# Chapter 6 Chemical Composition

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1 Chapter 6 Chemical Composition washers g 1 washer = 11.0 g (assuming 100 washers is exact) 100. g 1 washer g = 909 washers g 1 cork 1.63 g = = 307 corks 500. g 1 stopper 4.31 g = 116 stoppers 1 kg (1000 g) of corks contains (1000 g 1 cork ) = = 613 corks 1.63 g 613 stoppers would weigh (613 stoppers 4.31 g ) = 2644 g = 2640 g 1 stopper The ratio of the mass of a stopper to the mass of a cork is (4.31 g/1.63 g). So the mass of stoppers that contains the same number of stoppers as there are corks in 1000 g of corks is 1000 g 4.31 g = 2644 g = 2640 g 1.63 g 3. The atomic mass unit (amu) is a unit of mass defined by scientists to more simply describe relative masses on an atomic or molecular scale. One amu is equivalent to g. 4. We use the average atomic mass of an element when performing calculations because the average mass takes into account the individual masses and relative abundance of all the isotopes of an element. 5. a. 635 H atoms amu 1 H atom = 640. amu b W atoms amu 1 W atom = amu c. 42 K atoms amu 1 K atom = 1642 amu d N atoms amu 1 N atom = amu e. 891 Fe atoms amu 1 Fe atom = amu 37

2 38 Chapter 6 6. a amu 1 B atom amu = atom = 1 B atom b amu 1 S atom amu = 10 S atoms c. 19,691 amu d. 19,695 amu e amu 1 Au atom amu 1 Xe atom amu 1 Al atom amu = Au atoms = 100 Au atoms = Xe atoms = 150 Xe atoms = Al atoms = 133 Al atoms amu 1 S atom amu S atoms = 258 S atoms amu 1 S atom = amu Avogadro s number ( ) 10. The ratio of the atomic mass of Ca to the atomic mass of Mg is (40.08 amu/24.31 amu), and the masses of calcium are given by g Mg g Mg amu amu amu amu = g Ca = g Ca 11. The ratio of the atomic mass of Co to the atomic mass of F is (58.93 amu/19.00 amu), and the mass of cobalt is given by 57.0 g F amu amu = 177 g Co 12. O = = O atoms 1 O atom x O atoms = g O mol O atoms = 8.0 g O 4 mol H atoms g = 4 g H Half a mole of O atoms weighs more than 4 moles of H atoms.

3 Chemical Composition a g Au b g Ca Au g Ca g = mol Au = 1.04 mol Ca c. 335 mg Ba 1 g 10 3 mg Ba g = mol Ba d g Pd Pd g = mol Pd e g Ni 1 g 10 6 g f lb Fe g 1 lb g g C C g Ni g = mol Ni Fe = 8.12 mol Fe g = mol C 15. a mol Fe g = 112 g Fe b Ni g = 30.6 g Ni c mol Pt g = g Pt d mol Pb g e mol Mg g = g Pb = g Mg f mol Al g = g Al g mol Li g = 1468 g Li h mol Na g = g Na 16. a g Co x 1023 Co atoms g Co = Co atoms b mol Co x 1023 Co atoms = Co atoms

4 40 Chapter 6 c g Co g Co = mol Co d Co atoms x Co atoms = mol Co e mol Co g Co Co = 249 g Co f mol Co x 1023 Co atoms Co = Co atoms g g Co x 1023 Co atoms g Co = Co atoms 17. molar mass 18. adding together (summing) 19. a. mass of 3 mol Na = 3 (22.99 g) = g mass of N = 1 (14.01 g) = g molar mass of Na 3 N = g b. mass of C = g = g mass of 2 mol S = 2 (32.07 g) = g molar mass of CS 2 = g c. mass of N = g = g mass of 4 mol H = 4 (1.008 g) = g mass of Br = g = g molar mass of NH 4 Br = g = g d. mass of 2 mol C = 2 (12.01 g) = g mass of 6 mol H = 6 (1.008 g) = g mass of O = g = g molar mass of C 2 H 6 O = g = g e. mass of 2 mol H = 2 (1.008 g) = g mass of S = g = g mass of 3 mol O = 3 (16.00 g) = g molar mass of H 2 SO 3 = g = g f. mass of 2 mol H = 2 (1.008 g) = g mass of S = g = g mass of 4 mol O = 4 (16.00 g) = g molar mass of H 2 SO 4 = g = g

