Putting it together: chapter 7 (p.199, q. 1-22)
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13.
4.0 g of oxygen contain: (a) 1.5 x 1023 atoms O;
(b) 4.0 molar masses of O;
(c) 0.50 mol O
(d) 6.022 x 1023 atoms O.
One mole of H atoms contains:
(a) 2.0 g H
(b) 6.022 x 1023 atoms H
(c) 1 atom H (d) 12 g of 12C.
The mass of one atom of Mg is (a)24.31g (b)54.94g (c)12.00g (d) 4.037 x 1023 g.
Avogadro’s number of magnesium atoms: (a) has a mass of 1.0 g (b) has the
same mass as Avogadro’s number of sulfur atoms
(c) has a mass of 12.0 g
(d) is 1 mol of Mg atoms
Which of the following contains the largest number of moles?
(a) 1.0 g Li
(b) 1.0 g Na
(c) 1.0 g Al
(d) 1.0 g Ag
(the lightest element will have the largest number of moles in one gram)
The number of moles in 112 g of aspirin, C9H8O4, is:
(a) 1.61
(b) 0.622
(c) 112
(d) 0.161
molar mass = 9 x 12 + 8 x 1 + 4 x 16 = 180
1 mol aspirin
#moles = 112 g aspirin x 180 g aspirin = 0.622
How many moles of aluminum hydroxide are in one antacid tablet containing 400
mg of Al(OH)3? (a) 5.13 x 10-3
(b) 0.400
(c) 5.13
(d) 9.09 x 10-3
400 mg = 0.400 g.
molar mass of Al(OH)3 = 26.98 + 3 x (16.00 + 1.008) = 78.004 g
1 mol Al(OH)3
#moles = 0.400 g Al(OH)3 x 78.004 g Al(OH) = 5.13 x 10-3 mol
3
How many grams of Au2S can be obtained from 1.17 mol of Au?
(a) 182 g
(b) 249 g
(c) 364 g
(d) 499 g
The product, Au2S must contain the same number of moles of Au atoms as in the
reactant. However, one mol of Au2S requires two moles of Au.
Molar mass Au2S = 2 x 197.0 + 32.07 = 426.1
426.1 g Au2S 1 mol Au2S
#g Au2S = 1.17 mol Au x 1mol Au S x 2 mol Au = 249 g
2
The molar mass of Ba(NO3)2 is (a) 199.3 (b) 261.3 (c) 247.3
(d) 167.3
molar mass = 137.3 + 2 x (14.01 + 3 x 16.00) = 261.3
A 16-g sample of oxygen:
(a) is 1 mol of O2
(b) contains 6.022 x 1023 molecules of O2
(c) is 0.50 molecule of O2
(d) is 0.50 molar mass of O2.
What is the percent composition for a compound formed from 8.15 g of zinc and
2.00 g of oxygen?
(a) 80.3% Zn, 19.7% O
(b) 80.3% O, 19.7% O
(c) 70.3% Zn, 29.7% O
(d) 65.3% Zn, 34.7% O
total mass = 10.15 g, %Zn = 8.15 g / 10.15 g x 100 = 80.3 %
Which of these compounds contains the largest percentage of oxygen?
(a) SO2
(b) SO3
(c) N2O3
(d) N2O5
molar mass SO3 = 32.07 + 3 x 16.00 = 80.07 ; %O=48 / 80.07=60.0 %
molar mass of N2O5 = 14.01 x 2 + 5 x 16.00 = 108.02,
%O=74 %
2.00 mol CO2: (a) have a mass of 56.0 g (b) contain 1.20 x 1024 molecules
(c) have a mass of 44.0 g
(d) contain 6.00 mol. masses of CO2.
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In Ag2CO3, the percent by mass of:
(a) carbon is 43.5% (b) silver is 64.2% (c) oxygen is 17.4 (d) oxygen is 21.9%.
