Chapter 3 - Calculations with Chemical Formulas and Equations
1. The molecular formula of a particular solid is C8H6O8. Its molecular mass is
A) 352 amu.
B) 118 amu.
C) 364 amu.
D) 230 amu.
E) 226 amu.
ANS:
OBJ:
TOP:
MSC:
D
PTS: 1
DIF: easy
REF: 3.1
Calculate the formula mass from a formula. (Example 3.1)
stoichiometry | mass and moles of substance
KEY: molecular mass
general chemistry
2. The hydrocarbon heptane has the structural formula CH3(CH2)5CH3. What is the molecular
mass of this hydrocarbon?
A)
B)
C)
D)
E)
100.2 amu.
0.009980 amu.
86.17 amu.
0.01160 amu.
110.2 amu.
ANS: A
PTS: 1
DIF: easy
OBJ: Calculate the formula mass from a formula.
TOP: stoichiometry | mass and moles of substance
REF: 3.1
3. A single molecule of polystryene has the repeating unit -[CH2CH(C6H5)]n-, where n is the
number of repeating units. What is the value of n if the molecular mass of a single polymer
chain is 1.5782  10 6 amu?
A)
B)
C)
D)
E)
1.515  10 4
1.644  10 8
3.031  10 4
2.047  10 4
1.217  10 8
units.
units.
units.
units.
units.
ANS: A
PTS: 1
DIF: moderate
OBJ: Calculate the formula mass from a formula.
TOP: stoichiometry | mass and moles of substance
REF: 3.1
4. The formula mass of aluminum oxalate, Al2(C2O4)3, is
A) 140 amu.
B) 318 amu.
C) 283 amu.
D) 213 amu.
E) 185 amu.
ANS: B
Test Bank
PTS:
1
DIF:
easy
General Chemistry, 10th edition
REF: 3.1
1
OBJ: Calculate the formula mass from a formula. (Example 3.1)
TOP: stoichiometry | mass and moles of substance
KEY: formula mass
MSC: general chemistry
5. The formula mass of zinc acetate trihydrate, Zn(CH3COO)2 • 3H2O, is
A) 321 amu.
B) 184 amu.
C) 268 amu.
D) 238 amu.
E) 114 amu.
ANS:
OBJ:
TOP:
MSC:
D
PTS: 1
DIF: easy
REF: 3.1
Calculate the formula mass from a formula. (Example 3.1)
stoichiometry | mass and moles of substance
KEY: formula mass
general chemistry
6. The fully hydrated form of sodium sulfate is the decahydrate, Na2SO4 • 10H2O. When
heated the hydrated salt loses water. How many water molecules are found per formula unit
in a partially dehydrated sample of sodium sulfate with a formula mass of 160.1 amu (i.e.
find n for Na2SO4 • nH2O)?
A)
B)
C)
D)
E)
1 waters.
5 waters.
3 waters.
4 waters.
7 waters.
ANS: A
PTS: 1
DIF: moderate
OBJ: Calculate the formula mass from a formula.
TOP: stoichiometry | mass and moles of substance
REF: 3.1
7. What is the molecular mass of cycloheptane, C7H14?
A) 13.02 amu
B) 1191.19 amu
C) 85.08 amu
D) 98.19 amu
E) 26.12 amu
ANS:
OBJ:
TOP:
MSC:
D
PTS: 1
DIF: easy
REF: 3.1
Calculate the formula mass from a formula. (Example 3.1)
stoichiometry | mass and moles of substance
KEY: molecular mass
general chemistry
8. What is the formula mass of strontium phosphate, Sr3(PO4)2?
A) 357.83 amu
B) 421.83 amu
C) 182.59 amu
D) 715.66 amu
E) 452.80 amu
ANS: E
Test Bank
PTS:
1
DIF:
easy
General Chemistry, 10th edition
REF: 3.1
2
OBJ: Calculate the formula mass from a formula. (Example 3.1)
TOP: stoichiometry | mass and moles of substance
KEY: formula mass
MSC: general chemistry
9. What is the molecular mass of the hydrocarbon styrene (shown in the figure)?
A)
B)
C)
D)
E)
104.1 amu.
91.1 amu.
103.1 amu.
13.0 amu.
78.1 amu.
ANS:
OBJ:
TOP:
MSC:
A
PTS: 1
DIF: easy
REF: 3.1
Calculate the formula mass from molecular models. (Example 3.2)
stoichiometry | mass and moles of substance
KEY: molecular mass
general chemistry
10. What is the molar mass of ammonium sulfite, (NH4)2SO3?
A) 98 g/mol
B) 116 g/mol
C) 55 g/mol
D) 180 g/mol
E) 84 g/mol
ANS:
OBJ:
TOP:
MSC:
B
PTS: 1
DIF: easy
REF: 3.2
Understand how the molar mass is related to the formula weight of a substance.
stoichiometry | mass and moles of substance
KEY: formula mass
general chemistry
11. What is the molar mass of zinc sulfate heptahydrate, ZnSO4 • 7H2O?
A) 180. g/mol
B) 288 g/mol
C) 384 g/mol
D) 162 g/mol
E) 582 g/mol
ANS:
OBJ:
TOP:
MSC:
B
PTS: 1
DIF: easy
REF: 3.2
Understand how the molar mass is related to the formula weight of a substance.
stoichiometry | mass and moles of substance
KEY: formula mass
general chemistry
Test Bank
General Chemistry, 10th edition
3
12. Plastic wrap can be made from poly(vinylidene chloride). A single poly(vinylidene
chloride) strand has the general formula -(CH2CHCl)n-, where n ranges from 10,000 to
100,000. What is the molar mass of a single poly(vinylidene chloride) molecule containing
6.928  10 4 repeating units?
A)
B)
C)
D)
E)
4.330  10 6 g/mol
2.706  10 6 g/mol
2.310  10 7 g/mol
6.308  10 6 g/mol
1.585  10 7 g/mol
ANS: A
PTS: 1
DIF: easy
REF: 3.2
OBJ: Understand how the molar mass is related to the formula weight of a substance.
TOP: stoichiometry | mass and moles of substance
13. The dicarboxylic acid potassium hydrogen pthalate (shown in the figure) is used to
standardize solutions of strong base. What is the molar mass of this compound?
A)
B)
C)
D)
E)
204.2 g/mol
192.2 g/mol
248.9 g/mol
71.08 g/mol
172.2 g/mol
ANS: A
PTS: 1
DIF: easy
REF: 3.2
OBJ: Understand how the molar mass is related to the formula weight of a substance.
TOP: stoichiometry | mass and moles of substance
14. What is the molar mass of the solid C8H12N2O4?
A) 172 g/mol
B) 136 g/mol
C) 200 g/mol
D) 106 g/mol
E) 188 g/mol
ANS:
OBJ:
TOP:
MSC:
C
PTS: 1
DIF: easy
REF: 3.2
Understand how the molar mass is related to the formula weight of a substance.
stoichiometry | mass and moles of substance
KEY: molecular mass
general chemistry
15. A 1.067 g sample of an element contains 5.062  1021 atoms. What is the element symbol?
A) I
Test Bank
General Chemistry, 10th edition
4
B)
C)
D)
E)
Ag
La
Pd
Rh
ANS: A
PTS: 1
DIF: moderate
REF: 3.2
OBJ: Understand how the molar mass is related to the formula weight of a substance.
TOP: stoichiometry | determining chemical formulas
16. Monodisperse polyacrylonitrile contains molecules with the general formula -(CH2CHCN)n, where n is typically greater than 10,000. Given that a sample of monodisperse
polyacrilonitrile weighs 197.4 g and contains 1.046  10 20 molecules of -(CH2CHCN)n-,
calculate n.
A)
B)
C)
D)
E)
2.141  10 4
6.026  10 7
6.018  10 39
6.022  10 23
1.136  10 6
ANS: A
PTS: 1
DIF: difficult
REF: 3.2
OBJ: Understand how the molar mass is related to the formula weight of a substance.
TOP: stoichiometry | mass and moles of substance
17. Monodisperse polyacrylonitrile contains molecules with the general formula -(CH2CHCN)n, where n is typically greater than 10,000. Given that a sample of monodisperse
polyacrylonitrile weighs 252.6 g and contains 4.443  10 19 molecules of -(CH2CHCN)n-,
what is the molar mass of the polymer?
A)
B)
C)
D)
E)
6.452  10 4 g/mol
1.816  10 8 g/mol
1.997  10 39 g/mol
6.022  10 23 g/mol
3.423  10 6 g/mol
ANS: E
PTS: 1
DIF: difficult
REF: 3.2
OBJ: Understand how the molar mass is related to the formula weight of a substance.
TOP: stoichiometry | mass and moles of substance
18. An atom of an element weighs 3.00  10–23 g. What is the atomic mass of this element in
atomic mass units?
