= 11.0 g (assuming 100 washers is exact).
|
|
- Phyllis Kelly
- 7 years ago
- Views:
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
Moles, Molecules, and Grams Worksheet Answer Key
Moles, Molecules, and Grams Worksheet Answer Key 1) How many are there in 24 grams of FeF 3? 1.28 x 10 23 2) How many are there in 450 grams of Na 2 SO 4? 1.91 x 10 24 3) How many grams are there in 2.3
More informationMolar Mass Worksheet Answer Key
Molar Mass Worksheet Answer Key Calculate the molar masses of the following chemicals: 1) Cl 2 71 g/mol 2) KOH 56.1 g/mol 3) BeCl 2 80 g/mol 4) FeCl 3 162.3 g/mol 5) BF 3 67.8 g/mol 6) CCl 2 F 2 121 g/mol
More informationMoles. Moles. Moles. Moles. Balancing Eqns. Balancing. Balancing Eqns. Symbols Yields or Produces. Like a recipe:
Like a recipe: Balancing Eqns Reactants Products 2H 2 (g) + O 2 (g) 2H 2 O(l) coefficients subscripts Balancing Eqns Balancing Symbols (s) (l) (aq) (g) or Yields or Produces solid liquid (pure liquid)
More informationMonatomic Ions. A. Monatomic Ions In order to determine the charge of monatomic ions, you can use the periodic table as a guide:
Monatomic Ions Ions are atoms that have either lost or gained electrons. While atoms are neutral, ions are charged particles. A loss of electrons results in a positive ion or cation (pronounced cat-eye-on
More informationChapter 3 Mass Relationships in Chemical Reactions
Chapter 3 Mass Relationships in Chemical Reactions Student: 1. An atom of bromine has a mass about four times greater than that of an atom of neon. Which choice makes the correct comparison of the relative
More informationAnswers and Solutions to Text Problems
Chapter 7 Answers and Solutions 7 Answers and Solutions to Text Problems 7.1 A mole is the amount of a substance that contains 6.02 x 10 23 items. For example, one mole of water contains 6.02 10 23 molecules
More informationb. N 2 H 4 c. aluminum oxalate d. acetic acid e. arsenic PART 2: MOLAR MASS 2. Determine the molar mass for each of the following. a. ZnI 2 b.
CHEMISTRY DISCOVER UNIT 5 LOTS OF PRACTICE ON USING THE MOLE!!! PART 1: ATOMIC MASS, FORMULA MASS, OR MOLECULAR MASS 1. Determine the atomic mass, formula mass, or molecular mass for each of the following
More informationUnit 10A Stoichiometry Notes
Unit 10A Stoichiometry Notes Stoichiometry is a big word for a process that chemist s use to calculate amounts in reactions. It makes use of the coefficient ratio set up by balanced reaction equations
More informationSCH 4C1 Unit 2 Problem Set Questions taken from Frank Mustoe et all, "Chemistry 11", McGraw-Hill Ryerson, 2001
SCH 4C1 Unit 2 Problem Set Questions taken from Frank Mustoe et all, "Chemistry 11", McGraw-Hill Ryerson, 2001 1. A small pin contains 0.0178 mol of iron. How many atoms of iron are in the pin? 2. A sample
More informationStudy Guide For Chapter 7
Name: Class: Date: ID: A Study Guide For Chapter 7 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. The number of atoms in a mole of any pure substance
More information4. Balanced chemical equations tell us in what molar ratios substances combine to form products, not in what mass proportions they combine.