5 Chemical Composition a. mass of Ba = g = g mass of 2 mol Cl = 2 (35.45 g) = g mass of 8 mol O = 8 (16.00 g) = g molar mass of Ba(ClO 4 ) 2 = g b. mass of Mg = g = g mass of S = g = g mass of 4 mol O = 4 (16.00 g) = g molar mass of MgSO 4 = g c. mass of Pb = g = g mass of 2 mol Cl = 2 (35.45 g) = g molar mass of PbCl 2 = g d. mass of Cu = g = g mass of 2 mol N = 2 (14.01 g) = g mass of 6 mol O = 6 (16.00 g) = g molar mass of Cu(NO 3 ) 2 = g e. mass of Sn = g = g mass of 4 mol Cl = 4 (35.45 g) = g molar mass of SnCl 4 = g f. mass of 6 mol C = 6 (12.01 g) = g mass of 6 mol H = 6 (1.008 g) = g mass of O = g = g molar mass of C 6 H 6 O = g 21. a. molar mass of SO 3 = g 1 g 49.2 mg SO mg g = mol SO 3 b. molar mass of PbO 2 = g 7.44 x 10 4 kg PbO g 1 kg g = mol PbO 2 c. molar mass of CHCl 3 = g 59.1 g CHCl g = mol CHCl 3 d. molar mass of C 2 H 3 Cl 3 = g 3.27 g C 2 H 3 Cl 3 1 g 10 6 g e. molar mass of LiOH = g g = mol C 2 H 3 Cl g LiOH g = mol LiOH

6 42 Chapter a. molar mass of NaH 2 PO 4 = g g NaH 2 PO g = mol NaH 2 PO 4 b. molar mass of CuCl = g 521 g CuCl g = 5.26 mol CuCl c. molar mass of Fe = g 151 kg Fe 1000 g 1 kg g = mol Fe d. molar mass of SrF 2 = g 8.76 g SrF g = mol SrF 2 e. molar mass of Al = g g Al g = 467 mol Al 23. a. molar mass of AlI 3 = g 1.50 mol AlI g = 612 g AlI 3 b. molar mass of C 6 H 6 = g mol C 6 H g = g C 6 H 6 c. molar mass of C 6 H 12 O 6 = g 4.00 mol C 6 H 12 O g = 721 g C 6 H 12 O 6 d. molar mass of C 2 H 5 OH = g mol C 2 H 5 OH g = g C 2 H 5 OH e. molar mass of Ca(NO 3 ) 2 = g 2.27 mol Ca(NO 3 ) g = 373 g Ca(NO 3 ) 2

7 Chemical Composition a. molar mass of CO 2 = g 1.27 mmol 10 3 mmol g = g CO 2 b. molar mass of NCl 3 = g mol NCl g = g NCl 3 c. molar mass of NH 4 NO 3 = g NH 4 NO g = g NH 4 NO 3 d. molar mass of H 2 O = g 18.0 mol H 2 O g = 324 g H 2 O e. molar mass of CuSO 4 = g 62.7 mol CuSO g = g CuSO a mol CO x 1023 molecules = molecules CO b. molar mass of CO = g 6.37 g g x 1023 molecules = molecules CO c. molar mass of H 2 O = g g x 1023 molecules g = molecules H 2 O d mol x 1023 molecules = molecules H 2 O e. molar mass of C 6 H 6 = g 5.23 g x 1023 molecules = molecules C 6 H g

8 44 Chapter a. molar mass of Na 2 SO 4 = g 2.01 g Na 2 SO 4 Na 2 SO g b. molar mass of Na 2 SO 3 = g S Na 2 SO 4 = S 2.01 g Na 2 SO 3 Na 2 SO g S Na 2 SO 3 = mol S c. molar mass of Na 2 S = g 2.01 g Na 2 S Na S S g Na 2 S = mol S d. molar mass of Na 2 S 2 O 3 = g 2.01 g Na 2 S 2 O 3 Na 2 S 2 O g 2 mol S Na 2 S 2 O 3 = mol S 27. a. mass of Na present = 2 (22.99 g) = g mass of S present = g = g mass of O present = 4 (16.00 g) = g molar mass of Na 2 SO 4 = g % Na = g Na g 100 = 32.37% Na % S = g S g % O = g O g 100 = 22.58% S 100 = 45.05% O b. mass of Na present = 2 (22.99 g) = g mass of S present = g = g mass of O present = 3 (16.00 g) = g molar mass of Na 2 SO 3 = g % Na = g Na g 100 = 36.48% Na % S = g S g 100 = 25.44% S % O = g O g 100 = 38.08% O

