a. 351K-273 =78°C 1.80 (78°C) + 32 = 172.4 = 172°F b. [(72°F) - 32] / 1.80 =22.2°C 22.2°C +273 = 295.2 = 295 K 5, 40°C = -40°F; 1.8C + 32 = if C= F, 1.8C + 32 =C, thus 32 =-0.8C and -40=C 6. a, volume = 142.4 g/0.915 g/mL = 156 mL b. 4.2 1b=1.9 x 1 volume = 1.9 x 10! 93.75 g/m? = 507 em? =5.1 x 10% cm? 7. $2740.03 g/mL 128.1 g/24.3 mL=5.27 g/mL. We can get the extremes of the densities by dividing the ‘maximum 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 Worksheet 1, 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 usually given 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 = 10° molecules of the compound), and is calculated by summing the average atomic masses of all the atoms present in a molecule (or empirical formula unit for an ionic substance) of the compound, For example, a unit of the compound KO contains two potassiums and one oxygen: the molar mass is obtained by adding up the average atomic masses of these atoms: molar mass K20 = 2(39.10 g) + 1(16.00 g) = 94.20g World of Chemistry 128 Copyright © Houghton Miffin Company. All rights reserved.. The percent composition (by mass) of a compound shows the relative amount of each element present in the compound on a mass basis. For compounds whose formulas are known (and ‘whose molar masses are therefore known), the percentage of a given element present in the compound is given by mass of the element presentin} molof the compound 49 ‘mass of 1 mol of the compound ‘The percent composition of water, therefore is 11.2% hydrogen and 88.8% oxygen. 4, a. Cu (molar mass = 63.55 g) 1mol mol Cu=5.00 8 x SE = 0.0787 mol 6.02210” atoms atoms Cu = 0.0787 mol x mol = 4.74 x 10” atoms b. NH; (molar mass = 17.03 g) Imol 11 NH = 5.00 mol NH3 = 5.00 g x 17038 molecules NH; = 0.294 mol x =0.294 mol 22x10” molecules = 1.77 x 10” molecules Tmol atoms N* 1.77 x 102 molecules x N82 2 1.77 x 10? atoms N Tmolecule atoms H=1.77 x 108 molecules x 2H#OMS ~ 5.31 x 10% atoms H Tmolecule ©. KCIO; (molar mass = 122.6 g) mol KCI; = 5.00 g x —@21 = 0.0408 mol KCIOs 12268 e . 0.0408 mol KCIO3 x Sepa 0" formants 2.46 x 10” formula units KCIOs mo atoms K = 2.46 x 10? form. un. x aa =2.46 x 10” atoms K an, 1Clatoms atoms Cl = 2.46 x 107 form. un x =2.46 x 10” atoms Cl 1 form. atoms 0 = 2.46 x 10 form. units x 2282S 27.38 x 107 atoms O form. un. World of Chemistry 129 Copyright © Houghton Miffin Company. All rights reserved.4. Ca(OH); (molar mass = 74.096 g) 1mol 74.096 g mol Ca(OH), = 5.00 g x = 0,0675 mol Ca(OH), a . 0.0675 mol Ca(OH), x Sore rman = 4.06 x 10” formula units Ca(OH)2 mol 1Caatom Tform.un. atoms Ca = 4.06 x 10 form. un. x = 4.06 x 10” atoms Ca 20atoms =8.12 x 10” atoms O form. un. atoms O = 4.06 x 10” form. un x 2Hatoms _ 3.12 x 10 atoms H 1 form. un, atoms H = 4.06 x 10? form. units x 5. a 100% Cu b. NE: %N= AEN 100% = 92.27% N 17.03 g = (1.008 gH) 17.03 %H x 100% = 17.76% H x 100% = 31.89% K = 35.458C1 voci= 23458C1 190% = 28.92% C1 12268 * 316.0080) %0 = 316.0089) 199% =39.15%0 126g * 40.08 gCa d. Ca(OH): %Ca= x 100% = 54.09% Ca 74.096 g 0 = 216.008) 199% = 43.19% 0 74.096 g ett = 20.008 8B) 109% = 2.721% H 74.096 g World of Chemisiry 130 Copyright © Houghton Mifflin Company. All rights reserved.6. The empirical formula of a compound represents the smallest ratio of the relative number of atoms of each type present in a molecule of the compound, whereas the molecular formula represents the actual number of atoms of each type present in a real molecule of the compound. For example, both acetylene (molecular formula C;H,) and benzene (molecular formula CeHs) have the same relative number of carbon and hydrogen atoms (one hydrogen for each carbon atom), and so have the same empirical formula (CH). lmol 7. 