Download Empirical and Molecular Formulas

CHEM 110
BEAMER
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Date
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Practice Work 28: Empirical and Molecular Formulas
•
You should do this on your own paper.
•
I suggest a sideways format to the paper.

This assignment is not due.
•
Appendix 11 (available on the website) will be helpful.

Remember that you will not have the appendix with the steps for the Final.

Also remember that the first step of finding the molecular formula is to find the empirical
formula.
I apologize about this blank page. It was the only way to get all of the solutions to fit nicely on
pages of their own.
Also, the solution sets are highly color coded. I hope that this is helpful, instead of distracting.
I put the answer key on the first page to save space.
-- beamz
Answers (Solutions start on page 3)
Question
Question
Question
Question
51)
52)
145)
150)
C6H6O2
C4H10
C13H18O2 (The empirical formula is identical to the molecular formula)
C3H8O3
Page 1 of 6
CHEM 110
BEAMER
PAGE 335, QUESTION 51
51)
Analysis of a chemical compound used in photographic developing fluid indicates a chemical composition of 65.45% C, 5.48% H,
and 29.08% O. The molecular mass is found to be 110.0 g/mol. Determine the molecular formula. 10 min maximum
PAGE 335, QUESTION 52
52)
A compound was found to contain 49.98 g carbon and 10.47 g hydrogen. The molecular mass of the compound is 58.12 g/mol.
Determine the molecular formula. 8 min max
PAGE 349, QUESTION 145
145)
Determine the molecular formula for ibuprofen, a common headache remedy. Analysis of ibuprofen yields a molar mass of 206
g/mol and a percent composition of 75.7% C, 8.80% H, and 15.5% O. Determine the molecular formula. 10 minutes maximum.
PAGE 349, QUESTION 150
150)
Glycerol is a thick, sweet liquid obtained as a byproduct of the manufacture of soap. Its percent composition is 39.12% C,
8.75% hydrogen, and 52.12% oxygen. The molecular mass is 92.11 g/mol. What is the molecular formula for glycerol?
Page 2 of 6
CHEM 110
BEAMER
Part 1: Molecular Formulas
PAGE 335, QUESTION 51
51)
Analysis of a chemical compound used in photographic developing fluid indicates a chemical composition of 65.45% C, 5.48% H,
and 29.08% O. The molecular mass is found to be 110.0 g/mol. Determine the molecular formula. 10 min maximum
DETERMINING MOLECULAR FORMULA
Step 1:
The molecular mass was given.  (110.0 g/mol)
Step 2:
Determine the empirical formula
DETERMINING EMPIRICAL FORMULA
Step 1:
Mass wasn’t given. Go to Step 2.
Step 2
(
Step 3
65.45 g C
)
1
(
5.48 g H
)
1
(
(
29.08 g O
(
)
1
Step 9:
Step 4
1 mol C
)
12.01 g C
=
1 mol H
)
1.01 g H
=
1 mol O
(
)
16.00 g O
=
2.998 mol C
2.99 mol H
1.000 mol O
Step 5
(
Step 6:
2.998 mol C
)
1.000 mol

2.99 mol H
)
1.000 mol

(
1.000 mol O
(
)
1.000 mol
Step 8
All values
2.998 C

3

3

1
are very
close to
2.99 H
whole
numbers.

1.000 O
Skip Step 7
C3H3O1
Step 3:
Empirical Mass = (3 × 12.01 g/mol) + (3 × 1.01 g/mol) + (1 × 16.00 g/mol)
Step 4:
(
molecular mass
110.0 g/mol
) = (
) = 1.998 ≈
empirical mass
55.06 g/mol
0
20
1.998 is very close to a whole
number. Continue.
= 55.06 g/mol
Step 5:
(C3H3O1) 2
0
0

C6H6O2
answer
Page 3 of 6
CHEM 110
BEAMER
PAGE 335, QUESTION 52
52)
A compound was found to contain 49.98 g carbon and 10.47 g hydrogen. The molecular mass of the compound is 58.12 g/mol.
Determine the molecular formula. 8 min max
DETERMINING MOLECULAR FORMULA
Step 1:
The molecular mass was given.  (58.12 g/mol)
Step 2:
Determine the empirical formula
DETERMINING EMPIRICAL FORMULA
Step 1:
Mass was given. Skip to Step 2.
Mass Values
(
49.98 g C
)
1
Step 3
(
1 mol C
)
12.01 g C
Step 4
=
4.162 mol C
Step 5
(
4.162 mol C
)
4.162 mol
Step 6:

1.000 C
Step 7:
Values are
not close to
Step 8
× 2

2
× 2

5
whole
(
10.47 g H
)
1
(
1 mol H
)
1.01 g H
=
10.37 mol H
(
10.37 mol H
)
4.162 mol

2.492 H
numbers.
Must use a
coefficient.
Step 9:
C2H5
Step 3:
Empirical Mass = (2 × 12.01 g/mol) + (5 × 1.01 g/mol) = 29.07 g/mol
Step 4:
(
molecular mass
58.12 g/mol
) = (
) = 1.999 ≈
empirical mass
29.07 g/mol
0
20
1.999 is very close to a whole
number. Continue.
Step 5:
(C2H5) 2
0
0

C4H10
answer
Page 4 of 6
CHEM 110
BEAMER
PAGE 349, QUESTION 145
145)
Determine the molecular formula for ibuprofen, a common headache remedy. Analysis of ibuprofen yields a molar mass of 206
g/mol and a percent composition of 75.7% C, 8.80% H, and 15.5% O. Determine the molecular formula. 10 minutes maximum.
DETERMINING MOLECULAR FORMULA
Step 1:
The molecular mass was given.  (206 g/mol)
Step 2:
Determine the empirical formula
DETERMINING EMPIRICAL FORMULA
Step 1:
Mass wasn’t given. Go to Step 2.
Step 2
75.7 g C
)
1
(
8.80 g H
)
1
(
(
(
(
Step 9:
Step 3
15.5 g O
)
1
(
Step 4
1 mol C
)
12.01 g C
=
1 mol H
)
1.01 g H
=
8.71 mol H
1 mol O
)
16.00 g O
=
0.969 mol O
6.30 mol C
Step 5
6.30 mol C
)
0.969 mol

8.71 mol H
)
0.969 mol

0.969 mol O
)
0.969 mol

(
(
(
Step 6:
6.51 C
Step 7:
Values are
× 2

13
× 2

18
× 2

2
not close to
8.99 H
whole
numbers.
Must use a
1.000 O
coefficient.
Step 8
C13H18O2
Step 3:
Empirical Mass = (13 × 12.01 g/mol) + (18 × 1.01 g/mol) + (2 × 16.00 g/mol)
Step 4:
(
molecular mass
206 g/mol
) = (
) = 0.998 ≈
empirical mass
206.31 g/mol
0
10
0.998 is very close to a whole
number. Continue.
= 206.31 g/mol
Step 5:
(C13H18O2) 1
0
0

C13H18O2
answer
Remember: The molecular formula can be the same as the empirical formula. Don’t let that throw you.
Page 5 of 6
CHEM 110
BEAMER
PAGE 349, QUESTION 150
150)
Glycerol is a thick, sweet liquid obtained as a byproduct of the manufacture of soap. Its percent composition is 39.12% C,
8.75% hydrogen, and 52.12% oxygen. The molecular mass is 92.11 g/mol. What is the molecular formula for glycerol?
DETERMINING MOLECULAR FORMULA
Step 1:
The molecular mass was given.  (92.11 g/mol)
Step 2:
Determine the empirical formula
DETERMINING EMPIRICAL FORMULA
Step 1:
Mass wasn’t given. Go to Step 2.
Step 2
(
39.12 g C
)
1
(
8.75 g H
)
1
(
(
(
Step 9:
Step 3
52.12 g O
)
1
(
Step 4
1 mol C
)
12.01 g C
=
1 mol H
)
1.01 g H
=
8.66 mol H
1 mol O
)
16.00 g O
=
3.257 mol O
3.257 mol C
Step 5
3.257 mol C
)
3.257 mol

8.66 mol H
)
3.257 mol

3.257 mol O
)
3.257 mol

(
(
(
Step 6:
1.000 C
Step 7:
Values are
Step 8
× 3

3
× 3

8
× 3

3
not close to
2.66 H
whole
numbers.
Must use a
1.000 O
coefficient.
C3H8O3
Step 3:
Empirical Mass = (3 × 12.01 g/mol) + (8 × 1.01 g/mol) + (3 × 16.00 g/mol)
Step 4:
(
molecular mass
92.11 g/mol
) = (
) =
empirical mass
92.11 g/mol
0
10
0.998 is very close to a whole
number. Continue.
= 92.11 g/mol
Step 5:
(C3H8O3) 1
0
0

C3H8O3
answer
Page 6 of 6