CHAPTER 9 1. The coefficients of the balanced chemical equation for a reaction give the relative numbers of molecules of reactants and products that are involved in the reaction.. The coefficients of the
More informationStoichiometry Review
Stoichiometry Review There are 20 problems in this review set. Answers, including problem set-up, can be found in the second half of this document. 1. N 2 (g) + 3H 2 (g) --------> 2NH 3 (g) a. nitrogen
More informationNET IONIC EQUATIONS. A balanced chemical equation can describe all chemical reactions, an example of such an equation is:
NET IONIC EQUATIONS A balanced chemical equation can describe all chemical reactions, an example of such an equation is: NaCl + AgNO 3 AgCl + NaNO 3 In this case, the simple formulas of the various reactants
More information1. When the following equation is balanced, the coefficient of Al is. Al (s) + H 2 O (l)? Al(OH) 3 (s) + H 2 (g)
1. When the following equation is balanced, the coefficient of Al is. Al (s) + H 2 O (l)? Al(OH) (s) + H 2 (g) A) 1 B) 2 C) 4 D) 5 E) Al (s) + H 2 O (l)? Al(OH) (s) + H 2 (g) Al (s) + H 2 O (l)? Al(OH)
More informationDavid A. Katz Chemist, Educator, Science Communicator, and Consultant Department of Chemistry, Pima Community College
WRITING CHEMICAL EQUATIONS 2004, 2002, 1989 by David A. Katz. All rights reserved. Permission for classroom used provided original copyright is included. David A. Katz Chemist, Educator, Science Communicator,
More informationChem 115 POGIL Worksheet - Week 4 Moles & Stoichiometry Answers
Key Questions & Exercises Chem 115 POGIL Worksheet - Week 4 Moles & Stoichiometry Answers 1. The atomic weight of carbon is 12.0107 u, so a mole of carbon has a mass of 12.0107 g. Why doesn t a mole of
More informationReactions in Aqueous Solution
CHAPTER 7 1. Water is the most universal of all liquids. Water has a relatively large heat capacity and a relatively large liquid range, which means it can absorb the heat liberated by many reactions while
More informationSteps for balancing a chemical equation
The Chemical Equation: A Chemical Recipe Dr. Gergens - SD Mesa College A. Learn the meaning of these arrows. B. The chemical equation is the shorthand notation for a chemical reaction. A chemical equation
More informationThe Mole. Chapter 2. Solutions for Practice Problems
Chapter 2 The Mole Note to teacher: You will notice that there are two different formats for the Sample Problems in the student textbook. Where appropriate, the Sample Problem contains the full set of
More informationneutrons are present?
AP Chem Summer Assignment Worksheet #1 Atomic Structure 1. a) For the ion 39 K +, state how many electrons, how many protons, and how many 19 neutrons are present? b) Which of these particles has the smallest
More informationMoles. Balanced chemical equations Molar ratios Mass Composition Empirical and Molecular Mass Predicting Quantities Equations
Moles Balanced chemical equations Molar ratios Mass Composition Empirical and Molecular Mass Predicting Quantities Equations Micro World atoms & molecules Macro World grams Atomic mass is the mass of an
More informationChem 115 POGIL Worksheet - Week 4 Moles & Stoichiometry
Chem 115 POGIL Worksheet - Week 4 Moles & Stoichiometry Why? Chemists are concerned with mass relationships in chemical reactions, usually run on a macroscopic scale (grams, kilograms, etc.). To deal with
More informationDecomposition. Composition
Decomposition 1. Solid ammonium carbonate is heated. 2. Solid calcium carbonate is heated. 3. Solid calcium sulfite is heated in a vacuum. Composition 1. Barium oxide is added to distilled water. 2. Phosphorus
More informationName: Class: Date: 2 4 (aq)
Name: Class: Date: Unit 4 Practice Test Multiple Choice Identify the choice that best completes the statement or answers the question. 1) The balanced molecular equation for complete neutralization of
More informationUNIT (4) CALCULATIONS AND CHEMICAL REACTIONS
UNIT (4) CALCULATIONS AND CHEMICAL REACTIONS 4.1 Formula Masses Recall that the decimal number written under the symbol of the element in the periodic table is the atomic mass of the element. 1 7 8 12
More informationChapter 1: Moles and equations. Learning outcomes. you should be able to:
Chapter 1: Moles and equations 1 Learning outcomes you should be able to: define and use the terms: relative atomic mass, isotopic mass and formula mass based on the 12 C scale perform calculations, including
More informationNomenclature and Formulas of Ionic Compounds. Section I: Writing the Name from the Formula
Purpose: Theory: Nomenclature and Formulas of Ionic Compounds 1. To become familiar with the rules of chemical nomenclature, based on the classification of compounds. 2. To write the proper name of the
More informationName: Block: Date: Test Review: Chapter 8 Ionic Bonding
Name: Block: Date: Test Review: Chapter 8 Ionic Bonding Part 1: Fill-in-the-blank. Choose the word from the word bank below. Each word may be used only 1 time. electron dot structure metallic electronegativity
More informationSample Exercise 2.1 Illustrating the Size of an Atom
Sample Exercise 2.1 Illustrating the Size of an Atom The diameter of a US penny is 19 mm. The diameter of a silver atom, by comparison, is only 2.88 Å. How many silver atoms could be arranged side by side
More informationCalculation of Molar Masses. Molar Mass. Solutions. Solutions
Molar Mass Molar mass = Mass in grams of one mole of any element, numerically equal to its atomic weight Molar mass of molecules can be determined from the chemical formula and molar masses of elements
More informationFormulae, stoichiometry and the mole concept
3 Formulae, stoichiometry and the mole concept Content 3.1 Symbols, Formulae and Chemical equations 3.2 Concept of Relative Mass 3.3 Mole Concept and Stoichiometry Learning Outcomes Candidates should be
More informationNomenclature of Ionic Compounds
Nomenclature of Ionic Compounds Ionic compounds are composed of ions. An ion is an atom or molecule with an electrical charge. Monatomic ions are formed from single atoms that have gained or lost electrons.