A) 17.4 amu
B) 16.5 amu
C) 15.9 amu
D) 18.1 amu
E) 15.3 amu
ANS: D
PTS: 1
DIF: easy
REF: 3.2
OBJ: Calculate the mass of atoms and molecules. (Example 3.3)
Test Bank
General Chemistry, 10th edition
5
TOP: stoichiometry | mass and moles of substance
KEY: mole | mole calculations
MSC: general chemistry
19. What is the mass in grams of one propene, C3H6, molecule?
A) 6.99  10–23 g
B) 2.53  1025 g
C) 44.0 g
D) 42.0 g
E) 1.99  10–23 g
ANS:
OBJ:
TOP:
KEY:
A
PTS: 1
DIF: easy
REF: 3.2
Calculate the mass of atoms and molecules. (Example 3.3)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
20. What is the mass of oxygen in grams found in one molecule of the compound C7H8O4?
A)
B)
C)
D)
E)
1.06  10 22 g
3.86  10 21 g
1.34  10 23 g
1.40  10 22 g
156 g
ANS: A
PTS: 1
DIF: moderate
REF: 3.2
OBJ: Calculate the mass of atoms and molecules. (Example 3.3)
TOP: stoichiometry | mass and moles of substance
21. Which of the following compounds contains the largest number of atoms?
A) 4.0 mol of K2S
B) 3.0 mol of NH3
C) 2.0 mol of H2SO4
D) 5.0 mol of HCl
E) 1.0 mol of CH3COCl
ANS:
OBJ:
TOP:
KEY:
C
PTS: 1
DIF: easy
REF: 3.2
Perform calculations using the mole.
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
22. How many atoms of carbon are there in 0.51 mol of procaine, C13H20N2O2, a “pain killer”
used by dentists?
A) 6.6  1023
B) 4.3  1024
C) 4.0  1024
D) 6.1  1023
E) 4.6  1024
ANS: C
PTS: 1
DIF: easy
OBJ: Perform calculations using the mole.
TOP: stoichiometry | mass and moles of substance
Test Bank
General Chemistry, 10th edition
REF: 3.2
6
KEY: mole | mole calculations
MSC: general chemistry
23. A sample of ammonium phosphite, (NH4)3PO3, contains 0.909 mol of hydrogen atoms. The
number of moles of oxygen atoms in the sample is
A)
B)
C)
D)
E)
0.227 mol.
0.0909 mol.
0.300 mol.
0.227 mol.
3.00 mol.
ANS: A
PTS: 1
DIF: easy
OBJ: Perform calculations using the mole.
TOP: stoichiometry | mass and moles of substance
REF: 3.2
24. A sample of gallium(III) sulfite, Ga2(SO3)3, contains 1.95 mol of sulfite ions. The number of
moles of gallium(III) ions in the sample is
A)
B)
C)
D)
E)
1.30 mol.
2.92 mol.
1.95 mol.
5.84 mol.
3.90 mol.
ANS: A
PTS: 1
DIF: easy
OBJ: Perform calculations using the mole.
TOP: stoichiometry | mass and moles of substance
REF: 3.2
25. Sorbose, C6H12O6, is used in making vitamin C. A sorbose sample containing 96.0 g of
carbon atoms also contains ____ g of hydrogen atoms.
A) 32.2
B) 16.1
C) 1.15  103
D) 96.0
E) 7.99
ANS:
OBJ:
TOP:
KEY:
B
PTS: 1
DIF: moderate
REF: 3.2
Perform calculations using the mole.
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
26. Consider the following three samples.
A. A sample containing 180 g glucose (C6H12O6)
B. A sample containing 90 g glucose and 90 g fructose (C6H12O6)
C. A sample containing 180 g fructose
Which statement is correct?
A) All three samples have the same number of hydrogen atoms.
B) Both samples A and C have the same number of hydrogen atoms, but more than in
sample B.
C) Sample B has more hydrogen atoms than sample A or sample C.
Test Bank
General Chemistry, 10th edition
7
D) Sample C has more hydrogen atoms than sample A or sample B
E) Sample A has more hydrogen atoms than sample B or sample C.
ANS:
OBJ:
TOP:
KEY:
A
PTS: 1
DIF: difficult
REF: 3.2
Perform calculations using the mole.
stoichiometry | determining chemical formulas
mole | mole calculations
MSC: general chemistry
27. Styrene's empirical formula is CH. What mass of styrene contains 7.89  1021 atoms of
hydrogen? The molar mass of styrene is 104 g/mol.
A) 0.105 g
B) 0.170 g
C) 1.36 g
D) 0.157 g
E) 0.0131 g
ANS:
OBJ:
TOP:
KEY:
B
PTS: 1
DIF: moderate
REF: 3.2
Convert from moles of substance to grams of substance. (Example 3.4)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
28. What is the mass of oxygen atoms in 0.305 mol Fe(CO)5?
A) 17.0 g
B) 59.7 g
C) 24.4 g
D) 4.88 g
E) 18.3 g
ANS:
OBJ:
TOP:
KEY:
C
PTS: 1
DIF: moderate
REF: 3.2
Convert from moles of substance to grams of substance. (Example 3.4)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
29. What is the mass in grams of 0.699 mol of glucose, C6H12O6?
A) 0.00388 g
B) 67.1 g
C) 126 g
D) 21.0 g
E) 258 g
ANS:
OBJ:
TOP:
KEY:
C
PTS: 1
DIF: easy
REF: 3.2
Convert from moles of substance to grams of substance. (Example 3.4)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
30. Which one of the following samples has the greatest mass?
A) 0.39 mol of camphor, C10H16O
B) 4.1 mol of ammonia, NH3
C) 9.2 mol of krypton, Kr
D) 4.6 mol of iodine vapor, I2
E) 1.8 mol of formaldehyde, CH2O
Test Bank
General Chemistry, 10th edition
8
ANS:
OBJ:
TOP:
KEY:
D
PTS: 1
DIF: moderate
REF: 3.2
Convert from moles of substance to grams of substance. (Example 3.4)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
31. Which of the following contains the greatest mass of oxygen atoms?
A) 0.6 mol CoSO4 • 7H2O
B) 2.3 mol KHSO4
C) 1.2 mol K2Cr2O7
D) 2.3 mol H2O2
E) 2.3 mol Na2S2O3
ANS:
OBJ:
TOP:
KEY:
B
PTS: 1
DIF: easy
REF: 3.2
Convert from moles of substance to grams of substance. (Example 3.4)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
32. Calculate the number of moles of bromine present in 14.5 mL of Br2(l), whose density is
3.12 g/mL.
A) 3.53 mol
B) 0.181 mol
C) 0.566 mol
D) 0.091 mol
E) 0.283 mol
ANS:
OBJ:
TOP:
KEY:
E
PTS: 1
DIF: moderate
REF: 3.2
Convert from grams of substance to moles of substance. (Example 3.5)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
33. How many moles of hexachlorobenzene, C6Cl6, are in 4.45 g of C6Cl6?
A) 0.0208 mol
B) 1.27  103 mol
C) 0.0618 mol
D) 0.0156 mol
E) 0.0322 mol
ANS:
OBJ:
TOP:
KEY:
D
PTS: 1
DIF: easy
REF: 3.2
Convert from grams of substance to moles of substance. (Example 3.5)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
34. How many moles of iron atoms are contained in 4.39 g of iron?
A) 245 mol
B) 0.0579 mol
C) 0.0786 mol
D) 0.122 mol
E) 0.169 mol
ANS: C
Test Bank
PTS:
1
DIF:
easy
General Chemistry, 10th edition
REF: 3.2
9
OBJ: Convert from grams of substance to moles of substance. (Example 3.5)
TOP: stoichiometry | mass and moles of substance
KEY: mole | mole calculations
MSC: general chemistry
35. How many moles of pentane, C5H12, are contained in a 11-g sample?
A) 0.18 mol
B) 0.15 mol
C) 0.26 mol
D) 1.4 mol
E) 1.1 mol
ANS:
OBJ:
TOP:
KEY:
B
PTS: 1
DIF: easy
REF: 3.2
Convert from grams of substance to moles of substance. (Example 3.5)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
36. Sodium cyclamate, C6H11NHSO3Na, was used at one time as an artificial sweetener.
C6H11NHSO3Na has a molecular mass of 201.2 g/mol. How many moles of sodium
cyclamate are contained in a 67.6-g sample?
A) 0.211 mol
B) 0.193 mol
C) 0.336 mol
D) 0.307 mol
E) 13,600 mol
ANS:
OBJ:
TOP:
KEY:
C
PTS: 1
DIF: easy
REF: 3.2
Convert from grams of substance to moles of substance. (Example 3.5)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
37. How many moles of silver are contained in 7.00 kg of silver?
A) 64.9 mol
B) 64.9  101 mol
C) 64.9  10–3 mol
D) 64.9  103 mol
E) 64.9  10–1 mol
ANS:
OBJ:
TOP:
KEY:
A
PTS: 1
DIF: easy
REF: 3.2
Convert from grams of substance to moles of substance. (Example 3.5)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
38. A 0.0103-mol sample of urea, NH2CONH2, contains
A) 6.02  1023 molecules.
B) 2.48  1022 molecules.
C) 4.96  1022 atoms.
D) 2.48  1023 atoms.
E) 1.03  1024 atoms.
ANS: C
Test Bank
PTS:
1
DIF:
easy
General Chemistry, 10th edition
REF: 3.2
10
OBJ: Calculate the number of molecules in a given mass of substance. (Example 3.6)
TOP: stoichiometry | mass and moles of substance
KEY: mole | mole calculations
MSC: general chemistry
39. How many molecules are there in 90.0 g of butylene glycol, HO(CH2)4OH?
A) 1
B) (6.02  1023)/90.0
C) 90.0
D) 90.0  (6.02  1023)
E) 6.02  1023
ANS:
OBJ:
TOP:
KEY:
E
PTS: 1
DIF: easy
REF: 3.2
Calculate the number of molecules in a given mass of substance. (Example 3.6)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
40. How many molecules are there in 192 g of citric acid, C6H8O7?
A) (6.02  1023) / 192
B) 192
C) 6.02  1023
D) 96.0
E) 192  (6.02  1023)
ANS:
OBJ:
TOP:
KEY:
C
PTS: 1
DIF: easy
REF: 3.2
Calculate the number of molecules in a given mass of substance. (Example 3.6)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
41. In 0.358 mol of trimellitic acid, C6H3(COOH)3, there are
A) 2.16  1022 molecules.
