AP Chemistry - Problem Drill 09: The Mole
Question No. 1 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
1. Avogadro’s number is defined as a number equal to the number of atoms in 12
grams of the carbon-12. This number is 6.02x1023. This is one of the most
important numbers in chemistry. It is used to count atoms or molecules in any
substance. How many moles are equal to 2.85  1026 molecules?
Question
(A)
(B)
(C)
(D)
0.00211 mole
473 mole
1.72  1050 mole
4.73  1048 mole
A. Incorrect!
You divided the numbers the wrong way. Check your set-up and recalculate the
division. It is a one-step process.
B. Correct!
Good job! Take the given number of molecules and divide by the Avogadro’s
number. The mole number is usually within a few orders of magnitude of 1. You
converted molecules to moles correctly.
Feedback
C. Incorrect!
This number is too large for moles. Divide the given number with Avogadro’s
number needs. Use the exponent key (EE or EXP) when calculating with scientific
notation.
D. Incorrect!
This number is too large for moles. Divide the given number with Avogadro’s
number needs. Use the exponent key (EE or EXP) when calculating with scientific
notation.
1 mole = 6.02  1023 molecules. Set up the dimensional analysis and cancel the
unit. Run the calculation.
2.85  1026
molecules
Solution
1 mole
6.02  1023
molecules
= _473__ mole
There are total of 473 moles in 2.85 x 1026 molecules, whatever the molecule is.
The correct answer is (B).
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Question No. 2 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
2. The molar mass is defined as the mass of an element or compound divided by its
moles, typically in the unit of g/mol. The molar mass of a compound is equal to the
sum of the atomic masses of the atoms in a molecule. Find the molar mass of
Fe2(SO4)3.
(A)
(B)
(C)
(D)
Question
351.98
344.06
207.77
399.91
g/mol
g/mol
g/mol
g/mol
A. Incorrect!
You multiply the subscripts outside and inside the parenthesis, not add them.
B. Incorrect!
There are two iron atoms, not just one. The sum is miscalculated. Add the atomic
masses of all atoms together. Multiply the subscript outside the parenthesis by the
subscripts inside the parenthesis to get the total atom count.
Feedback
C. Incorrect!
You need to multiply the subscript outside the parenthesis by the subscripts inside
the parenthesis to get the total atom count.
D. Correct!
Good job! You correctly calculated the molar mass of the compound. Add the
atomic masses of all atoms together. Multiply the subscript outside the parenthesis
by the subscripts inside the parenthesis to get the total atom count.
Molar mass is the sum of all atomic masses. Add the atomic masses of all atoms
together. Multiply the subscript outside the parenthesis by the subscripts inside the
parenthesis to get the total atom count.
Fe 2  55.85 = 111.70
S
3  32.07 =
96.21
O 12  16.00 = 192.00
399.91 g/mol
The correct answer is (D).
Solution
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Question No. 3 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
3. The most common conversion in chemistry is mass-to-mole. The mole is defined
as the ratio of the mass (g) and molar mass (g/mol). How many molecules are in
15.75 g of H2O?
Question
(A)
(B)
(C)
(D)
0.874 molecules
4.71  10-22 molecules
5.26  1023 molecules
5.57  1023 molecules
A. Incorrect!
You converted grams to moles, but you need to take it one step farther to get to
molecules. This is a two-step conversion.
B. Incorrect!
Check your calculations again. Convert the mass to mole and apply the Avogadro’s
number to calculate the number of molecules.
Feedback
C. Correct!
Good job! You successfully converted grams to molecules. It is a two-step process.
Convert the mass to mole and apply the Avogadro’s number to calculate the
number of molecules.
D. Incorrect!
There are two hydrogen atoms in a water molecule. Convert the mass to mole and
apply the Avogadro’s number to calculate the number of molecules.
This takes two steps. Convert the mass to mole and apply the Avogadro’s number
to calculate the number of molecules.
Molar mass = grams for 1 mole
1 mole = 6.02  1023 molecules
Hydrogen … 2 molecules  1.01 g/mole H
Oxygen
=
2.02 g/mole of H2
… 1 molecule  16.00 g/mole = 16.00 g/mole
18.02 g/mole
Solution
15.75 g
H2O
1 mole
H2O
18.02 g
H2O
6.02  1023
molecules
H2O
1 mole H2O
= 5.2621023
molecules
H2O
The correct answer is (C).
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Question No. 4 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
4. The percent composition of a component in a compound is the percent of the
total mass of the compound that is due to that component. Find the percent
composition, by mass, of copper in Cu(CH3COO)2?