More informationTutorial 2 FORMULAS, PERCENTAGE COMPOSITION, AND THE MOLE
T-6 Tutorial 2 FORMULAS, PERCENTAGE COMPOSITION, AND THE MOLE FORMULAS: A chemical formula shows the elemental composition of a substance: the chemical symbols show what elements are present and the numerical
More informationCHAPTER 3 Calculations with Chemical Formulas and Equations. atoms in a FORMULA UNIT
CHAPTER 3 Calculations with Chemical Formulas and Equations MOLECULAR WEIGHT (M. W.) Sum of the Atomic Weights of all atoms in a MOLECULE of a substance. FORMULA WEIGHT (F. W.) Sum of the atomic Weights
More informationAtomic Structure. Name Mass Charge Location Protons 1 +1 Nucleus Neutrons 1 0 Nucleus Electrons 1/1837-1 Orbit nucleus in outer shells
Atomic Structure called nucleons Name Mass Charge Location Protons 1 +1 Nucleus Neutrons 1 0 Nucleus Electrons 1/1837-1 Orbit nucleus in outer shells The number of protons equals the atomic number This
More informationBalancing Chemical Equations Worksheet
Balancing Chemical Equations Worksheet Student Instructions 1. Identify the reactants and products and write a word equation. 2. Write the correct chemical formula for each of the reactants and the products.
More information1. How many hydrogen atoms are in 1.00 g of hydrogen?
MOLES AND CALCULATIONS USING THE MOLE CONCEPT INTRODUCTORY TERMS A. What is an amu? 1.66 x 10-24 g B. We need a conversion to the macroscopic world. 1. How many hydrogen atoms are in 1.00 g of hydrogen?
More informationW1 WORKSHOP ON STOICHIOMETRY
INTRODUCTION W1 WORKSHOP ON STOICHIOMETRY These notes and exercises are designed to introduce you to the basic concepts required to understand a chemical formula or equation. Relative atomic masses of
More informationSTOICHIOMETRY UNIT 1 LEARNING OUTCOMES. At the end of this unit students will be expected to:
STOICHIOMETRY LEARNING OUTCOMES At the end of this unit students will be expected to: UNIT 1 THE MOLE AND MOLAR MASS define molar mass and perform mole-mass inter-conversions for pure substances explain
More informationPART I: MULTIPLE CHOICE (30 multiple choice questions. Each multiple choice question is worth 2 points)
CHEMISTRY 123-07 Midterm #1 Answer key October 14, 2010 Statistics: Average: 74 p (74%); Highest: 97 p (95%); Lowest: 33 p (33%) Number of students performing at or above average: 67 (57%) Number of students
More informationPeriodic Table, Valency and Formula
Periodic Table, Valency and Formula Origins of the Periodic Table Mendelѐѐv in 1869 proposed that a relationship existed between the chemical properties of elements and their atomic masses. He noticed
More informationChemical Calculations: Formula Masses, Moles, and Chemical Equations
Chemical Calculations: Formula Masses, Moles, and Chemical Equations Atomic Mass & Formula Mass Recall from Chapter Three that the average mass of an atom of a given element can be found on the periodic
More informationChapter 3: Stoichiometry
Chapter 3: Stoichiometry Key Skills: Balance chemical equations Predict the products of simple combination, decomposition, and combustion reactions. Calculate formula weights Convert grams to moles and
More informationH 2 + O 2 H 2 O. - Note there is not enough hydrogen to react with oxygen - It is necessary to balance equation.