B) 8.62  1024 molecules.
C) 6.47  1023 oxygen atoms.
D) 1.94  1024 carbon atoms.
E) 3.59  1023 hydrogen atoms.
ANS:
OBJ:
TOP:
KEY:
D
PTS: 1
DIF: moderate
REF: 3.2
Calculate the number of molecules in a given mass of substance. (Example 3.6)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
42. How many molecules are there in 2.80 kg of hydrazine, N2H4?
A) 8.75  1023
B) 5.27  1025
C) 1.88  1022
D) 2.80  1026
E) 4.65  1021
ANS: B
PTS: 1
DIF: moderate
REF: 3.2
OBJ: Calculate the number of molecules in a given mass of substance. (Example 3.6)
TOP: stoichiometry | mass and moles of substance
Test Bank
General Chemistry, 10th edition
11
KEY: mole | mole calculations
MSC: general chemistry
43. How many molecules are there in 2.43 mg of mannose, C6H12O6, which is a sweet-tasting
sugar that has a bitter aftertaste?
A) 4.46  1021
B) 2.48  1021
C) 7.31  1018
D) 4.04  1024
E) 8.13  1018
ANS:
OBJ:
TOP:
KEY:
E
PTS: 1
DIF: moderate
REF: 3.2
Calculate the number of molecules in a given mass of substance. (Example 3.6)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
44. How many atoms are present in 463 g of KPF6 (MM = 184.1 g/mol)?
A) 2.23  1025
B) 1.51  1021
C) 2.51  1021
D) 1.13  1026
E) 1.21  1025
ANS:
OBJ:
TOP:
KEY:
E
PTS: 1
DIF: moderate
REF: 3.2
Calculate the number of molecules in a given mass of substance. (Example 3.6)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
45. Styrene's empirical formula is CH. When it is heated to 200°C, it is converted into a
polymer, polystyrene, which has excellent insulating properties. What mass of styrene
contains 9.75  1021 molecules of styrene? The molar mass of styrene is 104 g/mol.
A) 0.0162 g
B) 1.68 g
C) 3.24 g
D) 0.210 g
E) 0.130 g
ANS:
OBJ:
TOP:
KEY:
B
PTS: 1
DIF: difficult
REF: 3.2
Calculate the number of molecules in a given mass of substance. (Example 3.6)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
46. In 0.500 mol of dimethylhydrazine, (CH3)2N2H2, there are
A) 3.01  1024 molecules.
B) 1.51  1023 atoms.
C) 3.01  1022 molecules.
D) 3.61  1024 atoms.
E) 1.81  1024 atoms.
ANS: D
PTS: 1
DIF: moderate
REF: 3.2
OBJ: Calculate the number of molecules in a given mass of substance. (Example 3.6)
Test Bank
General Chemistry, 10th edition
12
TOP: stoichiometry | mass and moles of substance
KEY: mole | mole calculations
MSC: general chemistry
47. Which is a reasonable mass corresponding to 1020 molecules of a substance?
A) 100 g
B) 100 µg
C) 100 ng
D) 100 mg
E) 100 kg
ANS:
OBJ:
TOP:
KEY:
D
PTS: 1
DIF: moderate
REF: 3.2
Calculate the number of molecules in a given mass of substance. (Example 3.6)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
48. How many aluminum atoms are there in 52 g of Al2S3?
A) 4.2  1023
B) 1.6  1021
C) 2.1  1023
D) 1.1  1021
E) 6.3  1023
ANS:
OBJ:
TOP:
KEY:
A
PTS: 1
DIF: moderate
REF: 3.2
Calculate the number of molecules in a given mass of substance. (Example 3.6)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
49. Which one of the following contains 2.41  1024 atoms?
A) 52.0 g C2H2
B) 96.0 g O2
C) 32.0 g CH4
D) 168 g N2
E) 16.0 g He
ANS:
OBJ:
TOP:
KEY:
E
PTS: 1
DIF: moderate
REF: 3.2
Calculate the number of molecules in a given mass of substance. (Example 3.6)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
50. A sample of 496 g of white phosphorus, P4, contains the same number of atoms as
A) 192 g of ozone (O3).
B) 56.0 g of nitrogen (N2).
C) 92.0 g of sodium.
D) 120 g of formaldehyde (CH2O).
E) 128 g of oxygen (O2).
ANS:
OBJ:
TOP:
KEY:
D
PTS: 1
DIF: moderate
REF: 3.2
Calculate the number of molecules in a given mass of substance. (Example 3.6)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
Test Bank
General Chemistry, 10th edition
13
51. The total number of oxygen atoms in 1.76 g of CaCO3 (MM = 100.0 g/mol) is
A) 2.05  1023.
B) 4.24  1022.
C) 3.18  1022.
D) 1.75  1023.
E) 5.30  1022.
ANS:
OBJ:
TOP:
KEY:
C
PTS: 1
DIF: moderate
REF: 3.2
Calculate the number of molecules in a given mass of substance. (Example 3.6)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
52. A sample of 336 g of ozone, O3, contains the same number of atoms as
A) 336 g of oxygen (O2).
B) 28.2 g of hydrogen (H2).
C) 266 g of fluorine (F2).
D) 189 g of aluminum (Al).
E) 411 g of nickel (Ni).
ANS:
OBJ:
TOP:
KEY:
A
PTS: 1
DIF: moderate
REF: 3.2
Calculate the number of molecules in a given mass of substance. (Example 3.6)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
53. Which of the following samples contains the smallest number of molecules?
A) 8.00 g of TNT, C7H5N3O6
B) 8.00 g of benzene, C6H6
C) 8.00 g of glucose, C6H12O6
D) 8.00 g of naphthalene, C10H8
E) 8.00 g of formaldehyde, CH2O
ANS:
OBJ:
TOP:
KEY:
A
PTS: 1
DIF: moderate
REF: 3.2
Calculate the number of molecules in a given mass of substance. (Example 3.6)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
54. Which of the following samples contains the largest number of atoms?
A) 1 g N2
B) 1 g Li
C) 1 g Cl2
D) 1 g P4
E) 1 g Mg
ANS:
OBJ:
TOP:
KEY:
B
PTS: 1
DIF: moderate
REF: 3.2
Calculate the number of molecules in a given mass of substance. (Example 3.6)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
55. Which of the following samples contains the largest number of molecules?
Test Bank
General Chemistry, 10th edition
14
A)
B)
C)
D)
E)
10. g Pb
10. g Cl2
10. g Kr
10. g O2
10. g S8
ANS:
OBJ:
TOP:
KEY:
D
PTS: 1
DIF: moderate
REF: 3.2
Calculate the number of molecules in a given mass of substance. (Example 3.6)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
56. In 1928, 1.0 g of rhenium, Re, was isolated from 660 kg of the ore molybenite. The percent
by mass of this element in the molybenite was
A) 0.66 %.
B) 0.15 %.
C) 3.5  10–4 %.
D) 6.6  10–3 %.
E) 1.5  10–4 %.
ANS:
OBJ:
TOP:
MSC:
E
PTS: 1
DIF: easy
REF: 3.3
Calculate the percentage composition of the elements in a compound. (Example 3.7)
stoichiometry | determining chemical formulas
KEY: mass percentage
general chemistry
57. An ore sample is found to contain 24.1 g of mercury and 50.7 g waste rock (gangue). What
is the percent by mass of mercury in the ore?
A)
B)
C)
D)
E)
32.2 %
47.4 %
0.322 %
0.474 %
4.74 %
ANS: A
PTS: 1
DIF: easy
REF: 3.3
OBJ: Calculate the percentage composition of the elements in a compound. (Example 3.7)
TOP: stoichiometry | determining chemical formulas
58. An ore sample with a mass of 68.0 g is found to contain 15.5% by mass nickel. What mass
of nickel is contained in the ore?
A)
B)
C)
D)
E)
10.5 g
81.7 g
1.55 g
84.5 g
546 g
ANS: A
PTS: 1
DIF: easy
REF: 3.3
OBJ: Calculate the percentage composition of the elements in a compound. (Example 3.7)
TOP: stoichiometry | determining chemical formulas
59. What is the percent by mass oxygen in (NH4)2SO3?
Test Bank
General Chemistry, 10th edition
15
A)
B)
C)
D)
E)
41.3 %
20.7 %
54.0 %
42.0 %
1.00 %
ANS: A
PTS: 1
DIF: easy
REF: 3.3
OBJ: Calculate the percentage composition of the elements in a compound. (Example 3.7)
TOP: stoichiometry | determining chemical formulas
60. What is the percentage by mass of hydrogen in the insecticide Lindane, C6H6Cl6?
A)
B)
C)
D)
E)
20.0 %
1.20 %
47.2 %
8.80 %
2.08 %
ANS: E
PTS: 1
DIF: easy
REF: 3.3
OBJ: Calculate the percentage composition of the elements in a compound. (Example 3.7)
TOP: stoichiometry | determining chemical formulas
61. The mineral leadhillite, which is essentially Pb4(SO4)(CO3)2(OH)2 (FW = 1079 g/mol),
contains ____% hydrogen by mass.