199 = 0.665 mol C BC * Dog me 2.01 gH x 12° 1.99 mol H 1.008 g 1.99/0.665 = 2.99. Thus, there are 3 hydrogen atoms for every one carbon atom. The empirical formula is CH. 8. The molar mass of the empirical formula CHs is 15.034 (12.01 + 3(1.008)). This is half the molar mass of the compound, thus the compound must have the molecular formula CzHs. Chapter 6: Review Worksheet 1. 1 amu= 1.66 x 10™ g. For example, the average atomic mass of sodium is 22.99 amu, which represents the average mass of all the sodium atoms in the world, (including all the various isotopes and their relative abundances). So that we will be able to use the mass of a sample of sodium to count the number of atoms of sodium present in the sample, we consider that every sodium atom in a sample has exactly the same mass (the average atomic mass). The average atomic mass of an element is typically not a whole number of amu’s because of the presence of the different isotopes of the element, each with its own relative abundance. Since the relative abundance of an element can be any number, when the weighted average atomic mass of the element is calculated, the average is unlikely to be a whole number. 2. Ona microscopic basis, one mole of a substance represents Avogadro’s number (6.022 x 10) of individual units (atoms or molecules) of the substance. On a macroscopic basis, one mole ofa substance represents the amount of substance present when the molar mass of the substance in grams is taken (for example 12.01 g of carbon will be one mole of carbon). 3. The molar mass of a compound is the mass in grams of one mole of the compound (6.022 x 10” molecules of the compound), and is calculated by summing the average atomic masses of all the atoms present in a molecule of the compound. For example, a molecule of the compound HjPO, contains three hydrogen atoms, one phosphorus atom, and four oxygen atoms: the molar mass is obtained by adding up the average atomic masses of these atoms: molar mass HjPO, = 3(1.008 g) + 130.97 g) + 4(16.00 g) = 97.99 g 4, When a new compound is prepared (the formula is not known) the percent composition must be determined on an experimental basis. An elemental analysis must be done of a sample of the new compound to see what mass of each element is present in the sample. For example, if a 1.000 g sample of a hydrucarbon is analyzed, and it is found that the sample contains 0.7487 g of C, then the percent by mass of carbon present in the compound is World of Chemistry 131 Copyright © Houghton Mifflin Company. All rights reserved.—2TAB7 BC 100% = 74.87 %C 1.000g sample ¢ Since we can use the formula of a known compound to calculate the percent composition by mass of the compound, it is not surprising that we can go in the opposite direction — from experimentally determined percent compositions for an unknown compound, we can calculate the formula of the compound. 5. a. K2CrO, (molar mass = 194.2 g) ' 1 mol 1 mol K;CrO, = 5.00 g x 0.0257 mol KzCrO, 194.28 0.0257 mol x £92210 formula units 55 x 10” formula units KxCrOs Tmol atoms K = 1.55 x 10? form. units x 2X210ms _ 3.19 x 107 atoms K 1 form. un. atoms Cr= 1.55 x 10? form. units x 1CtOM _ 4 55 5 16? atoms Cr ' 7 form.un. ’ atoms O= 1.55 x 10” form. units x 4O8OMS _ 6.99 x 107 atoms 0 Cr: 1 form. un. b. AuCl; (molar mass = 303.4 g) mol i mol AuCh = 5.00 = 0.0165 mol AuCl a 4g : eae 0.0165 mol AuCly x pote omens =9.94 x 10" formula units AuCI, mol atoms Au=9.94 x 10" form. un, x 1A¥SOM _ 9.94. 