Question
(A)
(B)
(C)
(D)
50.56
67.55
51.84
34.98
%
%
%
%
A. Incorrect!
Check the calculation carefully. To find percent by mass, divide the mass of copper
by the mass of the entire compound.
B. Incorrect!
Check the calculation carefully. To find percent by mass, divide the mass of copper
by the mass of the entire compound.
Feedback
C. Incorrect!
Check the calculation carefully. You need to distribute the subscript outside the
parenthesis to the atoms inside the parenthesis when finding the whole molar
mass.
D. Correct!
You correctly calculated the percent copper by mass. To find percent by mass,
divide the mass of copper by the mass of the entire compound.
% = (part ÷ whole)  100%
The “part” is the mass of Cu in the compound.
The “whole” is the molar mass of the whole compound
Cu 1  63.55 = 63.55
C 4  12.01 = 48.04
H 6
1.01 = 6.06
O 4  16.00 = 64.00
181.65 g/mole
Solution
% Cu = (63.55  181.65)  100% = 34.98%
The correct answer is (D).
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Question No. 5 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
5. The empirical formula is the simplest formula of a compound. The molecular
formula is the same as or a multiple of the empirical formula. Find the empirical
formula and molecular formula if the sample is composed only of nitrogen and
oxygen. There is 30.4% of nitrogen. The sample molecule has a molar mass of 92
g/mole?
(A) NO2, N2O4
(B) NO2, NO2
Question
(C) NO2, N3O6
(D) N2O4, NO2
A. Correct!
Good job! Calculate the moles of each element and take the simplest ratio for the
empirical formula. Divide the molar mass of the two formulas to get the multiple.
B. Incorrect!
You correctly determined the empirical formula, but the molecular formula's mass is
double that, not the same.
Feedback
C. Incorrect!
You correctly determined the empirical formula, but the molecular formula's mass is
double that, not triple.
D. Incorrect!
You have the empirical and molecular formulas flipped.
Empirical formula is the lowest whole number ratio of the moles of each atom
If percent composition is given, use the percents as grams.
The molecular formula is the actual ratio of atoms—found by finding the ratio of the
empirical formula’s molar mass to the molecular formula’s molar mass.
Solution
30.4 g N
1 mole N
14.01 g N
= 2.17 mole N
69.6 g O
1 mole
16.00 g O
= 4.35 mole O
2.17 mole N
2.17 mole
=1N
4.35 mole N
2.17 mole
=2O
Empirical formula = NO2
Empirical formula’s molar mass:
N 1  14.01 = 14.01
O 2  16.00 = 32.00
46.01 g/mole
92 g/mole
=2
46 g/mole
Molecular formula = NO2  2 = N2O4
The correct answer is (A).
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Question No. 6 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
6. Oxygen constitutes most of the mass of living organisms, because it is their
major constituent. Oxygen, by mass, is the third most abundant element in the
universe, after hydrogen and helium. How many moles of oxygen atoms are in 4
moles of O2?
Question
(A) 2
(B) 4
(C) 6
(D) 8
A. Incorrect!
There are 2 moles of oxygen atoms in each mole of O2, but this problem is asking
about 4 moles of O2.
B. Incorrect!
There are 2 moles of oxygen atoms in each mole of O2.
Feedback
C. Incorrect!
There are 2 moles of oxygen atoms in each mole of O2.
D. Correct!
Good job! There are 2 moles of oxygen atoms in each mole of O2 and this problem
asked about 4 moles of O2 (2*4 = 8).
There are 2 moles of oxygen atoms in each mole of O2 and this problem asked
about 4 moles of O2, which would have 8 oxygen atoms.
The correct answer is (D).
Solution
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Question No. 7 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
7. Dinitrogen pentaoxide is a white crystalline oxide of nitrogen and regarded as
the anhydride (removal of water from an acid) of nitric acid, 2HNO3  N2O5 + H2O.
Calculate the number of moles of oxygen atoms in 1.20 x1025 N2O5 molecules?
Question
(A)
(B)
(C)
(D)
19.9 mol
39.8 mol
6.0 x 1025 mol
99.5 mol
A. Incorrect!
This is the number of moles for the molecule, not O atom. It is a two-step
calculation. Convert the molecular moles to atomic moles.
B. Incorrect!
Almost there! You multiplied the molecular moles by 2. This in fact is the number of
moles for N atoms, not O atoms.
Feedback
C. Incorrect!
Read the question carefully. This is the number of N atoms, not the moles. Check
the reasonableness of the answer from the order of magnitude.
D. Correct!
Good job! This is a two-step process. Convert the number of molecules into the
number of moles for the molecules. Multiply by 5 to get the number of O atoms.
This is a two-step process.
First, you need to calculate the mole number of the molecule N2O5 using the
Avogadro’s number.