CEMICAL REACTIONS 1 ydrogen + Oxygen Water 2 + O 2 2 O reactants product(s) reactant substance before chemical change product substance after chemical change Conservation of Mass During a chemical reaction,
More informationAPPENDIX B: EXERCISES
BUILDING CHEMISTRY LABORATORY SESSIONS APPENDIX B: EXERCISES Molecular mass, the mole, and mass percent Relative atomic and molecular mass Relative atomic mass (A r ) is a constant that expresses the ratio
More informationOXIDATION REDUCTION. Section I. Cl 2 + 2e. 2. The oxidation number of group II A is always (+) 2.
OXIDATION REDUCTION Section I Example 1: Na Example 2: 2C1 Example 3: K + + e Na + + e Cl 2 + 2e K Example 4: C1 2 + 2e 2Cl 1. The oxidation number of group I A is always (+) 1. 2. The oxidation number
More informationChapter 6 Notes Science 10 Name:
6.1 Types of Chemical Reactions a) Synthesis (A + B AB) Synthesis reactions are also known as reactions. When this occurs two or more reactants (usually elements) join to form a. A + B AB, where A and
More informationStoichiometry. 1. The total number of moles represented by 20 grams of calcium carbonate is (1) 1; (2) 2; (3) 0.1; (4) 0.2.
Stoichiometry 1 The total number of moles represented by 20 grams of calcium carbonate is (1) 1; (2) 2; (3) 01; (4) 02 2 A 44 gram sample of a hydrate was heated until the water of hydration was driven
More informationMOLECULAR MASS AND FORMULA MASS
1 MOLECULAR MASS AND FORMULA MASS Molecular mass = sum of the atomic weights of all atoms in the molecule. Formula mass = sum of the atomic weights of all atoms in the formula unit. 2 MOLECULAR MASS AND
More informationChemical Calculations: The Mole Concept and Chemical Formulas. AW Atomic weight (mass of the atom of an element) was determined by relative weights.
1 Introduction to Chemistry Atomic Weights (Definitions) Chemical Calculations: The Mole Concept and Chemical Formulas AW Atomic weight (mass of the atom of an element) was determined by relative weights.
More informationNAMING QUIZ 3 - Part A Name: 1. Zinc (II) Nitrate. 5. Silver (I) carbonate. 6. Aluminum acetate. 8. Iron (III) hydroxide
NAMING QUIZ 3 - Part A Name: Write the formulas for the following compounds: 1. Zinc (II) Nitrate 2. Manganese (IV) sulfide 3. Barium permanganate 4. Sulfuric acid 5. Silver (I) carbonate 6. Aluminum acetate
More informationSolution. Practice Exercise. Concept Exercise
Example Exercise 9.1 Atomic Mass and Avogadro s Number Refer to the atomic masses in the periodic table inside the front cover of this textbook. State the mass of Avogadro s number of atoms for each of
More informationChapter 8: Chemical Equations and Reactions
Chapter 8: Chemical Equations and Reactions I. Describing Chemical Reactions A. A chemical reaction is the process by which one or more substances are changed into one or more different substances. A chemical
More informationChapter 4 Compounds and Their Bonds
Chapter 4 Compounds and Their Bonds 4.1 Octet Rule and Ions Octet Rule An octet is 8 valence electrons. is associated with the stability of the noble gases. He is stable with 2 valence electrons (duet).
More informationCh. 6 Chemical Composition and Stoichiometry
Ch. 6 Chemical Composition and Stoichiometry The Mole Concept [6.2, 6.3] Conversions between g mol atoms [6.3, 6.4, 6.5] Mass Percent [6.6, 6.7] Empirical and Molecular Formula [6.8, 6.9] Bring your calculators!
More informationMolarity of Ions in Solution
APPENDIX A Molarity of Ions in Solution ften it is necessary to calculate not only the concentration (in molarity) of a compound in aqueous solution but also the concentration of each ion in aqueous solution.
