= 11.0 g (assuming 100 washers is exact).

Transcription

1 CHAPTER washers g 1 washer 100. g 1 washer g = 11.0 g (assuming 100 washers is exact). = 909 washers 2. The empirical formula is CFH from the structure given. The empirical formula represents the smallest whole number ratio of the number and types of atoms present. 3. The atomic mass unit (amu) is defined by scientists to more simply describe relative masses on an atomic or molecular scale. One amu is equivalent to g. 4. The average atomic mass takes into account the various isotopes of an element and the relative abundances in which those isotopes are found. 5. a. 125 C atoms b K atoms c Li atoms d. 1 Mg atom amu 1 C atom = amu amu 1 K atom = amu = amu amu 1 Mg atom amu 1 Li atom = amu = amu e I atoms amu 1 I atom = amu 6. a amu Ca 1 Ca atom = 1 Ca atom amu b amu W 1 W atom amu = 5 W atoms c amu Mn d amu I e amu 1 Mn atom = 10 Mn atoms amu 1 I atom amu = 50 I atoms 1 Pb atom = 10 Pb atoms amu 117

2 7. One average iron atom has a mass of amu amu 299 iron atoms would weigh: 299 atoms 1 atom = amu amu represents: amu 8. One tin atom has a mass of amu. 1 atom amu A sample containing 35 tin atoms would weigh: amu of tin would represent: amu 9. Avogadro s number ( atoms; 1.00 mol) g (1.00 mol) 11. molar masses: Na, g; K, g Na g Na = mol Na g mol Na Na mol K g K = g K = atoms = 99 atoms amu 1 atom = 4155 amu; 1 tin atom = 25 tin atoms amu 12. Since g of neon represents half the molar mass of neon (20.18 g), then g of neon contains 1 2 ( ) = neon atoms (0.500 mol). Also, mol of helium gas would contain the same number of atoms as mol of neon gas: this would be g = g of helium gas. 13. The ratio of the atomic mass of H to the atomic mass of N is (1.008 amu/14.01 amu), and the mass of hydrogen is given by 7.00 g N amu = g H amu 14. 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 amu = 177 g Co amu 118

3 15. mass of a magnesium atom = g amu Mg amu Na = g 16. mass of a carbon atom = g amu C amu Al = g 17. of He atoms = g 4 mol of H atoms g H = g 4 mol of H atoms has a mass slightly larger than of He atoms mol Ne atoms g = g Ne = 10. g Ne (two significant figures) B atoms g B = g B = 11 g B (two significant figures) The neon sample weighs less. 19. a g Ne b g Ni g g = mol Ne = 1.24 mol Ni c. 115 mg Ag 1 g = mol Ag 1000 mg g 1 g d. 6.22μg U = mol U 6 10 μg 8.0 g e. 135 g I g = 1.06 mol I 20. a g F g = mol of F atoms b mg Hg c g Si 1 mmol = mmol Hg (1 mmol = 1/1000 mol) mg = mol Si g d g Pt g = mol Pt e g Mg g = mol Mg 119

4 f g g = mol Mo 21. a Li b. 1.5 Al g Li Li g Al Al c mol Pb d. 125 mol Cr g Cr Cr e mol Fe f mol Mg = 1.74 g Li = 40.7 g Al g Pb Pb g Fe 22. a mol Ca g = 18.1 g Pb = g Cr = g Fe g Mg Mg = g Mg = g Ca g b millimol B = g B 3 10 millmol c. 135 mol Al g d mol Ba g e mol P g = g Al = 86.4 g P f mol As g = g Ba = g As. a g Ag Ag atoms g Ag = Ag atoms b mol Cu c g Cu d kg = g Cu atoms Cu atoms g Cu = Cu atoms = Cu atoms g Mg Mg atoms g Mg = Mg atoms 120

5 e oz = g 2.34 oz g Ca atoms = Ca atoms oz g Ca f g Ca Ca atoms g Ca = Ca atoms g mol Ca Ca atoms Ca = Ca atoms g Fe 24. a. 125 Fe atoms Fe atoms b. 125 Fe atoms c. 125 g Fe d. 125 mol Fe e. 125 g Fe amu 1 Fe atom = amu Fe g Fe g Fe Fe = 2.24 mol Fe = g Fe Fe atoms g Fe = g = Fe atoms f. 125 mol Fe 25. molar mass Fe atoms Fe = Fe atoms 26. The molar mass is calculated by summing the individual atomic masses of the atoms in the formula. 27. a. H 3 PO 4 phosphoric acid mass of 3 mol H = 3(1.008 g) = g mass of P = g mass of 4 mol O = 4(16.00 g) = g molar mass of H 3 PO 4 = (3.024 g g g) = g b. Fe 2 O 3 ferric oxide, iron(iii) oxide mass of 2 mol Fe = 2(55.85 g) = g mass of 3 mol O = 3(16.00 g) = g molar mass of Fe 2 O 3 = (111.7 g g) = g c. NaClO 4 sodium perchlorate mass of Na = g mass of Cl = g 121