A) 76.81
B) 0.1868
C) 2.226
D) 17.79
E) 2.972
ANS:
OBJ:
TOP:
MSC:
B
PTS: 1
DIF: easy
REF: 3.3
Calculate the percentage composition of the elements in a compound. (Example 3.7)
stoichiometry | determining chemical formulas
KEY: mass percentage
general chemistry
62. Which of the following compounds has the highest percentage of hydrogen atoms by mass?
A) CH3COOH
B) C2H5OH
C) CH3OH
D) H2CO3
E) H2C2O4
ANS:
OBJ:
TOP:
MSC:
B
PTS: 1
DIF: easy
REF: 3.3
Calculate the percentage composition of the elements in a compound. (Example 3.7)
stoichiometry | determining chemical formulas
KEY: mass percentage
general chemistry
63. Which of the following compounds has the highest percentage of nitrogen by mass?
A) (NH4)2SO3
B) NaNO3
Test Bank
General Chemistry, 10th edition
16
C) N2Cl4
D) NH4NO2
E) HNO3
ANS:
OBJ:
TOP:
MSC:
D
PTS: 1
DIF: easy
REF: 3.3
Calculate the percentage composition of the elements in a compound. (Example 3.7)
stoichiometry | determining chemical formulas
KEY: mass percentage
general chemistry
64. Which of the following compounds has the same percentage of carbon and hydrogen by
mass as cyclohexane, C6H12?
A) C6H14, hexane
B) C5H10, pentene
C) C6H10, cyclohexene
D) C6H6, benzene
E) C6H12O6, glucose
ANS:
OBJ:
TOP:
MSC:
B
PTS: 1
DIF: easy
REF: 3.3
Calculate the percentage composition of the elements in a compound. (Example 3.7)
stoichiometry | determining chemical formulas
KEY: mass percentage
general chemistry
65. What is the mass percentage of carbon in the compound C6H6O2?
A) 5.5 %
B) 70.9 %
C) 65.5 %
D) 29.1 %
E) 14.3 %
ANS:
OBJ:
TOP:
MSC:
C
PTS: 1
DIF: easy
REF: 3.3
Calculate the percentage composition of the elements in a compound. (Example 3.7)
stoichiometry | determining chemical formulas
KEY: mass percentage
general chemistry
66. What is the percentage by mass of hydrogen in ammonium phosphate, (NH4)3PO4?
A)
B)
C)
D)
E)
8.11 %
4.06 %
52.0 %
40.0 %
3.00 %
ANS: A
PTS: 1
DIF: easy
REF: 3.3
OBJ: Calculate the percentage composition of the elements in a compound. (Example 3.7)
TOP: stoichiometry | determining chemical formulas
67. A crystal of the mineral troegerite, (UO2)3(AsO4)2 • 12H2O (FM = 1304 amu), contains
____% arsenic by mass.
A) 15.4
B) 39.8
C) 61.0
Test Bank
General Chemistry, 10th edition
17
D) 26.4
E) 11.5
ANS:
OBJ:
TOP:
MSC:
E
PTS: 1
DIF: easy
REF: 3.3
Calculate the percentage composition of the elements in a compound. (Example 3.7)
stoichiometry | determining chemical formulas
KEY: mass percentage
general chemistry
68. How many grams of hydrogen atoms are present in 18.4 g of water?
A) 37.1 g
B) 1.02 g
C) 2.06 g
D) 1.96 g
E) 12.3 g
ANS:
OBJ:
TOP:
KEY:
C
PTS: 1
DIF: easy
REF: 3.3
Calculate the mass of an element in a given mass of compound. (Example 3.8)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
69. How many grams of potassium are present in 21.6 g of K2Cr2O7?
A) 5.74 g
B) 1.105 g
C) 2.87 g
D) 78.2 g
E) 10.8 g
ANS:
OBJ:
TOP:
MSC:
A
PTS: 1
DIF: moderate
REF: 3.3
Calculate the mass of an element in a given mass of compound. (Example 3.8)
stoichiometry | determining chemical formulas
KEY: mass percentage
general chemistry
70. NaHCO3 is the active ingredient in baking soda. How many grams of oxygen are present in
0.67 g of NaHCO3?
A) 0.024 g
B) 0.128 g
C) 7.98  103 g
D) 0.043 g
E) 0.38 g
ANS:
OBJ:
TOP:
MSC:
E
PTS: 1
DIF: easy
REF: 3.3
Calculate the mass of an element in a given mass of compound. (Example 3.8)
stoichiometry | determining chemical formulas
KEY: mass percentage
general chemistry
71. Which of the following contains the greatest mass of bromine atoms?
A) 11.0 g of KBr
B) 23.0 g of Br2
C) 0.076 mol of KBr
D) 0.092 mol of Br2
E) 32.0 g of NaBrO3
Test Bank
General Chemistry, 10th edition
18
ANS:
OBJ:
TOP:
MSC:
B
PTS: 1
DIF: moderate
REF: 3.3
Calculate the mass of an element in a given mass of compound. (Example 3.8)
stoichiometry | determining chemical formulas
KEY: mass percentage
general chemistry
72. The amount of calcium in a 15.0-g sample was determined by converting the calcium to
calcium oxalate, CaC2O4. The CaC2O4 weighed 12.6 g. What is the percent of calcium in
the original sample?
A) 10.8 %
B) 26.3 %
C) 14.8 %
D) 33.7 %
E) 84.0 %
ANS:
OBJ:
TOP:
KEY:
B
PTS: 1
DIF: difficult
REF: 3.3
Calculate the mass of an element in a given mass of compound. (Example 3.8)
stoichiometry | mass and moles of substance
mole | mole calculations
MSC: general chemistry
73. A compound containing only carbon, hydrogen, and oxygen is subjected to elemental
analysis. Upon complete combustion, a 0.1804-g sample of the compound produced
0.3051 g of CO2 and 0.1249 g of H2O. What is the empirical formula of the compound?
A) C3H6O3
B) C3H3O
C) C4H8O3
D) C2H2O
E) CH2O3
ANS:
OBJ:
TOP:
MSC:
C
PTS: 1
DIF: moderate
REF: 3.4
Calculate the percentage of C, H, and O from combustion data. (Example 3.9)
stoichiometry | determining chemical formulas
KEY: elemental analysis
general chemistry
74. A 3.075 g sample of a compound containing only carbon, hydrogen, and oxygen is burned
in an excess of dioxygen, producing 6.990 g CO2 and 2.862 g H2O. What mass of oxygen is
contained in the original sample?
A)
B)
C)
D)
E)
0.8472 g
1.167 g
3.915 g
4.129 g
0.2134 g
ANS: A
PTS: 1
DIF: moderate
REF: 3.4
OBJ: Calculate the percentage of C, H, and O from combustion data. (Example 3.9)
TOP: stoichiometry | determining chemical formulas
75. A 4.043 g sample of a compound containing only carbon, hydrogen, and oxygen is burned
in an excess of dioxygen, producing 9.191 g CO2 and 3.762 g H2O. What percent by mass
of oxygen is contained in the original sample?
Test Bank
General Chemistry, 10th edition
19
A)
B)
C)
D)
E)
27.54 %
37.96 %
12.73 %
13.43 %
6.939 %
ANS: A
PTS: 1
DIF: moderate
REF: 3.4
OBJ: Calculate the percentage of C, H, and O from combustion data.
TOP: stoichiometry | determining chemical formulas
76. A 2.841 g sample of a hydrocarbon is burned in an excess of dioxygen, producing 7.794 g
CO2 and water. What mass of hydrogen is contained in the original sample?
A)
B)
C)
D)
E)
0.7140 g
4.953 g
10.64 g
2.826 g
1.421 g
ANS: A
PTS: 1
DIF: moderate
REF: 3.4
OBJ: Calculate the percentage of C, H, and O from combustion data. (Example 3.9)
TOP: stoichiometry | determining chemical formulas
77. A 2.445 g sample of a hydrocarbon is burned in an excess of dioxygen, producing 6.708 g
CO2 and 5.492 g H2O. What is the empirical formula of the hydrocarbon?
A)
B)
C)
D)
E)
CH4
CH2
C2H3
CH3
CH
ANS: A
PTS: 1
DIF: moderate
OBJ: Determine the empirical formula of a binary compound.
TOP: stoichiometry | determining chemical formulas
REF: 3.5
78. A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen
was burned in dioxygen to yield 0.01962 mol of CO2 and 0.01961 mol of H2O. What is the
empirical formula of the compound?
A)
B)
C)
D)
E)
CHO
C3H3O2
C2H2O
C3H6O2
C6H3O2
ANS: D
PTS: 1
DIF: moderate
REF: 3.4
OBJ: Calculate the percentage of C, H, and O from combustion data. (Example 3.9)
TOP: stoichiometry | determining chemical formulas
Test Bank
General Chemistry, 10th edition
20
79. A sample containing only carbon, hydrogen, phosphorus, and oxygen is subjected to
elemental analysis. After complete combustion, a 0.4946-g sample of the compound yields
0.7092 g of CO2, 0.4355 g of H2O, and 0.3812 g of P4O10. What is the empirical formula of
the compound?
A) CH3PO
B) C2H3PO
C) C2H6P2O4
D) C3H9PO
E) CH2P4O13
ANS:
OBJ:
TOP:
MSC:
D
PTS: 1
DIF: difficult
REF: 3.4
Calculate the percentage of C, H, and O from combustion data. (Example 3.9)
stoichiometry | determining chemical formulas
KEY: elemental analysis
general chemistry
80. A sample containing only carbon, hydrogen, and silicon is subjected to elemental analysis.
After complete combustion, a 0.1099-g sample of the compound yields 0.2193 g of CO2,
0.1346 g of H2O, and 0.07485 g of SiO2. What is the empirical formula of the compound?