More informationCHAPTER 5: MOLECULES AND COMPOUNDS
CHAPTER 5: MOLECULES AND COMPOUNDS Problems: 1-6, 9-13, 16, 20, 31-40, 43-64, 65 (a,b,c,e), 66(a-d,f), 69(a-d,f), 70(a-e), 71-78, 81-82, 87-96 A compound will display the same properties (e.g. melting
More informationTOPIC 7. CHEMICAL CALCULATIONS I - atomic and formula weights.
TOPIC 7. CHEMICAL CALCULATIONS I - atomic and formula weights. Atomic structure revisited. In Topic 2, atoms were described as ranging from the simplest atom, H, containing a single proton and usually
More informationHOMEWORK 4A. Definitions. Oxidation-Reduction Reactions. Questions
HOMEWORK 4A Oxidation-Reduction Reactions 1. Indicate whether a reaction will occur or not in each of following. Wtiring a balcnced equation is not necessary. (a) Magnesium metal is added to hydrochloric
More informationChapter 1 The Atomic Nature of Matter
Chapter 1 The Atomic Nature of Matter 6. Substances that cannot be decomposed into two or more simpler substances by chemical means are called a. pure substances. b. compounds. c. molecules. d. elements.
More informationChapter 8 - Chemical Equations and Reactions
Chapter 8 - Chemical Equations and Reactions 8-1 Describing Chemical Reactions I. Introduction A. Reactants 1. Original substances entering into a chemical rxn B. Products 1. The resulting substances from
More informationProblem Solving. Mole Concept
Skills Worksheet Problem Solving Mole Concept Suppose you want to carry out a reaction that requires combining one atom of iron with one atom of sulfur. How much iron should you use? How much sulfur? When
More information602X10 21 602,000,000,000, 000,000,000,000 6.02X10 23. Pre- AP Chemistry Chemical Quan44es: The Mole. Diatomic Elements
Pre- AP Chemistry Chemical Quan44es: The Mole Mole SI unit of measurement that measures the amount of substance. A substance exists as representa9ve par9cles. Representa9ve par9cles can be atoms, molecules,
More informationCh. 10 The Mole I. Molar Conversions
Ch. 10 The Mole I. Molar Conversions I II III IV A. What is the Mole? A counting number (like a dozen) Avogadro s number (N A ) 1 mole = 6.022 10 23 representative particles B. Mole/Particle Conversions
More information1332 CHAPTER 18 Sample Questions
1332 CHAPTER 18 Sample Questions Couple E 0 Couple E 0 Br 2 (l) + 2e 2Br (aq) +1.06 V AuCl 4 + 3e Au + 4Cl +1.00 V Ag + + e Ag +0.80 V Hg 2+ 2 + 2e 2 Hg +0.79 V Fe 3+ (aq) + e Fe 2+ (aq) +0.77 V Cu 2+
More informationExercise 3.5 - Naming Binary Covalent Compounds:
Chapter Exercise Key 1 Chapter Exercise Key Exercise.1 Classifying Compounds: Classify each of the following substances as either a molecular compound or an ionic compound. a. formaldehyde, CH 2 O (used
More informationChapter 11. Electrochemistry Oxidation and Reduction Reactions. Oxidation-Reduction Reactions. Oxidation-Reduction Reactions
Oxidation-Reduction Reactions Chapter 11 Electrochemistry Oxidation and Reduction Reactions An oxidation and reduction reaction occurs in both aqueous solutions and in reactions where substances are burned
More informationChemical Equations. Chemical Equations. Chemical reactions describe processes involving chemical change
Chemical Reactions Chemical Equations Chemical reactions describe processes involving chemical change The chemical change involves rearranging matter Converting one or more pure substances into new pure
More informationRedox and Electrochemistry
Name: Thursday, May 08, 2008 Redox and Electrochemistry 1. A diagram of a chemical cell and an equation are shown below. When the switch is closed, electrons will flow from 1. the Pb(s) to the Cu(s) 2+
More informationThe Mole. 6.022 x 10 23
The Mole 6.022 x 10 23 Background: atomic masses Look at the atomic masses on the periodic table. What do these represent? E.g. the atomic mass of Carbon is 12.01 (atomic # is 6) We know there are 6 protons
More information6 Reactions in Aqueous Solutions
6 Reactions in Aqueous Solutions Water is by far the most common medium in which chemical reactions occur naturally. It is not hard to see this: 70% of our body mass is water and about 70% of the surface
More informationElement of same atomic number, but different atomic mass o Example: Hydrogen
Atomic mass: p + = protons; e - = electrons; n 0 = neutrons p + + n 0 = atomic mass o For carbon-12, 6p + + 6n 0 = atomic mass of 12.0 o For chlorine-35, 17p + + 18n 0 = atomic mass of 35.0 atomic mass
More informationB) atomic number C) both the solid and the liquid phase D) Au C) Sn, Si, C A) metal C) O, S, Se C) In D) tin D) methane D) bismuth B) Group 2 metal
1. The elements on the Periodic Table are arranged in order of increasing A) atomic mass B) atomic number C) molar mass D) oxidation number 2. Which list of elements consists of a metal, a metalloid, and
More informationEXPERIMENT 8: Activity Series (Single Displacement Reactions)
EPERIMENT 8: Activity Series (Single Displacement Reactions) PURPOSE a) Reactions of metals with acids and salt solutions b) Determine the activity of metals c) Write a balanced molecular equation, complete
More informationAll answers must use the correct number of significant figures, and must show units!