6 mass of 4 mol O = 4(16.00 g) = g molar mass of NaClO 4 = (22.99 g g g) = g d. PbCl 2 plumbous chloride, lead(ii) chloride mass of Pb = g mass of 2 mol Cl = 2(35.45 g) = g molar mass of PbCl 2 = (207.2 g g) = g e. HBr hydrogen bromide mass of H = g mass of Br = g molar mass of HBr = (1.008 g g) = g f. Al(OH) 3 aluminum hydroxide mass of Al = g mass of 3 mol H = 3(1.008) = g mass of 3 mol O = 3(16.00 g) = g molar mass of Al(OH) 3 = (26.98 g g g) = g 28. a. KHCO 3 potassium hydrogen carbonate, potassium bicarbonate mass of K = g mass of H = g mass of C = g mass of 3 mol O = 3(16.00 g) = g molar mass of KHCO 3 = ( ) = g b. Hg 2 Cl 2 mercurous chloride, mercury(i) chloride mass of 2 mol Hg = 2(200.6 g) = g mass of 2 mol Cl = 2(35.45 g) = g molar mass of Hg 2 Cl 2 = (401.2 g g) = g c. H 2 O 2 hydrogen peroxide mass of 2 mol H = 2(1.008 g) = g mass of 2 mol O = 2(16.00 g) = g molar mass of H 2 O 2 = (2.016 g g) = g d. BeCl 2 beryllium chloride mass of Be = g mass of 2 mol Cl = 2(35.45 g) = g molar mass of BeCl 2 = (9.012 g g) = g 122

7 e. Al 2 (SO 4 ) 3 aluminum sulfate mass of 2 mol Al = 2(26.98 g) = g mass of 3 mol S = 3(32.07 g) = g mass of 12 mol O = 12(16.00 g) = g molar mass of Al 2 (SO 4 ) 3 = (53.96 g g g) = g f. KClO 3 potassium chlorate mass of K = g mass of Cl = g mass of 3 mol O = 3(16.00 g) = g molar mass of KClO 3 = g 29. a. BaCl 2 mass of Ba = g mass of 2 mol Cl = 2(35.45 g) = g molar mass of BaCl 2 = (137.3 g g) = g b. Al(NO 3 ) 3 mass of Al = g mass of 3 mol N = 3(14.01 g) = g mass of 9 mol O = 9(16.00 g) = g molar mass of Al(NO 3 ) 3 = (26.98 g g g) = g c. FeCl 2 mass of Fe = g mass of 2 mol Cl = 2(35.45 g) = g molar mass of FeCl 2 = (55.85 g g) = g d. SO 2 mass of S = g mass of 2 mol O = 2(16.00 g) = g molar mass of SO 2 = (32.07 g g) = g e. Ca(C 2 H 3 O 2 ) 2 mass of Ca = g mass of 4 mol C = 4(12.01 g) = g mass of 6 mol H = 6(1.008 g) = g mass of 4 mol O = 4(16.00 g) = g molar mass of Ca(C 2 H 3 O 2 ) 2 = (40.08 g g g g) = g 1

8 30. a. LiClO 4 mass of Li = g mass of Cl = g mass of 4 mol O = 4(16.00 g) = g molar mass of LiClO 4 = (6.941 g g g) = g b. NaHSO 4 mass of Na = g mass of H = g mass of S = g = g mass of 4 mol O = 4(16.00 g) = g molar mass of NaHSO 4 = (22.99 g g g g) = g c. MgCO 3 mass of Mg = g mass of C = g mass of 3 mol O = 3(16.00 g) = g molar mass of MgCO 3 = (24.31 g g g) = g d. AlBr 3 mass of Al = g mass of 3 mol Br = 3(79.90 g) = 9.7 g molar mass of AlBr 3 = (26.98 g g) = g e. Cr 2 S 3 mass of 2 mol Cr = 2(52.00 g) = g mass of 3 mol S = 3(32.07 g) = g molar mass of Cr 2 S 3 = ( g) = g 31. a. molar mass of NO 2 = g; 21.4 mg = g g NO g = mol NO 2 b. molar mass of Cu(NO 3 ) 2 = g 1.56 g Cu(NO 3 ) g = mol Cu(NO 3 ) 2 c. molar mass of CS 2 = g 2.47 g CS 2 = mol g 124

9 d. molar mass of Al 2 (SO 4 ) 3 = g 5.04 g Al 2 (SO 4 ) g = mol Al 2(SO 4 ) 3 e. molar mass of PbCl 2 = g 2.99 g g = mol PbCl 2 f. molar mass of CaCO 3 = g 62.4 g CaCO g = 0.6 mol CaCO a. molar mass Al 2 O 3 = g 47.2 g = mol g b. molar mass KBr = g 1000 g 1.34 kg = 11.3 mol 1 kg g c. molar mass Ge = g 1 g 521 mg = mol 1000 mg g d. molar mass of U = 8.0 g 1 g 56.2 μg = mol 6 10 μg 8.0 g e. molar mass of NaC 2 H 3 O 2 = g 29.7 g = 1.69 mol = mol g f. molar mass of SO 3 = g 1.03 g = mol g 33. a. molar mass MgCl 2 = g 41.5 g = mol g b. molar mass Li 2 O = g; 135 mg = g g g = mol = 4.52 mmol 125

10 c. molar mass Cr = g; 1.21 kg = 1210 g 1210 g =.3 mol g d. molar mass H 2 SO 4 = g 62.5 g = mol g e. molar mass C 6 H 6 = g 42.7 g = mol g f. molar mass H 2 O 2 = g 135 g = 3.97 mol g 34. a. molar mass of Li 2 CO 3 = g g g = mol b. molar mass of CaCl 2 = g 1000 g 4. kg = kg g c. molar mass of SrCl 2 = g 1 g 1. mg = mol 1000 mg g d. molar mass of CaSO 4 = g 4.75 g g = mol e. molar mass of NO 2 = g 1 g 96.2 mg = mol 1000 mg g f. molar mass of Hg 2 Cl 2 = g 12.7 g = mol g 35. a. molar mass AlCl 3 = g 1.25 mol g = 167 g 126