A) CH3Si
B) C2H4Si
C) C4H12Si
D) C6H12Si2
E) CH2Si
ANS:
OBJ:
TOP:
MSC:
C
PTS: 1
DIF: difficult
REF: 3.4
Calculate the percentage of C, H, and O from combustion data. (Example 3.9)
stoichiometry | determining chemical formulas
KEY: elemental analysis
general chemistry
81. Of the following, the only empirical formula is
A) C4H10.
B) C4H6.
C) C5H14.
D) H2O2.
E) O2.
ANS:
OBJ:
TOP:
MSC:
C
PTS: 1
DIF: easy
Define empirical formula.
stoichiometry | determining chemical formulas
general chemistry
REF: 3.5
KEY: empirical formula
82. Which of the following is the empirical formula for the molecule below?
A)
B)
C)
D)
CHO
CH3COOH
C2H4O2
CH2O
Test Bank
General Chemistry, 10th edition
21
E) none of the above.
ANS:
OBJ:
TOP:
MSC:
D
PTS: 1
DIF: easy
Define empirical formula.
stoichiometry | determining chemical formulas
general chemistry
REF: 3.5
KEY: empirical formula
83. Analysis of a compound showed that it contained 76.0 % fluorine atoms and 24.0 % carbon
atoms by mass. What is its empirical formula?
A) CF2
B) C2F3
C) CF3
D) C2F5
E) CF
ANS: A
PTS: 1
DIF: moderate
REF: 3.5
OBJ: Determine the empirical formula of a binary compound from the masses of its
elements. (Example 3.10)
TOP: stoichiometry | determining chemical formulas
KEY: empirical formula
MSC: general chemistry
84. A sample of an oxide of antimony (Sb) contained 39.5 g of antimony combined with 13.0 g
of oxygen. What is the simplest formula for the oxide?
A) SbO2
B) SbO
C) Sb2O3
D) Sb2O
E) Sb2O5
ANS: E
PTS: 1
DIF: moderate
REF: 3.5
OBJ: Determine the empirical formula of a binary compound from the masses of its
elements. (Example 3.10)
TOP: stoichiometry | determining chemical formulas
KEY: empirical formula
MSC: general chemistry
85. Chlorine was passed over 1.30 g of heated titanium, and 4.20 g of a chloride-containing
compound of Ti was obtained. What is the empirical formula of the chloride-containing
compound?
A) TiCl2
B) TiCl4
C) TiCl
D) TiCl3
E) Ti2Cl3
ANS: D
PTS: 1
DIF: moderate
REF: 3.5
OBJ: Determine the empirical formula of a binary compound from the masses of its
elements. (Example 3.10)
TOP: stoichiometry | determining chemical formulas
KEY: empirical formula
MSC: general chemistry
Test Bank
General Chemistry, 10th edition
22
86. A 2.39-g sample of an oxide of chromium contains 1.48 g of chromium. Calculate the
simplest formula for the compound.
A) CrO5
B) Cr2O
C) CrO2
D) CrO
E) Cr2O3
ANS: C
PTS: 1
DIF: moderate
REF: 3.5
OBJ: Determine the empirical formula of a binary compound from the masses of its
elements. (Example 3.10)
TOP: stoichiometry | determining chemical formulas
KEY: empirical formula
MSC: general chemistry
87. A compound is composed of only C and H. It contains 92.26 % C. What is its empirical
formula?
A) C2H5
B) C2H3
C) C3H4
D) CH
E) CH2
ANS:
OBJ:
TOP:
MSC:
D
PTS: 1
DIF: easy
REF: 3.5
Determine the empirical formula from the percentage composition. (Example 3.11)
stoichiometry | determining chemical formulas
KEY: empirical formula
general chemistry
88. A compound composed of only C and F contains 17.39 % C by mass. What is its empirical
formula?
A) CF3
B) CF
C) C2F
D) CF4
E) CF2
ANS:
OBJ:
TOP:
MSC:
A
PTS: 1
DIF: easy
REF: 3.5
Determine the empirical formula from the percentage composition. (Example 3.11)
stoichiometry | determining chemical formulas
KEY: empirical formula
general chemistry
89. A hydrocarbon, subjected to elemental analysis, was found to contain 85.63 % carbon and
14.37 % hydrogen by mass. What is the empirical formula of the hydrocarbon?
A) CH4
B) C2H4
C) C6H
D) C10H
E) CH2
ANS: E
PTS: 1
DIF: easy
REF: 3.5
OBJ: Determine the empirical formula from the percentage composition. (Example 3.11)
TOP: stoichiometry | determining chemical formulas
KEY: empirical formula
Test Bank
General Chemistry, 10th edition
23
MSC: general chemistry
90. A particular compound contains, by mass, 41.4 % carbon, 3.47 % hydrogen, and 55.1 %
oxygen. A 0.050-mol sample of this compound weighs 5.80 g. The molecular formula of
this compound is
A) C3H3O3.
B) C3H3O.
C) CHO.
D) C4H4O4.
E) C5H5O5.
ANS:
OBJ:
TOP:
KEY:
D
PTS: 1
DIF: moderate
REF: 3.5
Determine the empirical formula from the percentage composition. (Example 3.11)
stoichiometry | determining chemical formulas
molecular formula
MSC: general chemistry
91. What is the empirical formula of an oxide of nitrogen that contains 25.93 % nitrogen by
mass?
A) NO2
B) N2O
C) N2O3
D) NO
E) N2O5
ANS:
OBJ:
TOP:
MSC:
E
PTS: 1
DIF: moderate
REF: 3.5
Determine the empirical formula from the percentage composition. (Example 3.11)
stoichiometry | determining chemical formulas
KEY: empirical formula
general chemistry
92. The analysis of an organic compound showed that it contained 1.386 mol of C, 0.0660 mol
of H, 0.924 mol of O, and 0.462 mol of N. How many nitrogen atoms are there in the
empirical formula for this compound?
A) 9
B) 7
C) 2
D) 4
E) 3
ANS:
OBJ:
TOP:
MSC:
B
PTS: 1
DIF: easy
REF: 3.5
Determine the empirical formula from the percentage composition. (Example 3.11)
stoichiometry | determining chemical formulas
KEY: empirical formula
general chemistry
93. A sample containing 0.700 mol of a compound is composed of 4.21  1023 atoms of sodium,
24.79 g of chlorine atoms, and 33.57 g of oxygen atoms. The formula of the compound is
A) NaClO3.
B) NaClO5.
C) NaClO.
D) NaClO4.
E) NaClO2.
Test Bank
General Chemistry, 10th edition
24
ANS:
OBJ:
TOP:
MSC:
A
PTS: 1
DIF: moderate
REF: 3.5
Determine the empirical formula from the percentage composition. (Example 3.11)
stoichiometry | determining chemical formulas
KEY: empirical formula
general chemistry
94. The analysis of an organic compound showed that it contained 0.0700 mol of C, 0.175 mol
of H, and 0.0350 mol of N. Its molecular mass is 86 amu. How many atoms of carbon are
there in the empirical formula for the compound, and how many are in the molecular
formula?
A) empirical = 2, molecular = 3
B) empirical = 2, molecular = 6
C) empirical = 2, molecular = 4
D) empirical = 5, molecular = 10
E) empirical = 3, molecular = 3
ANS:
OBJ:
TOP:
KEY:
C
PTS: 1
DIF: moderate
REF: 3.5
Determine the empirical formula from the percentage composition. (Example 3.11)
stoichiometry | determining chemical formulas
molecular formula
MSC: general chemistry
95. A given hydrocarbon is burned in the presence of oxygen gas and is converted completely to
water and carbon dioxide. The mole ratio of H2O to CO2 is 1.33:1.00. The hydrocarbon
could be
A) C2H2.
B) C2H6.
C) CH4.
D) C3H4.
E) C3H8.
ANS:
OBJ:
TOP:
MSC:
E
PTS: 1
DIF: difficult
REF: 3.5
Determine the empirical formula from the percentage composition. (Example 3.11)
stoichiometry | determining chemical formulas
KEY: empirical formula
general chemistry
96. A given hydrocarbon is burned in the presence of oxygen gas and is converted completely to
carbon dioxide and water. Equal numbers of moles of CO2 and H2O are produced. The
hydrocarbon could be
A) C3H4.
B) C5H10.
C) C2H3.
D) CH3.
E) C3H5.
ANS:
OBJ:
TOP:
MSC:
B
PTS: 1
DIF: difficult
REF: 3.5
Determine the empirical formula from the percentage composition. (Example 3.11)
stoichiometry | determining chemical formulas
KEY: empirical formula
general chemistry
97. An organic compound that has the empirical formula CHO has a molecular mass of 145
amu. Its molecular formula is
Test Bank
General Chemistry, 10th edition
25
A)
B)
C)
D)
E)
C9H9O9.
C4H4O4.
C3H3O3.
C5H5O5.
C12H12O12.
ANS: D
PTS: 1
DIF: easy
REF: 3.5
OBJ: Understand the relationship between the molecular mass of a substance and its
empirical formula mass.
TOP: stoichiometry | determining chemical formulas
KEY: molecular formula
MSC: general chemistry
98. A certain compound has a molar mass of 210 g/mol. Which is a possible empirical formula
for this compound?
A) CH2O
B) CHO
C) C2H2O2
D) C2HO
E) C2H2O
ANS: A
PTS: 1
DIF: moderate
REF: 3.5
OBJ: Understand the relationship between the molecular mass of a substance and its
empirical formula mass.