#Mole (N2O5) = 1.20x1025/6.02x1023 = 19.9 moles
Secondly, using the molecular formula given, you can relate the moles of the
molecule N2O5 to the moles of the atom O. There are 5 O atoms in every molecule,
i.e. for one mole of N2O5, there is 5 moles of O atoms.
#Mole O = 5 x #Mole (N2O5) = 5 x 19.9 = 99.5 moles
Note that the number of significant figures (3) of the result is the same in the
original data.
Solution
The correct answer is (D).
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Question No. 8 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
8. An empirical formula is a chemical formula that represents the relative
proportions of the elements in a molecule, rather than the actual number of atoms
of the elements. Which of the following is a possible molecular formula with the
empirical formula CH2O?
Question
(A)
(B)
(C)
(D)
C2H5OH
CH3CHO
C6H5COOH
C6H12O6
A. Incorrect!
The empirical formula’s atomic ratio is 1:2:1 for C:O:H. However, the ethanol
C2H5OH has 2:6:1 ratio instead.
B. Incorrect!
The empirical formula’s atomic ratio is 1:2:1 for C:O:H. However, the acetaldehyde
CH3CHO has 2:4:1 ratio instead.
C. Incorrect!
The empirical formula’s atomic ratio is 1:2:1 for C:O:H. However, the benzoic acid
C6H5COOH has 7:6:2 ratio instead.
Feedback
D. Correct!
Good job! This is a molecular formula of glucose. The empirical formula’s atomic
ratio given is 1:2:1 for C:O:H. The glucose C6H12O6 has 6:12:6 ratio, i.e. 1:2:1.
This matches the multiple of the empirical formula. Therefore glucose is the
molecular formula with the empirical formula of CH2O.
Empirical formula is the lowest possible ratio of atoms in a molecule. The C:H:O
ratio is 1:2:1. The molecular formula that is possible would have the same ratio as
1:2:1.
Sum up the total of each of three elements. Exam each option, C6H12O6 is the only
one with 1:2:1 ratio for C:H:O. The rest of the molecules listed do not fit the
empirical formula of CH2O.
The correct answer is (D).
Solution
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Question No. 9 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
9. Carbon monoxide is a colorless, odorless and tasteless gas, but it is toxic to
human and animals. What is the volume of 200.0 grams of carbon monoxide?
Question
(A)
(B)
(C)
(D)
200 L
160 L
714 L
22.4 mL
A. Incorrect!
Check your calculation. This needs two-step process. Convert the mass to moles
and then moles to volume using the molar volume at STP for gases.
B. Correct!
Good job! There are two steps to calculate this. Convert the mass to moles and
then moles to volume using the molar volume at STP for gases.
Feedback
C. Incorrect!
Check your calculation. This needs two-step process. Convert the mass to moles
and then moles to volume using the molar volume at STP for gases.
D. Incorrect!
Check your calculation. This needs two-step process. Convert the mass to moles
and then moles to volume using the molar volume at STP for gases.
This is a two-step process.
#moles (CO) = Mass/MolecularWeight = (200.0 g)/(28.00g/mol) = 7.14 mol
Molar volume Vm is 22.4 L/mol at STP.
V(CO) = #Moles x Vm = 7.14 mol x 22.4 L/mol = 160 L.
The correct answer is (B).
Solution
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Question No. 10 of 10
Instruction: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
10. Dinitrogen monoxide or nitrous oxide, is better known as laughing gas. It is
used as an anesthetic agent in dentistry. What is the mass of 300. liters of
dinitrogen monoxide?
Question
(A) 300 g
(B) 0.121 g
(C) 589.286 g
(D) 589 g
A. Incorrect!
Check your calculation. Use two steps to calculate the mass. Convert the number of
moles and then convert it to the mass using molecular mass.
B. Incorrect!
Check your calculation. Use two steps to calculate the mass. Convert the number of
moles and then convert it to the mass using molecular mass.
Feedback
C. Incorrect!
Check your calculation. Use two steps to calculate the mass. Convert the number of
moles and then convert it to the mass using molecular mass.
D. Correct!
Good job! Use two steps to calculate the mass. Convert the number of moles and
then convert it to the mass using molecular mass. Notice the number of significant
figures in 300. L.
This requires two steps. Be aware of the significant figures in calculation.
The molar volume is 22.4 L/mol at STP. With 300 liters of N2O, the total number of
moles = (300. L)/(22.4 L/mol) = 13.4 mol
Mass (N2O) = Moles x (Molecular Mass) = 13.4 mol x 44.0 g/mol = 589 g
There are three significant figures in the original data. Therefore, the answer should
have in three significant figures too.
The correct answer is (D).
Solution
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