11 b. molar mass NaHCO 3 = g 3.35 mol g = 281 g c. molar mass HBr = g mg 4.25 mmol = 344 mg = g 1 mmol d. molar mass U = 8.0 g mol 8.0 g = g e. molar mass CO 2 = g mol g = g f. molar mass Fe = g mol Fe g = g 36. a. molar mass of H 2 S = g g = 41.2 g b. molar mass of Li 2 S = g mol g = g c. molar mass of FeCl 3 = g 224 mol g = g d. molar mass of Na 2 CO 3 = g g 7.29 millmol = g 3 10 millimol e. molar mass of NaC 2 H 3 O 2 = g mol g = g f. molar mass of PH 3 = g mol g = g 127

12 37. a. molar mass of C 2 H 6 O = g g = 11.6 g C 2 H 6 O b. molar mass of CO 2 = g 1.26 mol g = 55.5 g CO 2 c. molar mass of AuCl 3 = g mol g = g AuCl 3 d. molar mass of NaNO 3 = g 7.74 mol g = 658 g NaNO 3 e. molar mass of Fe = g mol g = g Fe 38. a. molar mass C 6 H 6 = g mol g = 77.6 g b. molar mass CaH 2 = g g = 177 g c. molar mass H 2 O 2 = g mol g = g d. molar mass C 6 H 12 O 6 = g g 1.22 mmol = g 3 10 mmol e. molar mass Sn = g 10.6 mol g = g f. molar mass SrF 2 = g g = g 128

13 39. a millimol = mol mol molecules b. molar mass of PH 3 = g 4.75 g molecules g c. molar mass of Pb(C 2 H 3 O 2 ) 2 = g = ecules = molecules g formula units g = formula units formula units d mol = formula units e. If the sample contains a total of 5.40 mol of carbon, then because each benzene contains six carbons, there must be (5.40/6) = mol of benzene present molecules mol = molecules molecules 40. a mol CO = molecules CO b. molar mass of CO = g molecules 6.37 g = molecules CO g c. molar mass of H 2 O = g g molecules g = molecules H 2 O molecules d g = molec. H 2 O e. molar mass of C 6 H 6 = g molecules 5. g = molecules C 6 H g 41. a. molar mass of C 2 H 6 O = g g g = mol C 2H 6 O mol C 2 H 6 O 2 mol C = mol C C H O

14 b. molar mass of C 6 H 4 Cl 2 = g g g = mol C 6H 4 Cl 2 6 mol C mol C 6 H 4 Cl 2 C H Cl c. molar mass of C 3 O 2 = g g g = mol C 3O 2 3 mol C mol C 3 O 2 C O 3 2 d. molar mass of CH 2 Cl 2 = g g g = mol CH 2Cl 2 = mol C = mol C mol CH 2 Cl 2 C CH Cl 2 2 = mol C 42. a. molar mass of Na 2 SO 4 = g Na 2SO4 S 2.01 g Na 2 SO 4 = S g Na SO b. molar mass of Na 2 SO 3 = g 2 4 Na 2SO3 S 2.01 g Na 2 SO 3 = mol S g Na SO c. molar mass of Na 2 S = g 2 3 Na 2S S 2.01 g Na 2 S = mol S g Na S d. molar mass of Na 2 S 2 O 3 = g Na 2S2O3 S 2.01 g Na 2 S 2 O 3 = mol S g Na S 43. molar 44. less than 45. a. mass of H present = g = g mass of Cl present = g = g mass of O present = 3(16.00 g) = g molar mass of HClO 3 = g

15 % H = g H g 100 = %H % Cl = g Cl g 100 = %Cl % O = g O g 100 = %O b. mass of U present = 8.0 g = 8.0 g mass of F present = 4(19.00 g) = g molar mass of UF 4 = g % U = 8.0 g U 100 = 75.80% U g % F = g F g 100 = 24.20% F c. mass of Ca present = g = g mass of H present = 2(1.008 g) = g molar mass of CaH 2 = g % Ca = g C 100 = 95.21% Ca g % H = g H g 100 = 4.789% H d. mass of Ag present = 2(107.9 g) = g mass of S present = g = g molar mass of Ag 2 S = g g Ag % Ag = 100 = 87.06% Ag g % S = g S g 100 = 12.94% S e. mass of Na present = g = g mass of H present = g = g mass of S present = g = g mass of O present = 3(16.00 g) = g molar mass of NaHSO 3 = g g Na % Na = 100 = 22.09% Na g 131

16 % H = g H g 100 = % H % S = g S g 100 = 30.82% S % O = g O g 100 = 46.12% O f. mass of Mn present = g = g mass of O present = 2(16.00 g) = g molar mass of MnO 2 = g g Mn % Mn = 100 = 63.19% Mn g % S = g O g 100 = 36.81% O 46. a. mass of Zn present = g mass of O present = g molar mass of ZnO = g %Zn = g Zn 100 = 80.34% Zn g %O = g O g 100 = 19.66% O b. mass of Na present = 2(22.99 g) = g mass of S present = g molar mass of Na 2 S = g g Na %Na = 100 = 58.91% Na g %S = g S g 100 = 41.09% S c. mass of Mg present = g mass of O present = 2(16.00 g) = g mass of H present = 2(1.008 g) = g molar mass of Mg(OH) 2 = g g Mg %Mg = 100 = 41.68% Mg g 132

