CHEM 110 (BEAMER)
Solutions Set
PW46B
Last Name
Do Date
First Name
Section
M
T
W
R
PW46B: Mole Calculations – Solutions Set
Mass  Mole  Particles
General Information

Do Date: M (4/25)

Relevant Notes: Notes-52, Notes-53; Relevant Reading: Chapters 8.1 – 8.4
Mass
molar mass
⇔
Avogadro' s Number
Moles ⇔
Particles

Note:
For the first three questions, I will “break the question into individual pieces” for you. On tests, the questions will be written like Questions 4 – 6, and
you will be expected to see the pieces yourselves.
1)
A sample of carbon dioxide contains 5.6 × 1022 molecules of carbon dioxide. Calculate the mass (in grams) of carbon dioxide.
1A)
First, calculate the molar mass of carbon dioxide (you do not have to show your work):
44.01 g/mol CO2
1B)
Next, write your answer to Question 1A in conversion factor form:
44.01 g CO2 = 1 mol CO2
1C)
Next, write the conversion factor between moles of carbon dioxide and molecules of carbon dioxide:
1 mol CO2 = 6.022 × 1023 CO2 molec
1D)
Use the conversion factors that you have written above to solve the problem
(
particles
particles to moles
moles to mass
initial value
Avogadro’s Number
molar mass
5.6 × 10
22
CO2 molec
)
1
(
1 mol CO2
6.022 × 10
23
CO2 molec
)
(
44.01 g CO2
)
1 mol CO2
final answer, sigdigs, units
=
x4.1 g CO2 x
Page 1 of 5
CHEM 110 (BEAMER)
Solutions Set
PW46B
2)
A sample of lye contains 500.0 grams of sodium hydroxide. Calculate the number of particles of sodium hydroxide in the sample.
2A)
First, calculate the molar mass of sodium hydroxide (you do not have to show your work):
40.00 g/mol NaOH
2B)
Next, write your answer to Question 2A in conversion factor form:
40.00 g NaOH = 1 mol NaOH
2C)
Next, write the conversion factor between moles of carbon dioxide and molecules of carbon dioxide:
1 mol NaOH = 6.022 × 1023 NaOH form
2D)
Last, use the conversion factors that you have written above to solve the problem
(
Note:
mass
mass to moles
moles to particles
initial value
molar mass
Avogadro’s Number
500.0 g NaOH
)
1
(
1 mol NaOH
)
40.00 g NaOH
23
6.022 × 10 NaOH form
(
)
1 mol NaOH
final answer, sigdigs, units
=
x7.528 × 10
24
NaOH formx
I have broken Problems 3 through 6 into “parts” just like I showed for Questions 1 and 2. Unless I state otherwise, you do not have to write out the
individual conversion factors. For those of you who are more “methodical,” you might need to write them down in order to get to the correct answer.
Don’t skip steps if you are getting the wrong answers.
Page 2 of 5
CHEM 110 (BEAMER)
Solutions Set
PW46B
3)
Calculate the number of particles of CCl4 in 400. mL CCl4. The density of CCl4 is 1.59 g/mL.
3A)
First, calculate the molar mass of CCl4 (you do not have to show your work):
153.82 g/mol CCl4
3B)
Next, write your answer to Question 3A in conversion factor form:
153.82 g CCl4 = 1 mol CCl4
3C)
Second, write the conversion factor between moles of CCl4 and molecules of CCl4:
1 mol CCl4 = 6.022 × 1023 CCl4 molec
3D)
Third, write the conversion factor for the density of CCl4:
1.59 g CCl4 = 1 mL CCl4
3E)
Last, use the conversion factors that you have written above to solve the problem:
(
volume
volume to mass
mass to moles
moles to particles
initial value
density
molar mass
Avogadro’s Number
400. mL CCl4
)
1
(
1.59 g CCl4
)
1 mL CCl4
(
1 mol CCl4
)
153.82 g CCl4
23
(
6.022 × 10 CCl4 molec
)
1 mol CCl4
final answer, sigdigs, units
=
x2.49 × 10
24
CCl4 molecx
Page 3 of 5
CHEM 110 (BEAMER)
Solutions Set
PW46B
4)