TOP: stoichiometry | determining chemical formulas
KEY: empirical formula
MSC: general chemistry
99. The empirical formula for a group of compounds is CHCl. Lindane, a powerful insecticide,
is a member of this group. The molar mass of lindane is 290.8. How many atoms of carbon
does a molecule of lindane contain?
A) 3
B) 2
C) 4
D) 6
E) 8
ANS: D
PTS: 1
DIF: easy
REF: 3.5
OBJ: Understand the relationship between the molecular mass of a substance and its
empirical formula mass.
TOP: stoichiometry | determining chemical formulas
KEY: molecular formula
MSC: general chemistry
100. The empirical formula of styrene is CH; its molar mass is 104.1 g/mol. What is the
molecular formula of styrene?
A) C6H6
B) C2H4
C) C8H8
D) C10H12
E) none of these
ANS: C
PTS: 1
DIF: easy
REF: 3.5
OBJ: Understand the relationship between the molecular mass of a substance and its
Test Bank
General Chemistry, 10th edition
26
empirical formula mass.
TOP: stoichiometry | determining chemical formulas
KEY: molecular formula
MSC: general chemistry
101. The empirical formula of styrene is CH. An experimental determination of the molar mass
of styrene by a student yields the value of 104 g/mol. What is the molecular formula of
styrene?
A) C5H10
B) CH
C) C8H8
D) C3H8
E) C6H9
ANS: C
PTS: 1
DIF: easy
REF: 3.5
OBJ: Understand the relationship between the molecular mass of a substance and its
empirical formula mass.
TOP: stoichiometry | determining chemical formulas
KEY: molecular formula
MSC: general chemistry
102. An organic compound has a molecular mass of 279.2 amu and contains 85.96% carbon by
mass. How many carbon atoms are in each molecule of this compound?
A) 16
B) 21
C) 25
D) 29
E) 20
ANS: E
PTS: 1
DIF: easy
REF: 3.5
OBJ: Determine the molecular formula from the percentage composition and molecular
mass. (Example 3.12)
TOP: stoichiometry | determining chemical formulas
KEY: molecular formula
MSC: general chemistry
103. An unknown organic compound contains 41.4 % carbon, 3.47 % hydrogen, and 55.1 %
oxygen by mass. A 0.040-mol sample of this compound weighs 3.48 g. What is the
molecular formula of the organic compound?
A) C3H3O
B) C2H2O2
C) C3H3O3
D) C7H7O7
E) CHO
ANS: C
PTS: 1
DIF: moderate
REF: 3.5
OBJ: Determine the molecular formula from the percentage composition and molecular
mass. (Example 3.12)
TOP: stoichiometry | determining chemical formulas
KEY: molecular formula
MSC: general chemistry
104. A compound is composed of only C and I. It contains 5.935 % C by mass and has a molar
mass of 809.44 g/mol. What is its molecular formula?
Test Bank
General Chemistry, 10th edition
27
A)
B)
C)
D)
E)
CI
C4I6
C3I4
C2I2
C2I3
ANS: B
PTS: 1
DIF: moderate
REF: 3.5
OBJ: Determine the molecular formula from the percentage composition and molecular
mass. (Example 3.12)
TOP: stoichiometry | determining chemical formulas
KEY: molecular formula
MSC: general chemistry
105. An organic compound has a molar mass of 171.1 g/mol and contains 11.10 % hydrogen
atoms by mass. How many hydrogen atoms are in each molecule of this compound?
A) 19
B) 6
C) 21
D) 28
E) 11
ANS: A
PTS: 1
DIF: moderate
REF: 3.5
OBJ: Determine the molecular formula from the percentage composition and molecular
mass. (Example 3.12)
TOP: stoichiometry | determining chemical formulas
KEY: molecular formula
MSC: general chemistry
106. A compound has a molar mass of 171.6 g/mol and contains 55.94 % oxygen atoms by mass.
How many oxygen atoms are in each molecule of this compound?
A) 6
B) 4
C) 10
D) 2
E) 8
ANS: A
PTS: 1
DIF: moderate
REF: 3.5
OBJ: Determine the molecular formula from the percentage composition and molecular
mass. (Example 3.12)
TOP: stoichiometry | determining chemical formulas
KEY: molecular formula
MSC: general chemistry
107. A compound contains 43.84 % carbon atoms, 3.65 % hydrogen atoms, and 8.68 % fluorine
atoms by mass. Each molecule of this compound contains one fluorine atom. What is the
total number of carbon, hydrogen, and fluorine atoms in one molecule of this compound?
A) 17
B) 12
C) 9
D) 7
E) 14
ANS: A
PTS: 1
DIF: moderate
REF: 3.5
OBJ: Determine the molecular formula from the percentage composition and molecular
Test Bank
General Chemistry, 10th edition
28
mass. (Example 3.12)
TOP: stoichiometry | determining chemical formulas
KEY: molecular formula
MSC: general chemistry
108. A molecular compound contains 92.3 % carbon and 7.7 % hydrogen by mass. If 0.432 mol
of the compound weighs 22.46 g, what is its molecular formula?
A) C8H8
B) C6H10
C) C4H8
D) C4H4
E) CH
ANS: D
PTS: 1
DIF: moderate
REF: 3.5
OBJ: Determine the molecular formula from the percentage composition and molecular
mass. (Example 3.12)
TOP: stoichiometry | determining chemical formulas
KEY: molecular formula
MSC: general chemistry
109. Analysis of a compound containing only C and Br revealed that it contains 33.33 % C atoms
by number and has a molar mass of 515.46 g/mol. What is the molecular formula of this
compound?
A) CBr2
B) C2Br6
C) C2Br4
D) CBr3
E) C3Br6
ANS: E
PTS: 1
DIF: difficult
REF: 3.5
OBJ: Determine the molecular formula from the percentage composition and molecular
mass. (Example 3.12)
TOP: stoichiometry | determining chemical formulas
KEY: molecular formula
MSC: general chemistry
110. Complete combustion of a 0.30-mol sample of a hydrocarbon, CxHy, gives 1.20 mol of CO2
and 1.50 mol of H2O. The molecular formula of the original hydrocarbon is
A) C3H8.
B) C4H10.
C) C8H20.
D) C3H5.
E) C5H7.
ANS: B
PTS: 1
DIF: difficult
REF: 3.5
OBJ: Determine the molecular formula from the percentage composition and molecular
mass. (Example 3.12)
TOP: stoichiometry | determining chemical formulas
KEY: molecular formula
MSC: general chemistry
111. A chemical reaction has the equation: 2A + B  C. Which of the following figures best
illustrates a stoichiometric ratio of A and B?
Test Bank
General Chemistry, 10th edition
29
I.
A)
B)
C)
D)
E)
II.
III.
IV.
I only
III only
II only
Both I and IV
IV only
ANS: D
PTS: 1
DIF: easy
REF: 3.6
OBJ: Relate the coefficients in a balanced chemical equation to the number of molecules
or moles (molar interpretation).
TOP: stoichiometry |
stoichiometry calculation
KEY: molar interpretation
MSC: general chemistry
112. The balanced chemical equation for the combustion of methane is:
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
Which of the following statements concerning this chemical equation is/are correct?
1.
2.
3.
A)
B)
C)
D)
E)
One gram of methane gas reacts with two grams of dioxygen gas, producing
one gram of carbon dioxide gas and two grams of gaseous water.
One mole of methane gas reacts with two moles of dioxygen gas, producing
one mole of carbon dioxide gas and two moles of gaseous water.
One molecule of methane gas reacts with two molecules of dioxygen gas,
producing one molecule of carbon dioxide gas and two molecules of gaseous
water.
1 only
2 only
2 and 3
1 and 3
1,2 and 3
ANS: C
PTS: 1
DIF: easy
REF: 3.6
OBJ: Relate the coefficients in a balanced chemical equation to the number of molecules
or moles (molar interpretation).
TOP: stoichiometry |
stoichiometry calculation
KEY: molar interpretation
MSC: general chemistry
NOT: REVISED
113. Balance the following expression:
__ CH3CH2COOH + __ O2  __ CO2 + __ H2O
How many moles of O2 are required for the complete combustion of 8 mol of propanoic
acid?
A) 5 mol
B) 30 mol
Test Bank
General Chemistry, 10th edition
30
C) 28 mol
D) 37 mol
E) 2 mol
ANS: C
PTS: 1
DIF: easy
REF: 3.6
OBJ: Relate the coefficients in a balanced chemical equation to the number of molecules
or moles (molar interpretation).
TOP: stoichiometry |
stoichiometry calculation
KEY: molar interpretation
MSC: general chemistry
114. The products of the combustion of acetone with oxygen are shown in the following
equation:
__ CH3COCH3 + __ O2  __ CO2 + __ H2O
When properly balanced, the equation indicates that ____ mol of CO2 are produced for each
mole of CH3COCH3.
A) 3
B) 5
C) 4.5
D) 8
E) 1
ANS: A
PTS: 1
DIF: easy
REF: 3.6
OBJ: Relate the coefficients in a balanced chemical equation to the number of molecules
or moles (molar interpretation).
TOP: stoichiometry |
stoichiometry calculation
KEY: molar interpretation
MSC: general chemistry
115. Ammonia, NH3, and oxygen can be reacted together in the presence of a catalyst to form
only nitrogen monoxide and water. The number of moles of oxygen consumed for every
5.00 moles of NO produced is
.
A) 6.25
B) 25.0
C) 18.8
D) 3.13
E) 12.5
ANS: A
PTS: 1
DIF: difficult
REF: 3.6
OBJ: Relate the coefficients in a balanced chemical equation to the number of molecules
or moles (molar interpretation).