17 %O = g O g %H = g H g 100 = 54.86% O 100 = 3.456% H d. mass of H present = 2(1.008 g) = g mass of O present = 2(16.00 g) = g molar mass of H 2 O 2 = g %H = g H 100 = 5.926% H g %O = g O g 100 = 94.06% O e. mass of Ca present = g mass of H present = 2(1.008 g) = g molar mass of CaH 2 = g g Ca %Ca = 100 = 95.20% Ca g %H = g H g 100 = 4.789% H f. mass of K present = 2(39.10 g) = g mass of O present = g molar mass of K 2 O = g %K = g K 100 =83.01% K g %O = g O g 100 = 16.99% O 47. a. molar mass of CH 4 = g % C = g C 100 = 74.88% C g b. molar mass NaNO 3 = g g Na % Na = 100 = 27.05% Na g c. molar mass of CO = g % C = g C 100 = 42.88% C g 133

18 d. molar mass of NO 2 = g %N = g N 100 =30.45% N g e. molar mass of C 8 H 18 O = g % C = g C 100 = 73.79% C g f. molar mass of Ca 3 (PO 4 ) 2 = g g Ca % Ca = 100 = 38.76% Ca g g. molar mass of C 12 H 10 O = g % C = g C 100 = 84.67% C g h. molar mass of Al(C 2 H 3 O 2 ) 3 = g g Al % Al = 100 = 13.22% Al g 48. a. molar mass of BaO 2 = g g Ba % Ba = 100 = 81.10% Ba g b. molar mass of BaO = g g Ba % Ba = 100 = 89.56% Ba g c. molar mass of CoBr 2 = g g Co % Co = 100 = 26.94% Co g d. molar mass of CoBr 3 = g g Co % Co = 100 = 19.73% Co g e. molar mass of SnCl 2 = g g Sn % Sn = 100 = 62.61% Sn g f. molar mass of SnCl 4 = g g Sn % Sn = 100 = 45.57% Sn g 134

19 g. molar mass of LiH = g g Li % Li = 100 = 87.32% Li g h. molar mass of AlH 3 = g g Al % Al = 100 = 89.93% Al g 49. 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 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 50. a. molar mass of ICl = g %I = g I 100 = 78.16% I g 135

20 b. molar mass of N 2 O = g %N = g N 100 = 63.65% N g c. molar mass of NO = g %N = g N 100 = 46.68% N g d. molar mass of HgCl 2 = g g Hg %Hg = 100 = 73.89% Hg g e. molar mass of Hg 2 Cl 2 = g g Hg %Hg = 100 = 84.98% Hg g f. molar mass of SF 6 = g %S = g S 100 = 21.96% S g g. molar mass of XeF 2 = g g Xe %Xe = 100 = 77.55% Xe g h. molar mass of MnO 2 = g g Mn %Mn = 100 = 63.19% Mn g 51. a. molar mass of NH 4 I = g + NH4I NH g = mol NH g NH I molar mass of NH + 4 = g mol NH g NH 4 NH = g NH mol NH4 b. 6.3 (NH 4 ) 2 S (NH ) S = 12.6 mol NH 4 + molar mass of NH + 4 = g 12.6 mol NH g NH 4 NH = 227 g NH 4 + ion 136

21 c. molar mass of Ba 3 P 2 = g 2+ Ba3P2 3 mol Ba 9.71 g = mol Ba g Ba P molar mass of Ba 2+ = g mol Ba g Ba Ba mol Ca d mol Ca 3 (PO 4 ) 2 Ca (PO ) molar mass of Ca 2+ = g 22.9 mol Ca g Ca Ca = 8.44 g Ba = 918 g Ca 2+ = 22.9 mol Ca a. molar mass of (NH 4 ) 2 S = g; molar mass of S 2 ion = g % S g S = 100 = 47.06% S g NH S 4 b. molar mass of CaCl 2 = g; molar mass of Cl = g % Cl g Cl = g CaCl = 63.89% Cl c. molar mass of BaO = g; molar mass of O 2 ion = g % O g O = 100 = 10.44% O g BaO d. molar mass of NiSO 4 = g; molar mass of SO 2 4 ion = g % SO 4 2 = g SO g NiSO = 62.08% SO 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. 54. The empirical formula indicates the smallest whole number ratio of the number and type of atoms present in a molecule. For example, NO 2 and N 2 O 4 both have two oxygen atoms for every nitrogen atom and therefore have the same empirical formula 55. 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 137

22 56. 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) 57. Assume we have g of the compound so that the percentages become masses g Ba = mol Ba g g O g = mol O Dividing both of these numbers of moles by the smaller number of moles (0.65 mol Ba) gives 0.65 mol Ba = mol Ba 0.65 mol mol O = mol O 0.65 mol The empirical formula is BaO. 58. Assume we have g of the compound so that the percentages become masses g N = mol N g g Cl g = mol Cl Dividing both of these numbers of moles by the smaller number of moles gives mol N = mol N mol Cl = 3.00 Cl mol The empirical formula is NCl g C = mol C g g H g = mol H g O = mol O g Dividing each number of moles by the smallest number of moles ( mol C) gives 138

