Honors Chemistry
Unit 5 Sample Problems – Key
Page 1 of 3
These are typical questions similar to (or, maybe even the same as) those that will be used on the first semester exam.
1. The SI base unit for the amount of material is the _____.
c. mole
2. The molar mass of a chemical compound is the mass of
______ of the compound.
b. 6.02 × 1023 molecules
3. The atomic mass of a chemical element is the mass of
______ of the element.
b. 6.02 × 1023 atoms
4. The term “STP” as used in this section of the course
means ______.
c. “Standard Temperature and Pressure”
5. One mole of atoms of neon gas has a volume of ______
at STP.
d. 22.4 L
6. The value of temperature at STP is ______°C.
a. 0.00
7. The value of pressure at STP is ______ atm.
b. 1.00
8. What is the volume of 2.00 mole of oxygen gas, O2, at
STP?
d. 44.8 L
9. What is the molar mass of methane, CH4?
c. 16.0 g/mol
10. What is the molar mass of benzene, C6H6?
b. 78.0 g/mol
11. What is the molar volume of ethane gas, C2H6, at STP?
a. 22.4 L
12. The volume of a sample of hydrogen gas, H2, at STP is
measured to be 11.2 L. How many moles of H2 are
there?
c. 0.50
15. What is the volume of 1.20 mols of acetylene gas, C2H2,
at STP?
d. 27.8 L
16. What is the molar mass of acetylene gas, C2H2?
a. 26.0 g/mol
17. What is the mass of 0.50 mols of C2H2?
b. 13.0 g
18. How many mols are in 39.0 g of acetylene gas, C2H2?
c. 1.50 mol
19. To find the percent abundance of C in propane gas, C3H8,
which of the following calculations would we use?
𝟑×𝟏𝟐.𝟎𝟏𝟏 𝐠
a. %𝐂 = ((𝟑×𝟏𝟐.𝟎𝟏𝟏
) × 𝟏𝟎𝟎%
𝐠)+(𝟖×𝟏.𝟎𝟎𝟖 𝐠)
20. To find the percent abundance of O in glucose, C6H12O6,
which of the following calculations would we use?
b. %𝐂 = ((𝟔×𝟏𝟐.𝟎𝟏𝟏
𝟔×𝟏𝟓.𝟗𝟗𝟗 𝐠
21. A compound of unknown chemical formula is known to
have an empirical formula of CH2O. It is also known to
have a molar mass of 120.104 g/mol. What is its
molecular formula?
d. C4H8O4
22. A compound of unknown chemical formula is known to
have an empirical formula of KO. It is also known to have
a molar mass of 110.194 g/mol. What is its molecular
formula?
b. K2O2
23. In order to determine the molecular formula of a chemical
compound from its empirical formula, you would need to
know which of the following about the chemical
compound?
c. The molar mass of the compound
13. The volume of a sample of carbon dioxide gas, CO2, at
STP is measured to be 44.8 L. How many moles of CO2
are there?
a. 2.00
14. The volume of a sample of methane gas, CH4, at STP is
measured to be 33.6 L. How many moles of CH4 are
there?
b. 1.50
24.
How many particles are in 1 mole of particles?
6.02 × 1023 particles
25.
What is Avogadro’s Number and how is it used?
Avogadro’s Number is the number of particles in one mole of particles.
26.
) × 𝟏𝟎𝟎%
𝐠)+(𝟏𝟐×𝟏.𝟎𝟎𝟖 𝐠)+(𝟔×𝟏𝟓.𝟗𝟗𝟗𝒈)
Give the definition of the molar mass of a chemical compound.
Molar mass is the mass of a compound that is equal to the mass of 6.02 × 1023 particles of that compound.
Honors Chemistry
27.
Unit 5 Sample Problems – Key
Page 2 of 3
What is the relationship between the mass, the number of mols, and the molar mass of a compound?
mass = (# of mols)(molar mass), molar mass = (mass)/( # of mols); # of mols = (mass)/(molar mass)
28.
Give the definition of the molar volume of a gaseous chemical compound.
Molar volume is the volume of a gaseous compound that contains 6.02 × 1023 particles of that compound at STP.
29.
What is the relationship between the volume, the number of mols, and the molar volume of a gaseous compound?
volume = (# of mols)(molar volume), molar volume = (volume)/( # of mols); # of mols = (volume)/(molar volume)
30.
What are the conditions at STP?
The conditions at STP are a temperature of 0°C (273.16 K) and a pressure of 1 atm (760 mm Hg or 101.3 kPa)
31.
Distinguish between the empirical mass and the molar mass of a compound.
The empirical mass is the molar mass of the empirical formula and the molar mass is the mass of the molecular formula
of the compound.
32.
33.
34.
35.