TOP: stoichiometry |
stoichiometry calculation
KEY: molar interpretation
MSC: general chemistry
116. 2KHCO3(s)  K2CO3(s) + CO2(g) + H2O(l)
How many moles of potassium carbonate will be produced if 454 g of potassium hydrogen
carbonate are heated?
A) 2.27 mol
B) 3.29 mol
C) 11.4 mol
D) 227 mol
Test Bank
General Chemistry, 10th edition
31
E) 4.54 mol
ANS:
OBJ:
TOP:
KEY:
A
PTS: 1
DIF: moderate
REF: 3.7
Relate the quantities of reactant to the quantity of product. (Example 3.13)
stoichiometry | stoichiometry calculation
amounts of substances
MSC: general chemistry
117. Calculate the number of moles of O2 required to react with phosphorus to produce 4.76 g of
P4O6. (Molar mass P4O6 = 219.9 g/mol)
A) 0.0216 mol
B) 0.149 mol
C) 0.0649 mol
D) 0.0433 mol
E) 0.130 mol
ANS:
OBJ:
TOP:
KEY:
C
PTS: 1
DIF: moderate
REF: 3.7
Relate the quantities of reactant to the quantity of product. (Example 3.13)
stoichiometry | stoichiometry calculation
amounts of substances
MSC: general chemistry
118. Elemental sulfur can be converted to sulfur dioxide by combustion in air. Sulfur dioxide
will react with water to form sulfurous acid (see balanced equation below).
SO2(g) + H2O(l)  H2SO3(l)
What mass of sulfur dioxide is needed to prepare 36.86 g of H2SO3(l)?
A)
B)
C)
D)
E)
28.77 g
47.23 g
0.5754 g
0.4491 g
36.86 g
ANS: A
PTS: 1
DIF: moderate
REF: 3.7
OBJ: Relate the quantities of reactant to the quantity of product. (Example 3.13)
TOP: stoichiometry | stoichiometry calculation
119. One step in the isolation of pure rhodium metal (Rh) is the precipitation of rhodium(III)
hydroxide from a solution containing rhodium(III) sulfate according to the following
balanced chemical equation:
Rh2(SO4)3(aq) + 6NaOH(aq)

2Rh(OH)3(s) + 3Na2SO4(aq)
If 2.40 g of rhodium(III) sulfate reacts with excess sodium hydroxide, what mass of
rhodium(III) hydroxide may be produced?
A) 1.50 g
B) 4.80 g
C) 2.40 g
D) 0.374 g
E) 2.99 g
ANS: A
Test Bank
PTS:
1
DIF:
easy
General Chemistry, 10th edition
REF: 3.7
32
OBJ: Relate the quantities of reactant to the quantity of product. (Example 3.13)
TOP: stoichiometry | stoichiometry calculation
KEY: amounts of substances
MSC: general chemistry
120. Pure copper may be produced by the reaction of copper(I) sulfide with oxygen gas as
follows:
Cu2S(s) + O2(g)  2Cu(s) + SO2(g)
What mass of copper(I) sulfide is required in order to prepare 0.610 kg of copper metal?
A) 0.610 kg
B) 0.305 kg
C) 0.459 kg
D) 1.53 kg
E) 0.764 kg
ANS:
OBJ:
TOP:
KEY:
E
PTS: 1
DIF: easy
REF: 3.7
Relate the quantities of reactant to the quantity of product. (Example 3.13)
stoichiometry | stoichiometry calculation
amounts of substances
MSC: general chemistry
121. One step in the isolation of pure rhodium metal (Rh) is the precipitation of rhodium(III)
hydroxide from a solution containing rhodium(III) sulfate according to the following
balanced chemical equation:
Rh2(SO4)3(aq) + 6NaOH(aq)  2Rh(OH)3(s) + 3Na2SO4(aq)
What mass of sodium hydroxide is required to precipitate 74.0 g of rhodium(III) hydroxide
from a solution containing excess rhodium(III) sulfate?
A) 6.41 g
B) 57.7 g
C) 19.2 g
D) 222 g
E) 74.0 g
ANS:
OBJ:
TOP:
KEY:
B
PTS: 1
DIF: easy
REF: 3.7
Relate the quantities of reactant to the quantity of product. (Example 3.13)
stoichiometry | stoichiometry calculation
amounts of substances
MSC: general chemistry
122. Pure copper may be produced by the reaction of copper(I) sulfide with oxygen gas as
follows:
Cu2S(s) + O2(g)  2Cu(s) + SO2(g)
If 0.680 kg of copper(I) sulfide reacts with excess oxygen, what mass of copper metal may
be produced?
A) 0.680 kg
B) 0.136 kg
C) 0.271 kg
D) 0.543 kg
E) 1.36 kg
ANS: D
Test Bank
PTS:
1
DIF:
easy
General Chemistry, 10th edition
REF: 3.7
33
OBJ: Relate the quantities of two reactants or two products. (Example 3.14)
TOP: stoichiometry | stoichiometry calculation
KEY: amounts of substances
MSC: general chemistry
123. One step in the isolation of pure rhodium metal (Rh) is the precipitation of rhodium(III)
hydroxide from a solution containing rhodium(III) sulfate according to the following
balanced chemical equation:
Rh2(SO4)3(aq) + 6NaOH(aq)  2Rh(OH)3(s) + 3Na2SO4(aq)
If 0.620 g of rhodium(III) hydroxide is produced, what mass of sodium sulfate is also
produced?
A) 0.572 g
B) 0.930 g
C) 0.858 g
D) 0.620 g
E) 0.381 g
ANS:
OBJ:
TOP:
KEY:
C
PTS: 1
DIF: easy
REF: 3.7
Relate the quantities of two reactants or two products. (Example 3.14)
stoichiometry | stoichiometry calculation
amounts of substances
MSC: general chemistry
124. The balanced equation for the combustion of ethanol is
2C2H5OH(g) + 7O2(g)  4CO2(g) + 6H2O(g)
How many grams of dioxygen are required to burn 5.9 g of C2H5OH?
A)
B)
C)
D)
E)
14 g
21 g
4.1 g
38 g
55 g
ANS: A
PTS: 1
DIF: moderate
REF: 3.7
OBJ: Relate the quantities of two reactants or two products. (Example 3.14)
TOP: stoichiometry | stoichiometry calculation
125. 2Al(s) + 6HCl(aq)  2AlCl3(aq) + 3H2(g)
According to the equation above, how many grams of aluminum are needed to completely
react with 3.83 mol of hydrochloric acid?
A) 310 g
B) 46.6 g
C) 34.4 g
D) 3.83 g
E) 103.3 g
ANS:
OBJ:
TOP:
KEY:
C
PTS: 1
DIF: moderate
REF: 3.7
Relate the quantities of two reactants or two products. (Example 3.14)
stoichiometry | stoichiometry calculation
amounts of substances
MSC: general chemistry
Test Bank
General Chemistry, 10th edition
34
126. A chemical reaction has the equation: 2A + B C. In which case is B the limiting
reactant?
A)
B)
C)
D)
E)
II
I
III
IV
none of these
ANS: A
PTS: 1
DIF: easy
REF: 3.8
OBJ: Understand how a limiting reactant or limiting reagent determines the moles of
product formed during a chemical reaction and how much excess reactant remains.
TOP: stoichiometry | stoichiometry calculation
KEY: limiting reactant
MSC: general chemistry
127. Consider an initial mixture of CH4 and O2 represented in the container below:
Given the reaction CH4 + 2O2  CO2 + 2H2O, which of the following represents a
stoichiometric picture of the container after the reaction has gone to completion?
A)
Test Bank
General Chemistry, 10th edition
35
B)
C)
D)
E) none of the above
ANS: C
PTS: 1
DIF: easy
REF: 3.8
OBJ: Understand how a limiting reactant or limiting reagent determines the moles of
product formed during a chemical reaction and how much excess reactant remains.
TOP: stoichiometry | stoichiometry calculation
KEY: limiting reactant
MSC: general chemistry
128. Which of the following statements concerning the limiting reactant is/are correct?
1.
2.
3.
A)
B)
C)
D)
E)
The mass of the limiting reactant is the always the lowest mass of all reactant
masses.
The theoretical yield depends on the amount of limiting reactant.
The moles of limiting reactant is always the lowest moles of all reactants.
2 only
3 only
1 and 3
2 and 3
1, 2, and 3
ANS: A
PTS: 1
DIF: moderate
REF: 3.8
OBJ: Understand how a limiting reactant or limiting reagent determines the moles of
product formed during a chemical reaction and how much excess reactant remains.
TOP: stoichiometry | stoichiometry calculation
129. The limiting reactant is the reactant
Test Bank
General Chemistry, 10th edition
36
A)
B)
C)
D)
E)
that has the lowest coefficient in the balanced equation.
that has the lowest molar mass.
that is left over after the reaction has gone to completion.
for which there is the lowest mass in grams.
none of the above
ANS: E
PTS: 1
DIF: easy
REF: 3.8
OBJ: Understand how a limiting reactant or limiting reagent determines the moles of
product formed during a chemical reaction and how much excess reactant remains.
TOP: stoichiometry | stoichiometry calculation
KEY: limiting reactant
MSC: general chemistry
130. The commercial production of phosphoric acid, H3PO4, can be represented by the equation
1500 g
300 g
307 g
Ca3(PO4)2 + 3SiO2 + 5C +
1180 g
5O2 +
300 g
3H2O
310 g/mol
32.0 g/mol
18.0 g/mol
60.1 g/mol 12.0 g/mol
 3CaSiO3 + 5CO2 + 2H3PO4
The molar mass for each reactant is shown below the reactant, and the mass of each reactant
for this problem is given above. Which substance is the limiting reactant?