23 mol C mol mol H mol = 1.00 C = mol H mol O = mol O mol The empirical formula is CH 3 O. 60. Assume we have g of the compound, so that the percentages become masses g B = mol B g g H g = mol H Dividing each number of moles by the smaller number of moles gives mol B = mol B mol mol H = mol H mol The empirical formula is BH The mass of chlorine in the reaction is = g Cl g Al = Al g g Cl g = mol Cl Dividing each of these number of moles by the smaller ( Al) gives Al = mol Al mol Cl = mol Cl The empirical formula is AlCl Consider g of the compound so that percentages become masses g Sn = mol Sn g g Cl g = mol Cl 139

24 Dividing each number of moles by the smaller number of moles gives mol Sn mol = mol Sn mol Cl = mol Cl mol The empirical formula is SnCl g Zn g = mol Zn g O g = mol O Because the two components are present in equal amounts on a molar basis, the empirical formula must be simply ZnO. 64. Consider g of the compound g Co = mol Co g If the sulfide of cobalt is 55.06% Co, then it is 44.94% S by mass g S = 1.40 S g Dividing each number of moles by the smaller ( mol Co) gives mol Co = mol Co S = mol S mol Multiplying by two, to convert to whole numbers of moles, gives the empirical formula for the compound as Co 2 S The amount of fluorine that reacted with the aluminum sample must be (3.89 g 1.25 g) = 2.64 g of fluorine 1.25 g Al = mol Al g 2.64 g F g = mol F Dividing each number of moles by the smaller number of moles gives mol Al = mol Al mol 140

25 mol F mol = mol F The empirical formula is just AlF g Al g = mol Al 5.28 g F g = mol F Dividing each number of moles by the smaller number of moles gives mol Al = mol Al mol mol F = mol F mol The empirical formula is just AlF 3. Note the similarity between this problem and question 65: they differ in the way the data is given. In question 65, you were given the mass of the product, and first had to calculate how much fluorine had reacted. 67. Consider g of the compound g U = U 8.0 g g F g = mol F Dividing each number of moles by the smaller number of moles (0.284 U) gives U = mol U mol F = mol F The empirical formula is UF Consider g of the compound so that percentages become masses g Li = mol Li g g O g = mol O Dividing each number of moles by the smaller number of moles gives mol Li = 2.00 Li mol 141

26 3.346 mol O = mol O mol The empirical formula is Li 2 O 69. Consider g of the compound g Cu = Cu g g N g O g g = mol N = mol O Dividing each number of moles by the smallest number of moles (0.533 Cu) gives Cu = mol Cu mol N = mol N mol O = 6.00 O The empirical formula is CuN 2 O 6 [i.e., Cu(NO 3 ) 2 ]. 70. Consider g of the compound g Li = mol Li g g N g = 2.87 N Dividing each number of moles by the smaller number of moles (2.87 N) gives mol Li = mol Li N = mol N 2.87 The empirical formula is Li 3 N. 71. Consider g of the compound g Cu = mol Cu g g P g = mol P 142

27 22.41 g O g = 1.40 O Dividing each number of moles by the smallest number of moles ( mol P) gives mol Cu = mol Cu mol mol P = mol P 1.40 O = mol O mol The empirical formula is thus Cu 3 PO Consider g of the compound so that percentages become masses g Co = mol Co g g O g = mol O Dividing each number of moles by the smaller number of moles gives mol Co = mol Co mol mol O = mol O mol Multiplying these relative numbers of moles by 2 to give whole numbers gives the empirical formula as Co 2 O The compound must contain 1.00 mg of lithium and 2.73 mg of fluorine. 1 mmol 1.00 mg Li = mmol Li mg 2.73 mg F 1 mmol mg = mol F The empirical formula of the compound is LiF. 74. Compound 1: Assume g of the compound g P = P g g Cl g = mol Cl Dividing each number of moles by the smaller (0.728 P) indicates that the formula of Compound 1 is PCl

28 Compound 2: Assume g of the compound g P = P g g Cl g = 2.40 Cl Dividing each number of moles by the smaller (0.480 P) indicates that the formula of Compound 2 is PCl 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. 76. If only the empirical formula is known, the molar mass of the substance must be determined before the molecular formula can be calculated. 77. Assume that we have g of the compound: then g will be boron and g will be hydrogen g B = mol B g g H g = mol H Dividing each number of moles by the smaller number (7.228 mol B) gives the empirical formula as BH 3. The empirical molar mass of BH 3 would be [10.81 g + 3(1.008 g)] = g. This is approximately half of the indicated actual molar mass, and therefore the molecular formula must be B 2 H empirical formula mass of CH = 13 g molar mass n = empirical formula mass = 78 g 13 g = 6 The molecular formula is (CH) 6 or C 6 H 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 C 2 H 5 O = 46 g molar mass n = empirical formula mass = 90 g ~2 46 g = molecular formula is (C 2 H 5 O) 2 = C 4 H 10 O 2 144

29 81. Consider g of the compound g C = mol C g g H g O g N g g g = mol H = mol O = mol N Dividing each number of moles by the smallest number of moles (1.784 mol O or N) gives mol C = 2.00 C mol mol H mol mol O mol = 2.00 H = mol O mol N = mol N mol The empirical formula of the compound is C 2 H 2 ON and the empirical formula mass of C 2 H 2 ON = 56. n = molar mass empirical formula mass = 168 g 56 g = 3 The molecular formula is (C 2 H 2 ON) 3 = C 6 H 6 O 3 N For NO 2 : molar mass = (16.00) = g %N = g N 100 = %N g %O = 2(16.00 g O) 100 = %O g For N 2 O 4 : molar mass = 2(14.01 g) + 4(16.00 g) = g %N = 2(14.01 g N) 100 = %N g %O = 4(16.00 g O) 100 = %O g 145