9 Chemical Composition 45 c. mass of Na present = 2 (22.99 g) = g mass of S present = g = g molar mass of Na 2 S = g % Na = g Na g 100 = 58.91% Na % S = g S g 100 = 41.09% S d. mass of Na present = 2 (22.99 g) = g mass of S present = 2 (32.07 g) = g mass of O present = 3 (16.00 g) = g molar mass of Na 2 S 2 O 3 = g % Na = g Na g 100 = 29.08% Na % S = g S g % O = g O g 100 = 40.56% S 100 = 30.36% O e. mass of K present = 3 (39.10 g) = g mass of P present = g = g mass of O present = 4 (16.00 g) = g molar mass of K 3 PO 4 = g % K = g K g % P = g P g % O = g O g 100 = 55.25% K 100 = 14.59% P 100 = 30.15% O f. mass of K present = 2 (39.10 g) = g mass of H present = g = g mass of P present = g = g mass of O present = 4 (16.00 g) = g molar mass of K 2 HPO 4 = g = g % K = g K g % H = g H g 100 = 44.90% K 100 = % H

10 46 Chapter 6 % P = g P g % O = g O g 100 = 17.78% P 100 = 36.74% O g. mass of K present = g = g mass of H present = 2 (1.008 g) = g mass of P present = g = g mass of O present = 4 (16.00 g) = g molar mass of KH 2 PO 4 = g % K = g K g % H = g H g % P = g P g % O = g O g 100 = 28.73% K 100 = 1.481% H 100 = 22.76% P 100 = 47.03% O h. mass of K present = 3 (39.10) g = g mass of P present = g = g molar mass of K 3 P = g = g % K = g K g % P = g P g 100 = 79.10% K 100 = 20.88% P 28. a. molar mass of CuBr 2 = g g Cu % Cu = 100 = 28.45% Cu g b. molar mass of CuBr = g g Cu % Cu = 100 = 44.29% Cu g c. molar mass of FeCl 2 = g g Fe % Fe = 100 = 44.06% Fe g

11 Chemical Composition 47 d. molar mass of FeCl 3 = g g Fe % Fe = 100 = 34.43% Fe g e. molar mass of CoI 2 = g g Co % Co = 100 = 18.85% Co g f. molar mass of CoI 3 = g g Co % Co = 100 = 13.41% Co g g. molar mass of SnO = g g Sn % Sn = 100 = 88.12% Sn g h. molar mass of SnO 2 = g g Sn % Sn = 100 = 78.77% Sn g 29. a. molar mass of C 6 H 10 O 4 = g % C = g C 100 = 49.32% C g b. molar mass of NH 4 NO 3 = g % N = g N 100 = 35.00% N g c. molar mass of C 8 H 10 N 4 O 2 = g % C = g C 100 = 49.47% C g d. molar mass of ClO 2 = g g Cl % Cl = 100 = 52.56% Cl g e. molar mass of C 6 H 11 OH = g % C = g C 100 = 71.92% C g f. molar mass of C 6 H 12 O 6 = g % C = g C 100 = 39.99% C g

12 48 Chapter 6 g. molar mass of C 20 H 42 = g % C = g C 100 = 85.03% C g h. molar mass of C 2 H 5 OH = g % C = g C 100 = 52.14% C g 30. a. molar mass of NH 4 Cl = g molar mass of NH + 4 ion = g % NH + 4 = g NH g b. molar mass of CuSO 4 = molar mass of Cu 2+ ion = g = 33.73% NH 4 + % Cu 2+ = g Cu g 100 = 39.81% Cu 2+ c. molar mass of AuCl 3 = g molar mass of Au 3+ ion = g % Au 3+ = g Au g 100 = 64.93% Au 3+ d. molar mass of AgNO 3 = g molar mass of Ag + ion = g % Ag + = g Ag g 100 = 63.51% Ag To determine the empirical formula of a new compound, the composition of the compound by mass must be known. To determine the molecular formula of the compound, the molar mass of the compound must also be known. 32. The empirical formula represents the smallest whole number ratio of the elements present in a compound. The molecular formula indicates the actual number of atoms of each element found in a molecule of the substance. 33. a. NaO b. C 4 H 3 O 2 c. C 12 H 12 N 2 O 3 is already the empirical formula d. C 2 H 3 Cl