A) H2O
B) C
C) O2
D) Ca3(PO4)2
E) SiO2
ANS:
OBJ:
TOP:
MSC:
E
PTS: 1
DIF: moderate
REF: 3.8
Calculate with a limiting reactant involving masses. (Example 3.16)
stoichiometry | stoichiometry calculation
KEY: limiting reactant
general chemistry
131. SO2 reacts with H2S as follows:
2H2S + SO2  3S + 2H2O
When 7.50 g of H2S reacts with 12.75 g of SO2, which statement applies?
A) 6.38 g of sulfur is formed.
B) SO2 is the limiting reagent.
C) 0.0216 mol of H2S remains.
D) 10.6 g of sulfur is formed.
E) 1.13 g of H2S remains.
ANS:
OBJ:
TOP:
MSC:
D
PTS: 1
DIF: moderate
REF: 3.8
Calculate with a limiting reactant involving masses. (Example 3.16)
stoichiometry | stoichiometry calculation
KEY: limiting reactant
general chemistry
132. A 15-g sample of lithium is reacted with 15 g of fluorine to form lithium fluoride:
2Li + F2  2LiF. After the reaction is complete, what will be present?
A) 0.789 mol of lithium fluoride only
B) 2.16 mol of lithium fluoride only
C) 2.16 mol of lithium fluoride and 0.395 mol of fluorine
Test Bank
General Chemistry, 10th edition
37
D) 0.789 mol of lithium fluoride and 1.37 mol of lithium
E) none of these
ANS:
OBJ:
TOP:
MSC:
D
PTS: 1
DIF: difficult
REF: 3.8
Calculate with a limiting reactant involving masses. (Example 3.16)
stoichiometry | stoichiometry calculation
KEY: limiting reactant
general chemistry
133. When 20.0 g C2H6 and 60.0 g O2 react to form CO2 and H2O, how many grams of water are
formed?
A) 14.5 g
B) 58.0 g
C) 18.0 g
D) 20.0 g
E) none of these
ANS:
OBJ:
TOP:
MSC:
E
PTS: 1
DIF: moderate
REF: 3.8
Calculate with a limiting reactant involving masses. (Example 3.16)
stoichiometry | stoichiometry calculation
KEY: limiting reactant
general chemistry
134. If 48.8 g of O2 is mixed with 48.8 g of H2 and the mixture is ignited, what is the maximum
mass of water that may be produced?
A) 439 g
B) 54.9 g
C) 48.8 g
D) 98 g
E) 86.8 g
ANS:
OBJ:
TOP:
MSC:
B
PTS: 1
DIF: moderate
REF: 3.8
Define and calculate the theoretical yield of chemical reactions.
stoichiometry | stoichiometry calculation
KEY: limiting reactant
general chemistry
135. One step in the isolation of pure rhodium metal (Rh) is the precipitation of rhodium(III)
hydroxide from a solution containing rhodium(III) sulfate according to the following
balanced chemical equation:
Rh2(SO4)3(aq) + 6NaOH(aq)  2Rh(OH)3(s) + 3Na2SO4(aq)
What is the theoretical yield of rhodium(III) hydroxide from the reaction of 0.590 g of
rhodium(III) sulfate with 0.266 g of sodium hydroxide?
A) 0.341 g
B) 0.266 g
C) 0.184 g
D) 0.856 g
E) 0.368 g
ANS:
OBJ:
TOP:
MSC:
A
PTS: 1
DIF: moderate
REF: 3.8
Define and calculate the theoretical yield of chemical reactions.
stoichiometry | stoichiometry calculation
KEY: limiting reactant
general chemistry
Test Bank
General Chemistry, 10th edition
38
136. A 5.95-g sample of AgNO3 is reacted with BaCl2 according to the equation
2AgNO3(aq) + BaCl2(aq)  2AgCl(s) + Ba(NO3)2(aq)
to give 3.36 g of AgCl. What is the percent yield of AgCl?
A) 44.6 %
B) 33.5 %
C) 66.9 %
D) 56.5 %
E) 100 %
ANS: C
PTS: 1
DIF: moderate
OBJ: Determine the percentage yield of a chemical reaction.
TOP: stoichiometry | stoichiometry calculation
REF: 3.8
MSC: general chemistry
137. The reaction of 11.9 g of CHCl3 with excess chlorine produced 10.2 g of CCl4, carbon
tetrachloride:
2CHCl3 + 2Cl2  2CCl4 + 2HCl
What is the percent yield?
A) 85.7 %
B) 100 %
C) 66.5 %
D) 33.3 %
E) 44.3 %
ANS: C
PTS: 1
DIF: moderate
OBJ: Determine the percentage yield of a chemical reaction.
TOP: stoichiometry | stoichiometry calculation
REF: 3.8
MSC: general chemistry
138. Consider the following reaction:
2A + B  3C + D
3.0 mol A and 2.0 mol B react to form 4.0 mol C. What is the percent yield of this reaction?
A) 75 %
B) 67 %
C) 89 %
D) 50 %
E) 100 %
ANS:
OBJ:
TOP:
MSC:
C
PTS: 1
DIF: moderate
Determine the percentage yield of a chemical reaction.
stoichiometry | stoichiometry calculation
general chemistry
REF: 3.8
KEY: limiting reactant
139. Pure copper may be produced by the reaction of copper(I) sulfide with oxygen gas as
follows:
Cu2S(s) + O2(g)  2Cu(s) + SO2(g)
If the reaction of 0.630 kg of copper(I) sulfide with excess oxygen produces 0.190 kg of
copper metal, what is the percent yield?
Test Bank
General Chemistry, 10th edition
39
A)
B)
C)
D)
E)
75.5 %
39.9 %
30.2 %
151 %
37.8 %
ANS:
OBJ:
TOP:
MSC:
E
PTS: 1
DIF: difficult
Determine the percentage yield of a chemical reaction.
stoichiometry | stoichiometry calculation
general chemistry
REF: 3.8
KEY: limiting reactant
140. One step in the isolation of pure rhodium metal (Rh) is the precipitation of rhodium(III)
hydroxide from a solution containing rhodium(III) sulfate according to the following
balanced chemical equation:
Rh2(SO4)3(aq) + 6NaOH(aq)  2Rh(OH)3(s) + 3Na2SO4(aq)
If the reaction of 0.650 g of rhodium(III) sulfate with excess sodium hydroxide produces
0.320 g of rhodium(III) hydroxide, what is the percent yield?
A) 316 %
B) 158 %
C) 39.5 %
D) 49.2 %
E) 79.0 %
ANS:
OBJ:
TOP:
MSC:
E
PTS: 1
DIF: difficult
Determine the percentage yield of a chemical reaction.
stoichiometry | stoichiometry calculation
general chemistry
REF: 3.8
KEY: limiting reactant
141. Sulfur trioxide, SO3, is made from the oxidation of SO2, and the reaction is represented by
the equation:
2SO2 + O2  2SO3
A 21-g sample of SO2 gives 18 g of SO3. The percent yield of SO3 is
A) 11 %
B) 69 %
C) 17 %
D) 26 %
E) 100 %
ANS: B
PTS: 1
DIF: moderate
OBJ: Determine the percentage yield of a chemical reaction.
TOP: stoichiometry | stoichiometry calculation
.
REF: 3.8
MSC: general chemistry
142. Nitric oxide, NO, is made from the oxidation of NH3, and the reaction is represented by the
equation
4NH3 + 5O2  4NO + 6H2O
An 9.1-g sample of NH3 gives 12.0 g of NO. The percent yield of NO is
A) 94 %
B) 46 %
Test Bank
General Chemistry, 10th edition
.
40
C) 17 %
D) 75 %
E) 28 %
ANS: D
PTS: 1
DIF: moderate
OBJ: Determine the percentage yield of a chemical reaction.
TOP: stoichiometry | stoichiometry calculation
REF: 3.8
MSC: general chemistry
143. Consider the fermentation reaction of glucose:
C6H12O6  2C2H5OH + 2CO2
A 1.00-mol sample of C6H12O6 was placed in a vat with 100 g of yeast. If 67.8 g of C2H5OH
was obtained, what was the percent yield of C2H5OH?
A) 73.6 %
B) 36.8 %
C) 67.8 %
D) 100 %
E) none of these
ANS:
OBJ:
TOP:
MSC:
A
PTS: 1
DIF: difficult
Determine the percentage yield of a chemical reaction.
stoichiometry | stoichiometry calculation
general chemistry
REF: 3.8
KEY: limiting reactant
144. One commercial system removes SO2 emissions from smoke at 95.0°C by the following set
of balanced reactions:
SO2(g) + Cl2  SO2Cl2(g)
SO2Cl2 + 2H2O  H2SO4 + 2HCl
H2SO4 + Ca(OH)2  CaSO4(s) + 2H2O
Assuming the process is 95.0 % efficient, how many grams of CaSO4 may be produced
from 100. g of SO2? (molar masses: SO2, 64.1 g/mol; CaSO4, 136 g/mol)
A) 87.2 g
B) 202 g
C) 44.8 g
D) 47.1 g
E) 212 g
ANS:
OBJ:
TOP:
MSC:
B
PTS: 1
DIF: difficult
Determine the percentage yield of a chemical reaction.
stoichiometry | stoichiometry calculation
general chemistry
Test Bank
General Chemistry, 10th edition
REF: 3.8
KEY: limiting reactant
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