30 83. [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 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 ec atoms 86. mass of 2 mol X = 2(41.2 g) = 82.4 g mass of Y = 57.7 g = 57.7 g mass of 3 mol Z = 3(63.9 g) = g molar mass of X 2 YZ 3 = g %X = 82.4 g 100 = 24.8 %X g % Y = 57.7 g g = 17.4% Y % Z = g g = 57.8% Z If the molecular formula were actually X 4 Y 2 Z 6, the percentage composition would be the same, and the relative mass of each element present would not change. The molecular formula is always a whole number multiple of the empirical formula. 87. magnesium/nitrogen compound: mass of nitrogen = g = g N g Mg = Mg g g N g = mol N Dividing each number of moles by the smaller number of moles gives 146

31 Mg mol = mol Mg mol N = mol N mol Multiplying by two, to convert to whole numbers, gives the empirical formula as Mg 3 N 2. magnesium/oxygen compound: Consider g of the compound g Mg = 2.48 Mg g g O g = 2.48 O The empirical formula is MgO. 88. For the first compound (restricted amount of oxygen) g Cu = mol Cu g g O g = 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. For the second compound (stream of pure oxygen) g Cu = mol Cu g g O g = mol O Since the numbers of moles are the same, the empirical formula is CuO. 89. Compound Molar Mass % H HF g 5.037% HCl g 2.765% HBr g 1.246% HI g % 90. a. molar mass H 2 O = g 4.21 g molecules = molecules g The sample contains oxygen atoms and 2( ) = hydrogen atoms. 147

32 b. molar mass CO 2 = g 6.81 g molecules = molecules g The sample contains carbon atoms and 2( ) = oxygen atoms. c. molar mass C 6 H 6 = g molecules g = molec g The sample contains 6( ) = atoms of each element molecules d mol = molecules atoms C = 12( ) = atoms atoms H = 22( ) = atoms atoms O = 11( ) = atoms 91. a. Assuming 10,000,000,000 = g N2 molecules = g N molecules b g CO2 molecules = g CO molecules c mol NaCl g d mol C 2 H 4 Cl g = g NaCl = g C 2 H 4 Cl a. molar mass of C 3 O 2 = 3(12.01 g) + 2(16.00 g) = g % C = g C 100 = 52.96% C g g C 3 O g C g C O = g C g C molecules g C = C atoms b. molar mass of CO = g g = g % C = g C 100 = 42.88% C g 148

33 ecules CO 1 C atom ecule CO = C atoms g C C atoms = g C C atoms c. molar mass of C 6 H 6 O = 6(12.01 g) + 6(1.008 g) g = g % C = g C 100 = 76.57% C g mol C 6 H 6 O 6 mol C C H O 6 6 = 1.20 mol C 1.20 mol C g = 14.4 g C 14.4 g C C atoms g C = C atoms 93. [1] g [6] i [2] c [7] f [3] b [8] h [4] a [9] e [5] j [10] d g Co g Fe g Co = 2.12 g Fe g Fe g Co g Fe = 2.36 g Co g Te g Hg g Te = 7.86 g Hg kg = g g Li g Zr g Zr = 76.1 g Li g CCl 4 = molecules CCl g CCl4 ecule = g molecules 149

34 99. a. molar mass of C 6 H 6 = g g g H g = g H b. molar mass of CaH 2 = g g g H = g H g c. molar mass of C 2 H 5 OH = g g g H g = g H d. molar mass of C 3 H 7 O 3 N = g g g H = g H g 100. a. molar mass of C 2 H 5 O 2 N = 2(12.01 g) + 5(1.008 g) + 2(16.00 g) g = g g g N g = g N b. molar mass of Mg 3 N 2 = 3(24.31 g) + 2(14.01 g) = g g g N g = g N c. molar mass of Ca(NO 3 ) 2 = g + 2(14.01 g) + 6(16.00 g) = g g g N g = g N d. molar mass of N 2 O 4 = 2(14.01 g) + 4(16.00 g) = g g g N = g N g 101. Consider g of the compound g Cu = mol Cu g g S g H g g = mol S = mol H g O g = mol O 150

35 Dividing each number of moles by the smallest number of moles ( mol) gives mol Cu mol mol S mol mol H mol = mol Cu = mol S = mol H mol O = 9.00 O mol The empirical formula is CuSH 10 O 9 (which is usually written as CuSO 4 5H 2 O) Consider g of the compound g Mg = mol Mg g g N g O g g = mol N = mol O Dividing each number of moles by the smallest number of moles mol Mg = mol Mg mol mol N mol = mol N mol O = mol O mol The empirical formula is MgN 2 O 6 [i.e., Mg(NO 3 ) 2 ] atomic mass unit (amu) 104. We use the average mass because this average is a weighted average and takes into account both the masses and the relative abundances of the various isotopes a. 160,000 amu 1 O atom amu = O atoms (assuming exact) b amu 1 N atom amu c. 13,490 amu 1 Al atom amu = 581 N atoms = 500 Al atoms 151