Determine the molar mass for each of the following compounds.
a. FeI2
309.653 g/mol
b. CaCO3
100.086 g/mol
c. NaC2H4O2
83.042 g/mol
d. C6H6
78.114 g/mol
Determine the number of moles for each of the masses of the following compounds.
a. 100 g of FeI2
n = m/M = (100 g)/(309.653 g/mol)
n = 0.323 mol
b. 0.125 g of CaCO3
n = m/M = (0.125 g)/(100.086 g/mol)
n = 0.00125 mol
c. 500 g of NaC2H4O2
n = m/M = (500 g)/(83.042 g/mol)
n = 6.02 mol
d. 19.5 g of C6H6
n = m/M = (19.5 g)/(78.114 g/mol)
n = 0.250 mol
Determine the mass for the amounts of each of the following compounds.
a. 1.50 mol of FeI2
m = n×M = (1.50 mol)(309.653 g/mol)
m = 464 g
b. 0.125 mol of CaCO3
m = n×M = (0.125 mol)(100.086 g/mol)
m = 12.5 g
c. 5.00 mol of NaC2H4O2
m = n×M = (5.00 mol)(83.042 g/mol)
m = 415 g
d. 12.1 mol of C6H6
m = n×M = (12.1 mol)(78.114 g/mol)
m = 945 g
Determine the number of mols for the following volumes of gases at STP.
a. 44.8 L of CO2
n = V/V = (44.8 L)/(22.4 L/mol)
n = 2.00 mol
b. 16.2 L of O2
n = V/V = (16.2 L)/(22.4 L/mol)
n = 0.723 mol
c. 5.00 L of C2H4
n = V/V = (5.00 L)/(22.4 L/mol)
n = 0.223 mol
d. 175 L of C2H2
n = V/V = (175 L)/(22.4 L/mol)
n = 7.81 mol
Honors Chemistry
36.
36.
36.
Unit 5 Sample Problems – Key
Page 3 of 3
Determine the volume for the following number of mols of gases at STP.
a. 0.125 mol of CO2
V = n×V = (0.125 mol)(22.4 L/mol)
V = 2.80 L
b. 3.55 mol of O2
V = n×V = (3.55 mol)(22.4 L/mol)
V = 79.5 L
c. 2.72 × 10–3 mol of C2H4
V = n×V = (2.72 × 10–3 mol)(22.4 L/mol)
V = 0.0609 L = 6.09 × 10–2 L
d. 52.5 mol of C2H2
V = n×V = (52.5 mol)(22.4 L/mol)
V = 1170 L
Determine the percent abundance for each of the elements in the following compounds.
a. CO2
M = 44.009 g/mol
%C = [(1 × 12.011)/(44.009)] × 100 % = 27.292%
%O = [(2 × 15.999)/(44.009)] × 100 % = 27.292%
b. C6H12O6
M = 180.156 g/mol
%C = [(6 × 12.011)/(180.156)] × 100 % = 40.002%
%H = [(12 × 1.008)/(180.156)] × 100 % = 6.714%
%O = [(6 × 15.999)/(180.156)] × 100 % = 53.284%
c. CaCO3
M = 100.086 g/mol
%Ca = [(1 × 40.078)/(100.086)] × 100 % = 40.044%
%C = [(1 × 12.011)/(100.086)] × 100 % = 12.001%
%O = [(3 × 15.999)/(100.086)] × 100 % = 47.956%
d. C2H2
M = 26.038 g/mol
%C = [(2 × 12.011)/(26.038)] × 100 % = 92.215%
%H = [(2 × 1.008)/(26.038)] × 100 % = 7.743%
Determine empirical formulas for each of the following compounds with the indicated percent abundances.
a. 74.87% C and 25.13% H
nC = mC/MC = 74.87 g/12.011 g/mol = 6.233 mol
nH = mH/MH = 25.13 g/1.008 g/mol = 24.93 mol
b. 40.00% C, 6.71% H, and 53.29% O
nC = mC/MC = 40.00 g/12.011 g/mol = 3.330 mol
nH = mH/MH = 6.71 g/1.008 g/mol = 6.657 mol
nO = mO/MO = 53.29 g/15.999 g/mol = 3.330 mol
nH/nC = 24.93/6.233 = 4.000/1 = 4/1
nH/nC = 6.657/3.330 = 1.999/1 = 2/1
CH4
CH2O
c. 40.04% Ca, 12.00% C, and 47.96% O
nCa = mCa/MCa = 40.04 g/40.078 g/mol = 0.9991 mol
nC = mC/MC = 12.00 g/12.011 g/mol = 0.9991 mol
nO = mO/MO = 47.96 g/15.999 g/mol = 2.998 mol
d. 27.05% Na, 16.48% N, and 56.47% O
nNa = mNa/MNa = 27.05 g/22.990 g/mol = 1.177 mol
nN = mN/MN = 16.48 g/14.007 g/mol = 1.177 mol
nO = mO/MO = 56.47 g/15.999 g/mol = 3.530 mol
nO/nCa = 2.998/0.9991 = 3.000/1 = 3/1
nO/nNa = 3.530/1.177 = 2.999/1 = 3/1
CaCO3
NaNO3