CHEM 10113, Quiz 2 September 7, 2011 Name (please print) All answers must use the correct number of significant figures, and must show units! IA Periodic Table of the Elements VIIIA (1) (18) 1 2 1 H IIA
More informationChemical Composition Review Mole Calculations Percent Composition. Copyright Cengage Learning. All rights reserved. 8 1
Chemical Composition Review Mole Calculations Percent Composition Copyright Cengage Learning. All rights reserved. 8 1 QUESTION Suppose you work in a hardware store and a customer wants to purchase 500
More informationNaming Compounds Handout Key
Naming Compounds Handout Key p. 2 Name each of the following monatomic cations: Li + = lithium ion Ag + = silver ion Cd +2 = cadmium ion Cu +2 = copper (II) ion Al +3 = aluminum ion Mg +2 = magnesium ion
More informationChemical Proportions in Compounds
Chapter 6 Chemical Proportions in Compounds Solutions for Practice Problems Student Textbook page 201 1. Problem A sample of a compound is analyzed and found to contain 0.90 g of calcium and 1.60 g of
More informationUnit 9 Compounds Molecules
Unit 9 Compounds Molecules INTRODUCTION Compounds are the results of combinations of elements. These new substances have unique properties compared to the elements that make them up. Compounds are by far
More informationThe Mole Notes. There are many ways to or measure things. In Chemistry we also have special ways to count and measure things, one of which is the.
The Mole Notes I. Introduction There are many ways to or measure things. In Chemistry we also have special ways to count and measure things, one of which is the. A. The Mole (mol) Recall that atoms of
More informationChem 1100 Chapter Three Study Guide Answers Outline I. Molar Mass and Moles A. Calculations of Molar Masses
Chem 1100 Chapter Three Study Guide Answers Outline I. Molar Mass and Moles A. Calculations of Molar Masses B. Calculations of moles C. Calculations of number of atoms from moles/molar masses 1. Avagadro
More informationChapter 3! Stoichiometry: Calculations with Chemical Formulas and Equations. Stoichiometry
Chapter 3! : Calculations with Chemical Formulas and Equations Anatomy of a Chemical Equation CH 4 (g) + 2O 2 (g) CO 2 (g) + 2 H 2 O (g) Anatomy of a Chemical Equation CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2
More information7) How many electrons are in the second energy level for an atom of N? A) 5 B) 6 C) 4 D) 8
HOMEWORK CHEM 107 Chapter 3 Compounds Putting Particles Together 3.1 Multiple-Choice 1) How many electrons are in the highest energy level of sulfur? A) 2 B) 4 C) 6 D) 8 2) An atom of phosphorous has how
More informationTopic 4 National Chemistry Summary Notes. Formulae, Equations, Balancing Equations and The Mole
Topic 4 National Chemistry Summary Notes Formulae, Equations, Balancing Equations and The Mole LI 1 The chemical formula of a covalent molecular compound tells us the number of atoms of each element present
More informationChemistry 65 Chapter 6 THE MOLE CONCEPT
THE MOLE CONCEPT Chemists find it more convenient to use mass relationships in the laboratory, while chemical reactions depend on the number of atoms present. In order to relate the mass and number of
More informationName period Unit 3 worksheet
Name period Unit 3 worksheet Read chapter 8, 2.52.7 1. Explain the difference between metallic, ionic, and covalent bonding Metallic cations share a sea of electrons Ionic atoms give and take electrons
More informationUnit 9 Stoichiometry Notes (The Mole Continues)
Unit 9 Stoichiometry Notes (The Mole Continues) is a big word for a process that chemist s use to calculate amounts in reactions. It makes use of the coefficient ratio set up by balanced reaction equations