36 d amu 1 H atom amu = H atoms e. 367, amu amu Na atoms 1 Na atom amu = Na atoms 1 Na atom amu = Na atoms amu 1 Na atom = amu 107. a. 1.5 mg = g g Cr g = mol Cr b g Sr c g B d μg = g g = mol Sr g = mol B g Cf 251 g = mol Cf e lb g 1 lb g Fe = g g = mol Fe f g Ba g g Co g g = mol Ba = 1.07 mol Co 108. a. 5.0 mol K g b mol Hg g c mol Mn g d mol P g = 195 g = g K = 325 g P = g Hg = g Mn 152

37 e mol Fe g f. 125 mol Li g g mol F g = 868 g Li = g Fe = g F 109. a g Au Au atoms g Au = Au atoms b mol Pt c g Pt Pt atoms Pt atoms g Pt = Pt atoms = Pt atoms d. 2.0 lb g 1 lb = 908 g 908 g Mg Mg atoms g Mg = Mg atoms e ml 13.6 g 1 ml = 25.8 g Hg 25.8 g Hg Hg atoms g Hg = Hg atoms f mol W g g W W atoms W atoms g W 110. a. mass of Fe = 1(55.85 g) = g mass of S = 1(32.07 g) = g mass of 4 mol O = 4(16.00 g) = g molar mass of FeSO 4 = g b. mass of Hg = 1(200.6 g) = g mass of 2 mol I = 2(126.9 g) = g molar mass of HgI 2 = g c. mass of Sn = 1(118.7 g) = g mass of 2 mol O = 2(16.00 g) = g molar mass of SnO 2 = g = W atoms = W atoms 153

38 d. mass of Co = 1(58.93 g) = g mass of 2 mol Cl = 2(35.45 g) = g molar mass of CoCl 2 = g e. mass of Cu = 1(63.55 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 111. a. mass of 6 mol C = 6(12.01 g) = g mass of 10 mol H = 10(1.008 g) = g mass of 4 mol O = 4(16.00 g) = g molar mass of C 6 H 10 O 4 = g b. mass of 8 mol C = 8(12.01 g) = g mass of 10 mol H = 10(1.008 g) = g mass of 4 mol N = 4(14.01 g) = g mass of 2 mol O = 1(16.00 g) = g molar mass of C 8 H 10 N 4 O 2 = g c. mass of 20 mol C = 20(12.01 g) = g mass of 42 mol H = 42(1.008 g) = g molar mass of C 20 H 42 = g d. mass of 6 mol C = 6(12.01 g) = g mass of 12 mol H = 12(1.008 g) = g mass of O = g = g molar mass of C 6 H 11 OH = g e. mass of 4 mol C = 4(12.01 g) = g mass of 6 mol H = 6(1.008 g) = g mass of 2 mol O = 2(16.00 g) = g molar mass of C 4 H 6 O 2 = g f. mass of 6 mol C = 6(12.01 g) = g mass of 12 mol H = 12(1.008 g) = g mass of 6 mol O = 6(16.00 g) = g molar mass of C 6 H 12 O 6 = g 112. a. molar mass of (NH 4 ) 2 S = g 21.2 g g = 0.31 (NH 4) 2 S 154

39 b. molar mass of Ca(NO 3 ) 2 = g 44.3 g g = mol Ca(NO 3) 2 c. molar mass of Cl 2 O = 86.9 g 4.35 g 86.9 g = Cl 2O d. 1.0 lb = 454 g; molar mass of FeCl 3 = g g = 2.8 mol FeCl 3 e. 1.0 kg = g; molar mass of FeCl 3 = g g g = 6.2 mol FeCl a. molar mass of FeSO 4 = g 1.28 g g = mol FeSO 4 b mg = g; molar mass of HgI 2 = g g g = mol HgI 2 c μg = g; molar mass of SnO 2 = g g g = mol SnO 2 d lb = 1.26( g) = 572 g; molar mass of CoCl 2 = g 572 g g = 4.4 CoCl 2 e. molar mass of Cu(NO 3 ) 2 = g 4.25 g g = mol Cu(NO 3 ) a. molar mass of CuSO 4 = g mol g = 4.2 g CuSO 4 b. molar mass of C 2 F 4 = g mol g = g C 2 F 4 c mmol = mol; molar mass of C 5 H 8 = g 155

40 mol g = g C 5 H 8 d. molar mass of BiCl 3 = g 6.30 mol g = g BiCl 3 e. molar mass of C 12 H 22 O 11 = g 12.2 mol g = g C 12 H 22 O a. molar mass of (NH 4 ) 2 CO 3 = g 3.09 mol g = 297 g (NH 4 ) 2 CO 3 b. molar mass of NaHCO 3 = g mol g = g NaHCO 3 c. molar mass of CO 2 = g mol g = 3874 g CO 2 d mmol = mol; molar mass of AgNO 3 = g mol g = g AgNO 3 e. molar mass of CrCl 3 = g mol g = 0.38 g CrCl a. molar mass of C 6 H 12 O 6 = g 3.45 g molecules g = molecules C 6 H 12 O 6 b mol molecules c. molar mass of ICl 5 = g 25.0 g molecules g d. molar mass of B 2 H 6 = g = molecules C 6 H 12 O 6 = molecules ICl g molecules g = molecules B 2 H 6 156