13 Chemical Composition a. yes (each of these has empirical formula CH) b. no (the number of hydrogen atoms is wrong) c. yes (both have empirical formula NO 2 ) d. no (the number of hydrogen and oxygen atoms is wrong) g C C g C = mol C g H H g H g O O = mol H = mol O g Cl Cl g Cl = mol Cl Dividing each number of moles by the smallest number of moles gives: mol C = mol C mol H = mol H mol O = mol O Cl = mol Cl The empirical formula is C 3 H 2 OCl g Ca = mol Ca g Ca The increase in mass represents the oxygen with which the calcium reacted: g O O = mol O Since we have effectively the same number of moles of Ca and O, the empirical formula must be CaO. 37. Consider having g of the compound. Then the percentages of the elements present are numerically equal to their masses in grams g Ba = mol Ba g Ba g S = mol S g S

14 50 Chapter g O O = mol O Dividing each number of moles by the smallest number of moles ( mol S) gives mol Ba mol S = mol Ba = mol S mol O = 4.00 O The empirical formula is BaSO The mass of chlorine involved in the reaction is = g Cl g Al g Cl Al g Al Cl g Cl = Al = mol Cl Dividing each of these numbers of moles by the smaller ( Al) shows that the empirical formula is AlCl Consider g of the compound g Co g Co = mol Co If the sulfide of cobalt is 55.06% Co, then it is 44.94% S by mass g S = 1.40 S g S Dividing each number of moles by the smaller ( mol Co) gives mol Co = mol Co 1.40 S = mol S Multiplying by two, to convert to whole numbers of moles, gives the empirical formula for the compound as Co 2 S g Ca g Cl Ca g Ca Cl g Cl = mol Ca = mol Cl Dividing each of the number of moles by the smaller ( mol Ca) shows that the empirical formula is CaCl 2.

15 Chemical Composition g Cu g Cu = mol Cu 2.52 g O O = mol O The numbers are almost equal: the empirical formula is CuO. 42. Consider g of the compound. Cu g Cu = Cu g Cu g N N g N g O O = mol N = mol O Dividing each number of moles by the smaller number of moles (0.533 Cu) gives Cu mol N = mol Cu = mol N mol O = 6.00 O The empirical formula is CuN 2 O 6 [i.e., Cu(NO 3 ) 2 ] 43. Compound 1: Assume g of the compound. Na g Na = mol Na g Na g N N = mol Na g N Dividing each number of moles by the smaller (1.205 mol Na) indicates that the formula of Compound 1 is Na 3 N. Compound 2: Assume g of the compound. Na g Na = mol Na g Na g N N g N = mol N Dividing each number of moles by the smaller (1.538 mol Na) indicates that the formula of Compound 2 is NaN 3.

16 52 Chapter The empirical formula of a compound represents only the smallest whole number relationship between the number and type of atoms in a compound, whereas the molecular formula represents the actual number of atoms of each type in a true molecule of the substance. Many compounds (for example, H 2 O) may have the same empirical and molecular formulas. 45. If only the empirical formula is known, the molar mass of the substance must be determined before the molecular formula can be calculated. 46. empirical formula mass of CH 2 O = 30 g molar mass n = empirical formula mass = 90 g 30 g = 3 molecular formula is (CH 2 O) 3 = C 3 H 6 O empirical formula mass of CH 2 = 14 molar mass n = empirical formula mass = 84 g 14 g = 6 molecular formula is (CH 2 ) 6 = C 6 H empirical formula mass of CH 4 O = g molar mass n = empirical formula mass = 192 g g = 6 molecular formula is (CH 4 O) 6 = C 6 H 24 O Consider g of the compound g C C = mol C g C g H H g H g O O g N N g N = mol H = mol O = mol N Dividing each number of moles by the smallest number of moles (1.784 mol O) gives mol C = 2.00 C mol H = 2.00 H mol O = mol O

17 Chemical Composition mol N = mol N The empirical formula of the compound is C 2 H 2 ON, empirical formula mass of C 2 H 2 ON = 56 molar mass n = empirical formula mass = 168 g = 3 56 g The molecular formula is (C 2 H 2 ON) 3 = C 6 H 6 O 3 N Consider g of the compound g C C = mol C g C g H H = mol H g H g O O = mol O Dividing each number of moles by the smallest number of moles (1.816 mol O) gives mol C = 3.00 C mol H = mol H mol O = mol O The empirical formula is C 3 H 3 O, and the empirical formula mass is approximately 55 g. molar mass n = empirical formula mass = 110 g = 2 55 g The molecular formula is (C 3 H 3 O) 2 = C 6 H 6 O [1] c [6] d [2] e [7] a [3] j [8] g [4] h [9] i [5] b [10] f g Al mol atoms g Fe mol atoms g Cu 4.3 mol atoms g Mg mol atoms g Na mol atoms g U mol atoms