A sample of hydrogen gas contains 1.204 × 1028 molecules of hydrogen. Calculate the mass of hydrogen in kilograms. (You must show the conversion
of grams to kilograms in your dimensional analysis work.)
4A)
First, calculate the molar mass of hydrogen gas (you do not have to show your work):
2.02 g/mol H2
4B)
Next, write your answer to Question 4A in conversion factor form:
2.02 g H2 = 1 mol H2
4C)
Next, write the conversion factor between moles of hydrogen gas and molecules of hydrogen gas:
1 mol H2 = 6.022 × 1023 H2 molec
4D)
Last, use the conversion factors that you have written above to solve the problem
(
particles
particles to moles
moles to mass
g to kg
initial value
Avogadro’s Number
molar mass
g to kg
1.204 × 10
28
1
H2 molec
)
(
1 mol H2
6.022 × 10
23
H2 molec
)
(
2.02 g H2
)
1 mol H2
(
final answer, sigdigs, units
1 kg H2
)
1000 g H2
=
x40.39 kg H2 x
5)
The price of gold, as of April 16, 2006, is $50.80 per gram (treat this as a conversion factor). Determine the number of atoms of gold present in a
sample of gold that costs $285,000 dollars. (Assume three sigdigs).
5A)
First, find the molar mass of gold (you do not have to show your work):
197.0 g/mol Au
5B)
Next, write your answer to Question 5A in conversion factor form:
197.0 g Au = 1 mol Au
5C)
Second, write the conversion factor between moles of Au and atoms of Au:
1 mol Au = 6.022 × 1023 Au atoms
5D)
Third, write the conversion factor for the price of Au:
$285,000 = 1 g Au
5E)
Last, use the conversion factors that you have written above to solve the problem:
cost
cost to weight
mass to moles
moles to particles
initial value
price
molar mass
Avogadro’s Number
(
$285,000
)
1
(
1 g Au
)
$50.80
(
1 mol Au
)
197.0 g Au
23
(
6.022 × 10 Au atoms
)
1 mol Au
final answer, sigdigs, units
=
x1.71 × 10
22
Au atomsx
Page 4 of 5
CHEM 110 (BEAMER)
Solutions Set
PW46B
6)
This is an example of an “extra credit” question: A chemical pipeline is leaking ammonia gas at a rate of 4.27 × 1026 molecules per minute (assume this
is a measured value). Convert the rate of leakage from molecules per minute to metric tons (t) per day. Note: 1000 kg = 1 metric ton (t)
6A)
First, find the molar mass of ammonia (you do not have to show your work):
17.03 g/mol NH3
6B)
Next, write your answer to Question 6A in conversion factor form:
197.0 g NH3 = 1 mol NH3
6C)
Second, write the conversion factor between moles of NH3 and molecules of NH3:
1 mol NH3 = 6.022 × 1023 NH3 atoms
6D)
In the margins somewhere, you should write out the time conversion factors:
60 min = 1 hr
6E)
Last, use the conversion factors that you have written – and the given conversion factors – to solve the problem:
(
initial rate
particles to moles
moles to mass
g to kg
kg to t
initial value
Avogadro’s Number
molar mass
g to kg
kg to t
1 mol NH3
(
)
23
6.022 × 10 NH3 molec
17.03 g NH3
(
)
1 mol NH3
1 kg NH3
(
)
1000 g NH3
1 t NH3
(
)
1000 kg NH3
60 min
(
)
1 hr
24 hr
(
)
1 day
(4 pts)
(4 pts)
(4 pts)
(4 pts)
(4 pts)
(4 pts)
4.27 × 10
26
NH3 molec
)
min
(4 pts)
min to hr
24 hr = 1 day
hr to day
final answer, sigdigs, units
time conversions
=
x17.4 t NH3 /dayx
(3 pts)
If you set up the initial value like this, you will receive zero credit for this type of problem.
26
(
4.27 × 10 H2 /min
)
1

It’s the wrong way to set up this problem, and it’s important that you learn this.
This is one of the few instances that I do not give partial credit.
This will be on the test in the normal section…as well as in the extra credit section.
Page 5 of 5