41 e mmol = mol mol formula units = formula units 117. a. molar mass of NH 3 = g 2.71 g g = mol NH 3 mol H = 3(0.159 mol) = mol H b. mol H = 2(0.824 mol) = mol H = 1.65 mol H c mg = g; molar mass of H 2 SO 4 = g g g = mol H 2 SO 4 mol H = 2( mol) = mol H d. molar mass of (NH 4 ) 2 CO 3 = g 451 g g = 4.69 mol (NH 4) 2 CO 3 mol H = 8(4.69 mol) = 37.5 mol H 118. a. mass of Ca present = 3(40.08 g) = g mass of P present = 2(30.97 g) = g mass of O present = 8(16.00 g) = g molar mass of Ca 3 (PO 4 ) 2 = g g Ca % Ca = g 100 = 38.76% Ca % P = g P g 100 = 19.97% P % O = g O g 100 = 41.27% O b. mass of Cd present = g = g mass of S present = g = g mass of O present = 4(16.00 g) = g molar mass of CdSO 4 = g g Cd % Cd = 100 = 53.91% Cd g % S = g S g 100 = 15.38% S 157

42 % O = g O g 100 = 30.70% O c. mass of Fe present = 2(55.85 g) = g mass of S present = 3(32.07 g) = g mass of O present = 12(16.00 g) = g molar mass of Fe 2 (SO 4 ) 3 = g g Fe % Fe = 100 = 27.93% Fe g % S = g S g % O = g O g 100 = 24.06% S 100 = 48.01% O d. mass of Mn present = g = g mass of Cl present = 2(35.45 g) = g molar mass of MnCl 2 = g g Mn % Mn = 100 = 43.66% Mn g % Cl = g Cl g 100 = 56.34% Cl e. mass of N present = 2(14.01 g) = g mass of H present = 8(1.008 g) = g mass of C present = g = g mass of O present = 3(16.00 g) = g molar mass of (NH 4 ) 2 CO 3 = g % N = g N 100 = 29.16% N g % H = g H g % C = g C g 100 = 8.392% H 100 = 12.50% C % O = g O g 100 = 49.95% O f. mass of Na present = g = g mass of H present = g = g 158

43 mass of C present = g = g mass of O present = 3(16.00 g) = g molar mass of NaHCO 3 = g g Na % Na = 100 = 27.37% Na g % H = g H g % C = g C g % O = g O g 100 = 1.200% H 100 = 14.30% C 100 = 57.14% O g. mass of C present = g = g mass of O present = 2(16.00 g) = g molar mass of CO 2 = g % C = g C 100 = 27.29% C g % O = g O g 100 = 72.71% O h. mass of Ag present = g = g mass of N present = g = g mass of O present = 3(16.00 g) = g molar mass of AgNO 3 = g g Ag % Ag = 100 = 63.51% Ag g % N = g N g % O = g O g 100 = 8.246% N 100 = 28.25% O 119. a. molar mass of NaN 3 = g g Na % Na = 100 = 35.36% Na g b. molar mass of CuSO 4 = g g Cu % Cu = 100 = 39.81% Cu g 159

44 c. molar mass of AuCl 3 = g g Au % Au = 100 = 64.93% Au g d. molar mass of AgNO 3 = g g Ag % Ag = 100 = 63.51% Ag g e. molar mass of Rb 2 SO 4 = g g Rb % Rb = 100 = 64.01% Rb g f. molar mass of NaClO 3 = g g Na % Na = 100 = 21.60% Na g g. molar mass of NI 3 = g % N = g N 100 = 3.550% N g h. molar mass of CsBr = g g Cs % Cs = 100 = 62.45% Cs g 120. a. % Fe = b. % Ag = g Fe g g Ag 1.8 g 100 = 36.76% Fe 100 = 93.10% Ag c. % Sr = g Sr g 100 = 55.28% Sr d. % C = g C g e. % C = g C g 100 = 55.80% C 100 = 37.48% C f. % Al = g Al g 100 = 52.92% Al g. % K = g K g 100 = 36.70% K h. % K = g K g 100 = 52.45% K 160

45 g C g = mol C g H g = mol H g N g O g g = mol N = mol O Dividing each number of moles by the smallest number of moles ( mol) gives mol C = mol C mol N = mol N The empirical formula is C 6 H 7 NO mol H mol O = mol H = 2.00 O g C g = mol C g H g = mol H 0.18 g N g O g g = mol N = mol O Dividing each number of moles by the smallest number of moles ( mol) gives mol C = mol C mol N = mol N The empirical formula is C 3 H 7 N 2 O mol H mol O = mol H = mol O g Ca g N g g = mol Ca = mol N Dividing each number of moles by the smaller number of moles gives 161

46 mol Ca mol = mol Ca mol N = mol N mol Multiplying these relative numbers of moles by 2 to give whole numbers gives the empirical formula as Ca 3 N Mass of oxygen in compound = 4.33 g 4.01 g = 0.32 g O 4.01 g Hg g = mol Hg 0.32 g O g = mol O Since the numbers of moles are equal, the empirical formula is HgO Mass of chlorine in compound = g 1.00 g = g Cl 1.00 g Cr g = mol Cr g Cl g = mol Cl Dividing each number of moles by the smaller number of moles mol Cr = 1.00 mol Cr mol Cl = 3.0 Cl mol The empirical formula is CrCl Assume we have g of the compound g Ba g Cl g g = mol Ba = mol Cl Dividing each of these number of moles by the smaller number gives mol Ba mol = mol Ba mol Cl mol = mol Cl The empirical formula is then BaCl