18 54 Chapter g mol molec atoms 4.04 g mol molec atoms 1.98 g mol molec atoms 45.9 g 1.26 mol molec atoms 126 g 6.99 mol molec atoms g mol 5.58 x 10 2ec x atoms 54. magnesium/nitrogen compound: mass of nitrogen contained = g g = g N Mg g Mg = Mg g Mg g N N g N = mol N Dividing each number of moles by the smaller number of moles gives Mg = mol Mg mol N = mol N Multiplying by two, to convert to whole numbers, gives the empirical formula as Mg 3 N 2. magnesium oxygen compound: Consider g of this compound. Mg g Mg = 2.48 Mg g Mg g O O = 2.48 O Since the numbers of moles are the same, the compound contains the same relative number of Mg and O atoms: the empirical formula is MgO. 55. For the first compound (restricted amount of oxygen): g Cu Cu g Cu = mol Cu g O O = mol O Since the number of moles of Cu ( mol) is twice the number of moles of O ( mol), the empirical formula is Cu 2 O.

19 Chemical Composition 55 For the second compound (stream of pure oxygen): Cu g Cu = mol Cu g Cu g O O = mol O Since the numbers of moles are the same, the empirical formula is CuO. 56. We need to find the subscripts for C a H b O c S d, where a:b:c:d is the mole ratio of C:H:O:S atoms. We are given the following relationships: b = 2a a = c b = 8d We can see that d will be the smallest subscript. Thus, let d = x and we get b = 8x 2a = 8x or a = 4x since a = c, c = 4x Thus, we have C 4x H 8x O 4x S x. The empirical formula is C 4 H 8 O 4 S, which has a molar mass of about 152 g/mol [4(12.01) + 8(1.008) + 4(16.00) + 1(32.07)]. We are given that the molar mass of the compound is 152 g/mol. Thus, the molecular formula is C 4 H 8 O 4 S g Co g Fe g Co = 2.12 g Fe g Fe g Co g Fe = 2.36 g Co 59. Consider g of the compound. Cu g Cu = mol Cu g Cu g S S g S = mol S g H H g H = mol H g O O = mol O

20 56 Chapter 6 Dividing each number of moles by the smallest number of moles gives mol Cu mol S = mol Cu = mol S mol H = mol H mol O = 9.00 O The empirical formula is CuSH 10 O 9 (which is usually written as CuSO 4 5H 2 O) g C C g C = mol C g H H g H g N N g N = mol H = mol N g O O = mol O Dividing each number of moles by the smallest number of moles ( mol O) gives mol C mol H mol N = mol C = mol H = mol N mol O = mol O The empirical formula is C 3 H 7 N 2 O. 61. Mass of oxygen in compound = 4.33 g 4.01 g = 0.32 g O 4.01 g Hg Hg g Hg = mol Hg 0.32 g O O = mol O Since the numbers of moles are equal, the empirical formula is HgO.

21 Chemical Composition Assume we have g of the compound. Ba g Ba = mol Ba g Ba Cl g Cl = mol Cl g Cl Dividing each of these numbers of moles by the smaller number gives mol Ba = mol Ba mol Cl = mol Cl The empirical formula is then BaCl We need to find the subscripts for H x N y O z, where x:y:z is the mole ratio of H:N:O atoms. We are given that x = 4.0 moles H. To solve for y, we use 56.0 g N N = 4.00 moles N g N To solve for z, we use 7.2 x O atoms O = 12 moles O x atoms The mole ratio is 4:4:12 or 1:1:3. The empirical formula is HNO For every g of A 2 O, we have 63.7 g A and 36.3 g O g O O = 2.27 mol O The ratio between A and O is 2:1; with 2.27 mol O we must have 4.54 mol A g A = 14.0 g/mol 4.54 mol A Thus, A must be nitrogen. The compound is N 2 O. 65. [1] g [6] i [2] c [7] f [3] b [8] h [4] a [9] e [5] j [10] d

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