192 atoms Au ‘ form. un. ; atoms C1=9.94 x 10" form. un x 2CMOME ~ 9 98 x 1072 atoms Cl 1 form. un ©. Sil (molar mass = 32.12 g) mol SiH =5.00 g x 2°. — 0.156 mot Sits 32.128 ( World of Chemistry 132 Copyright © Houghton Mifflin Company. All rights reserved.6.022%10” molecules 0.156 mol SiH, x 9.39 x 10” molecules SiH, Tmol atoms Si=9.39 x 10% molecules x 1Si#0™ 9.39 x 10%? atoms Si Tmolecule atoms H=9.39 x 10% molecules x SHAMS - 3.76 x 107 atoms H Tmolecule 4. Cas(PO4): (molar mass = 310.18 g) 1 mol 310.188 mol Ca;(POs)2 = 5.00 g x = 0.0161 mol Cax(PO.)z 6.02210” form. units 0.0161 mol Ca(PO,)2 x =9.70 x 10” formula units Cas(POs)2 Imol atoms Ca=9.70 x 10" form, un, x 2C&8MS > 91 x 10%? atoms Ca Tform.un. atoms P=9.70 x 10" form. un x 2PMOMS 1.94 x 107 atoms P Tform.un. atoms O- 9.70 x 107! form. units x 20MOms _ 7.76 x 107 atoms O form. un. 6. a KiCr0e — %K = ZC2NOBKY 109% = 40.27% K 194.28 ogc w 52:008C8 . A pudg * 100% = 26.78% Cr _ 416.000) : Om rg * 100% =32.96% 0 b. Auch: — %Au= L2LOBAL 109% = 64.93% Au 303.48 35.45 gCl) ———— « 100! 35.05% Cl 303.48 76 = 35.05% World of Chemistry 133 Copyright © Houghton Miffin Company. All rights reserved.09 gSi . SiHlé: 100% = 87.45% Si Gets 32.128 40.008 8H) 100% = 12.55% H 32.128 4. Cay(PO,x: %Ca= 340-08 BC2). 199% = 38.76% Ca 310.188 20.97 gP) agp = 2GOSTEP) 100% = 19.97%P aor: 8(16.0020) 0 = BUS008O) 109% = 41.27% 0 310.18g * 7. The empirical formula of a compound represents the smallest ratio of the relative number of atoms of each type present in a molecule of the compound, whereas the molecular formula represents the actual number of atoms of each type present in a real molecule of the compound. ‘Once the empirical formula of a compound has been determined, itis also necessary to determine the molar mass of the compound before the actual molecular formula can be calculated, 8 Assume 100.0 g Imol 70.0% Fe = 70.0 g Fe x = 1.25 mol Fe on 70.0 gFe x Je mol Fe Imol 30.0% O = 30.0 88 mol O > 0=300 80 x ser 55 = 1.88 mo 1.88/1.25 = 1.50 = 3/2. Thus, the empirical formula is Fe,0s. 9. Assume 100.0 30.4% N=30.4gN x mol 1401g .17 mol N Imol 16.008 69.6% O = 69.6 gO x -35 mol O 4,35/2.17 = 2. The empirical formula is NOz. The molar mass of NO; is about 46 g/mol, which falls in the range of half the molar mass of the compound. The molecular formula must be N04. World of Chemistry 134 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 6: Challenge Review Worksheet 1. Chemists have chosen these definitions so that there will be simple relationship between measurable amounts of substances (grams) and the actual number of atoms or molecules present, and so that the number of particles present in samples of different substances can easily be compared. For example, it is known that carbon and oxygen react according to the equation Cs) + Oxfg) > CO2(8) Chemists understand this equation to mean that one carbon atom reacts with one oxygen molecule to produce one molecule of carbon dioxide, and also that one mole (12.01 g) of carbon will react with one mole (32.00 g) of oxygen to produce one mole (44.01 g) of carbon dioxide. 2. It’sall relative. The mass of each substance mentioned in this question happens to be the molar mass of that substance. Each of the three samples of elemental substances mentioned (O, C, and Na) contains Avogadro’s number (6.022 x 10") of atoms of its respective element. For the ‘compound Na;CO3 given, since each unit of Na;CO; contains two sodium atoms, one carbon atom, and three oxygen atoms, then it’s not surprising that a sample having a mass equal to the ‘molar mass of Na;CO} should contain one molar mass of carbon, two molar masses of sodium, and three molar masses of oxygen: 12.01 g + 2(22.99 g) + 3(16.00 g) = 106.08. 3. The molar mass of a compound is the mass in grams of one mole of the compound (6.022 x 10% molecules of the compound), and is calculated by summing the average atomic masses of all the atoms present in a molecule of the compound (or empirical formula unit of an ionic substance). For example, a formula unit of the compound Al;(SO,)s contains two aluminum, three sulfur, and twelve oxygen: the molar mass is obtained by adding up the average atomic masses of these atoms: molar mass Al,(SO4)s = 2(26.98 g) + 3(32.07 g) + 12(16.00 g)=342.2 4, The subscripts in an empirical formula represent the relative numbers of each type of atom in the molecule. The ratios of these numbers must be the same in the molecular formula (which represents the actual numbers of each type of atom in the molecule). Thus, the molecular formula is always a multiple of the empirical formula. The subscripts must be integers because ‘we cannot have fractions of atoms. 