More informationPERIODIC TABLE OF THE ELEMENTS
PERIODIC TABLE OF THE ELEMENTS Periodic Table: an arrangement of elements in horizontal rows (Periods) and vertical columns (Groups) exhibits periodic repetition of properties First Periodic Table: discovered
More informationNomenclature Packet. 1. Name the following ionic compounds: a. Al 2 O 3 Aluminum oxide. b. Cs 2 O Cesium oxide. c. Rb 3 N Rubidium nitride
Nomenclature Packet Worksheet I: Binary Ionic Compounds (representative metals) metals from groups 1A, 2A, and 3A (1, 2, and 13) have constant charges as ions and do NOT get Roman Numerals in their names
More informationSolution a homogeneous mixture = A solvent + solute(s) Aqueous solution water is the solvent
Solution a homogeneous mixture = A solvent + solute(s) Aqueous solution water is the solvent Water a polar solvent: dissolves most ionic compounds as well as many molecular compounds Aqueous solution:
More informationUnit 6 The Mole Concept
Chemistry Form 3 Page 62 Ms. R. Buttigieg Unit 6 The Mole Concept See Chemistry for You Chapter 28 pg. 352-363 See GCSE Chemistry Chapter 5 pg. 70-79 6.1 Relative atomic mass. The relative atomic mass
More informationDescription of the Mole Concept:
Description of the Mole Concept: Suppose you were sent into the store to buy 36 eggs. When you picked them up you would get 3 boxes, each containing 12 eggs. You just used a mathematical device, called
More informationChapter 5. Chemical Reactions and Equations. Introduction. Chapter 5 Topics. 5.1 What is a Chemical Reaction
Introduction Chapter 5 Chemical Reactions and Equations Chemical reactions occur all around us. How do we make sense of these changes? What patterns can we find? 1 2 Copyright The McGraw-Hill Companies,
More informationWriting and Balancing Chemical Equations
Name Writing and Balancing Chemical Equations Period When a substance undergoes a chemical reaction, chemical bonds are broken and new bonds are formed. This results in one or more new substances, often
More informationCHEMICAL REACTIONS. Chemistry 51 Chapter 6
CHEMICAL REACTIONS A chemical reaction is a rearrangement of atoms in which some of the original bonds are broken and new bonds are formed to give different chemical structures. In a chemical reaction,
More informationChapter 4. Chemical Composition. Chapter 4 Topics H 2 S. 4.1 Mole Quantities. The Mole Scale. Molar Mass The Mass of 1 Mole
Chapter 4 Chemical Composition Chapter 4 Topics 1. Mole Quantities 2. Moles, Masses, and Particles 3. Determining Empirical Formulas 4. Chemical Composition of Solutions Copyright The McGraw-Hill Companies,
More informationStoichiometry. What is the atomic mass for carbon? For zinc?
Stoichiometry Atomic Mass (atomic weight) Atoms are so small, it is difficult to discuss how much they weigh in grams We use atomic mass units an atomic mass unit (AMU) is one twelfth the mass of the catbon-12
More informationCalculating Atoms, Ions, or Molecules Using Moles
TEKS REVIEW 8B Calculating Atoms, Ions, or Molecules Using Moles TEKS 8B READINESS Use the mole concept to calculate the number of atoms, ions, or molecules in a sample TEKS_TXT of material. Vocabulary
More information10 The Mole. Section 10.1 Measuring Matter
Name Date Class The Mole Section.1 Measuring Matter In your textbook, read about counting particles. In Column B, rank the quantities from Column A from smallest to largest. Column A Column B 0.5 mol 1.
More informationChapter 8 How to Do Chemical Calculations
Chapter 8 How to Do Chemical Calculations Chemistry is both a qualitative and a quantitative science. In the laboratory, it is important to be able to measure quantities of chemical substances and, as
More information