molar mass = 2 x 107.9 + 12.01 + 3 x 16.00 = 275.8
%Ag = 2 x 107.9 / 275.8 = 78.2%, %C = 4.35%, %O = 17.4%
The empirical formula of the compound containing 31.0% Ti and 69.0% Cl is:
(a) TiCl
(b) TiCl2
(c) TiCl3
(d) TiCl4
1 mol Ti
1 mol Cl
31.0 g Ti x 47.87 g Ti = 0.648
69.0 g Cl x 35.45 g Cl = 1.946
1.946 / 0.648 = 3.00
A compound contains 54.3% C, 5.6% H, 40.1% Cl. The empirical formula is:
(a) CH3Cl
(b) C2H5Cl
(c) C2H4Cl2
(d) C4H5Cl
1 mol C
1 mol H
54.3 g C x 12.01 g C = 4.521
5.6 g H x 1 g H = 5.6
1 mol Cl
40.1 g Cl x 35.45 g Cl = 1.13
4.521/1.13 = 4
5.6/1.13 = 5
1.13/1.13 = 1
A compound contains 40.0% C, 6.7% H, and 53.3% O. The molar mass is 60.0
g/mol. The molecular formula is
(a) C2H3O2
(b) C3H8O
(c) C2HO
(d) C2H4O2
1 mol C
1 mol O
40.0 g C x 12.01 g C = 3.330 53.3 g
O x 16.00 g O = 3.331
H=6.7
6.7 /3.33 = 2
thus, empirical formula is CH2O.
molar mass of CH2O is 30.03,
n = 60.0 / 30.03 = 2.
Thus molecular formula is (CH2O)2 or C2H4O2.
How many chlorine atoms are in 4.0 mol of PCl3?
(a) 3 (b) 7.2 x 1024
(c) 12
(d) 2.4 x 1024
1 mol PCl3 has 3 mol of Cl atoms; 4 moles of PCl3 12 x 6.022 x 1023 atoms.
What is the mass of 4.53 mol of Na2SO4?
(a) 142.1 g
(b) 644 g
(c) 31.4 g
(d) 3.19 x 10-2 g
molar mass Na2SO4 = 22.99 x 2 + 32.07 + 16.00 x 4 = 142.1
142.1 g Na2SO4
4.53 mol Na2SO4 x 1 mol Na SO = 643.5 g
2
4
The percent composition of Mg3N2 is:
(a) 72.2% Mg, 27.8% N
(b) 63.4% Mg, 36.6% N
(c) 83.9% Mg, 16.1% N
(d) no correct answer given
molar mass = 24.31 x 3 + 14.01 x 2 = 100.95
%Mg = 24.31 x 3 / 100.95 = 72.2
%N = 14.01 x 2 / 100.95 = 27.8
How many grams of oxygen are contained in 0.500 mol of Na2SO4?
(a) 16.0 g
(b) 32.0 g
(c) 64.0 g
(d) no correct answer given
4 moles of O atoms in 1 mol Na2SO4; in 0.5 mol there is 2 moles of O, or 32 g.
The empirical formula of a compound is CH. If the molar mass of this compound
is 78.11, then the molecular formula is: (a) C2H2
(b) C5H18
(c) C6H6
Molecular formula of C2H2 is~ 26, of C6H6 is ~78. Molecular mass of empirical
formula is 13. Number of empirical units is n = 78 / 13 = 6. Thus, it is C6H6.
Putting it together: Chapters 9 (p.199, q.36-50)
36.
How many moles is 20.0 g Na2CO3? (a) 1.89, (b) 2.12 x 103, (c) 212,
(d) 0.189 mol
1 mol Na2CO3
Molar mass Na2CO3 = 105.99;
20.0 g * (105.9 g Na CO ) = 0.189 mol
2
3
37.
What is the mass of 0.30 mol BaSO4? (a) 7 x 103g, (b) 70. g, (c) 0.13 g, (d) 700.2
g
233.37 g BaSO4
Mol.mass BaSO4 = 233.37;
0.30 mol BaSO4 * 1 mol BaSO = 70. g
4
38.
How many molecules are in 5.8 g of acetone, C3H6O? (a) 0.10 molecule,
(b) 6.0 x 1022 molecules, (c) 3.5 x 1024 molecules, (d) 6.0 x 1023 molecules
Molar mass C3H6O = 58.01
1 mol C3H6O
1 mol C3H6O
5.8 g C3H6O * 58.01 g C H O * 6.022 x 1023 molecules = 6.0 x 1022 molecules
3 6
Problems 39-45 refer to the reaction: 2 C2H4 + 6 O2  4 CO2 + 4 H2O
39.
If 6.0 mol CO2 are produced, how many moles of O2 were reacted?
6 mol O2
6.0 mol CO2 * 4 mol CO = 9.0 mol O2
2
40.
How many moles of O2 are required for the complete reaction of 45 g C2H4?
Molar mass C2H4 = 28.052 g
1 mol C2H4
6 mol O2
45 g C2H4 * 28.05 g C H * 2 mol C H * = 4.8 mol O2
2 4
2 4
41.
If 18.0 g CO2 are produced, how many grams of H2O are produced?