5. Answers will vary, but consider the “known” compound phosphoric acid, HsPO.. First we will calculate the percentage composition (by mass) for HsPO,, and then we will use our results to calculate the empirical formula. ‘Molar mass = 3(1.008 g) + 1(30.97 g) + 4(16.00 g) = 97.99 g get = 3(1.008 8H) 97.99g x 100% = 3.086 %H 30.97gP x 100% = 31.60 %P 97.998 %0 = 416.0080) 199% = 65.31 %O 97.99 g World of Chemistry 135 Copyright © Houghion Mifflin Company. All rights reserved.‘Now we will use this percentage composition data to calculate the empirical formula, We will pretend that we did not know the formula, and were just presented with a question of the type: a compound contains 3.086 % hydrogen, 31.60 % phosphorus, and 65.31 % oxygen by mass; calculate the empirical formula.” First, we will assume, as usual, that we have 100.0 g of the compound, so that the percentages tum into masses in grams. So our sample will contain 3.086 g H, 31.60 g P, and 65.31 gO. ‘Next we can calculate the number of moles of each element these masses represent. 1 mol H mol H = 3.086 gH x 1H ~ 3,062 mol H Bu [008gH mol P=31.60gP x 1O1P_ 1.020 mol P 3097 gP ImolO 10=65.31 g0 x =a. = 4.082 mol O ca 8°* 750020 os To get the empirical formula, divide each of these numbers of moles by the smallest number of moles: this puts things on a relative basis. 3.062 mol H =3.002 mol H 1.020 1.020mol P ————— = 1.000 7,020 mol P 4.082 M010 _ 4.099 molO 20 Which gives us as the empirical formula—H3POs. 6. Itdoesn’t. The percent by mass of hydrogen in water, for example, is only dependent on the fact that there are 2 hydrogen atoms for every oxygen atom in a molecule of water, and that hydrogen hhas an average atomic mass of 1.008 compared to 16.00 for oxygen. Whether we have a cup of water, a gallon of water, or an ocean of water does not change the percent by mass of hydrogen and oxygen. Amiltimol _ 7. millimol C = 82.80 imo! ™B* 721mg 9.894 millimol C 1 millimol 008mg millimol H = 13.90 mg x 3.79 millimol H World of Chemistry 136 Copyright © Houghton Miffin Company. All rights reserved.Lmillimol 16.008 millimol O = 55.15 mg x 477 millimol O ing cach of these numbers of millimols by the smallest number of millimols (3.447 millimol O) gives the empirical formula as C2H,0. 12.01gC 8. 3.66 gCO, x 2018C_ <9.999 9 8002 x Tr o1gCO; . 2.016gH 150 gH,0 x 201684 16s gH BHO x TF OI6gHLO e 2.500 g— 0.999 g- 0.168 g= 1.333 g0 ImolC 0.999 BC x TOT = 0.0833 mol C 1mol H 0.168 gC x “22 = 0.167 mol H BC* F008 gH mo 1mol0 1.332 gO x =0.0 80x Tero 7008331 mol O Since 0.0833:0.167:0.08331 = 1:2:1, the empirical formula is CH20. 9. ‘The molar mass of CHO is about 30 g/mol. This is half of the molar mass of the compound. Thus the molecular formula is C2H,O2. Chapter 7: Basic Review Worksheet 1. There are numerous ways we can recognize that a chemical reaction has taken place. In some reactions there may be a color change. For example, the ions of many of the transition ‘metals are brightly colored in aqueous solution. If one of these ions undergoes a reaction in which the oxidation state changes, however, the characteristic color of the ion may be changed. For example, when a piece of zinc is added to an aqueous copper (II) ion solution (which is bright blue), the Cu”* ions are reduced to copper metal, and the blue color of the solution fades as the reaction takes place. Zn (s) + Cu* (aq) > Zn? (aq) + Cu (s) blue red/black solution solid World of Chemistry 137 Copyright © Houghton Mifflin Company. All rights reserved.