Molar masses: CO2 = 44.01 g, H2O = 18.02 g
1 mol CO2 4 mol H2O 18.02 g H2O
18.0 g CO2 * 44.01 g CO * 4 mol CO * 1 mol H O = 7.37 g H2O
2
2
2
42.
How many moles of CO2 can be produced by the reaction of 5.0 mol C2H4 and
12.0 mol O2?
From the balanced equation: 2 mol C2H4 react with 6 mol O2. Thus, complete
burning of 5.0 mol C2H4 would need 15 mol O2. Since there is only 12.0 mol O2,
oxygen is the limiting reactant. Proceed as if only mol O2 are given.
4 mol CO2
12.0 mol O2 * 6 mol O = 8.0 mol CO2
2
43.
How many moles of CO2 can be produced by the reaction of 0.480 mol of C2H4
and 1.08 mol O2?
Solution: O2 is again the limiting reactant (because complete burning of 0.480
mol C2H4 needs 2.40 mol O2). Proceed as if only O2 is given.
4 mol CO2
1.08 mol O2 * 6 mol O = 0.720 mol CO2
2
44.
How many grams of CO2 can be produced from 2.0 g of C2H4 and 5.0 g of O2?
1 mol C2H4
4 mol CO2 44.01 g CO2
2.0 g C2H4 * 28.05 g C H * 2 mol C H * 1 mol CO = 6.3 g CO2
2 4
2 4
2
1 mol O2 4 mol CO2 44.01 g CO2
5.0 g O2 * 32 g O * 6 mol O * 1 mol CO = 4.6 g CO2.
2
2
2
45.
If 14.0 g C2H4 is reacted and the actual yield of H2O is 7.84 g, the percent yield of
the reaction is?
theoretical yield, grams H2O is:
1 mol C2H4
4 mol H2O 18.02 g H2O
14.0 g C2H4 * 28.05 g C H * 2 mol C H * 1 mol H O = 18.0 g H2O
2 4
2 4
2
7.84 g H2O
Percent yield = 18.0 g H O * 100% = 43.6 %
2
Problems 46-48 refer to reaction: H3PO4 + MgCO3 --> Mg3(PO4)2 + CO2 + H2O
Balanced eq.: 2 H3PO4 + 3 MgCO3 --> Mg3(PO4)2 + 3 CO2 + 3 H2O
Molar masses: MgCO3 = 84.32 g, CO2 = 44.01 g
46.
The sequence of coefficients for the balanced eq. is: (a) 2,3,1,3,3; (b) 3,1,3,2,3;
(c) 2,2,1,2,3; (d) 2,3,1,3,2.
47.
If 20.0 g CO2 is produced, the number of moles of MgCO3 used is?
1 mol CO2 3 mol MgCO3
20.0 g CO2 * 44.01 g CO * 3 mol CO = 0.454 mol MgCO3
2
2
48.
If 50.0 g MgCO3 reacts completely with H3PO4, the number of grams of CO2
produced is?
1 mol MgCO3
3 mol CO2
44.01 g CO2
50.0 g MgCO3 * 84.32 g MgCO * 3 mol MgCO * 1 mol CO = 26.1 g CO2
3
3
2
49.
When 10.0 g MgCl2 and 10.0 g Na2CO3 are reacted in eq.
MgCl2 + Na2CO3 --> MgCO3 + 2 NaCl, the limiting reactant is: (a) MgCl2, (b)
Na2CO3, (c) MgCO3, (d) NaCl
Molar masses: MgCl2 = 95.15 g, Na2CO3 = 83.00 g
Reactants are MgCl2 and Na2CO3. To find which one is limiting, calculate number
of moles of a product, say MgCO3, which can be obtained from each reactant. The
reactant that gives smaller # of moles of the product is the limiting reactant.
1 mol MgCl2 1 mol MgCO3
10.0 g MgCl2 * 95.15 g MgCl * 1 mol MgCl = 0.105 mol MgCO3 limit react.
2
2
1 mol Na2CO3
1 mol MgCO3
10.0 g Na2CO3 * 83.00 g Na CO * 1 mol Na CO = 0.120 mol MgCO3
2
3
2
3
50.
When 50.0 g Cu is reacted with AgNO3, 148 g Ag is obtained. What is percent
yield if the reaction is:
Cu + 2 AgNO3 --> Cu(NO3)2 + 2 Ag?
Molar masses: Ag = 107.9, Cu = 63.55 g
1 mol Cu 2 mol Ag 1 mol Ag
Theoretical yield: 50.0 g Cu * 63.55 g Cu * 1 mol Cu * 107.9 g Ag = 170. g Ag
148 g
Percent yield = 170. g * 100% = 87.1 %.