Download Chemical Composition CW These problems are to be done in your

Chemical Composition CW
These problems are to be done in your notebook!
1. Determine the percent composition of propane, C3H8.
2. Write the molecular formula of a compound with an empirical formula of C2OH4
and a molar mass of 88.105 grams per mole.
3. The percentage composition of acetic acid is found to be 39.9% C, 6.7% H, and
53.4% O. Determine the empirical formula of acetic acid.
4. The molar mass for question #3 was determined by experiment to be 60.0 g/mol.
What is the molecular formula?
5. Calculate the mass percent of carbon, nitrogen and oxygen in acetamide,
C2H5NO.
6. A compound with an empirical formula of C4H4O and a molar mass of 136 grams
per mole.
7. A 50.51 g sample of a compound made from phosphorus and chlorine is
decomposed. Analysis of the products showed that 11.39 g of phosphorus atoms
were produced. What is the empirical formula of the compound?
8. When 2.5000 g of an oxide of mercury, (HgxOy) is decomposed into the elements
by heating, 2.405 g of mercury are produced. Calculate the empirical formula.
9. The compound benzamide has the following percent composition. What is the
empirical formula?
C = 69.40 % H= 5.825 % O = 13.21 % N= 11.57 %
10. A component of protein called serine has an approximate molar mass of 100
g/mole. If the percent composition is as follows, what is the empirical and
molecular formula of serine?
C = 34.95 % H= 6.844 % O = 46.56 % N= 13.59 %
Chemical Composition CW
1. Determine the percent composition of propane, C3H8.
3C
3(12.011)
36.033
(100) = 81.714% C
(100) =
(100) =
C3 H 8
3(12.011) + 8(1.0079)
44.0962
8H
8(1.0079)
(100) =
(100) = 18.285% H
C3 H 8
3(12.011) + 8(1.0079)
2. A compound with an empirical formula of C2OH4 and a molar mass of 88.105
grams per mole.
Mass of empirical formula = EF = 2(12.011) + 15.999 + 4(1.0079) = 44.0526 g/mol
88.105
2(C2OH4) = C4O2H8
=2
44.0526
3. The percentage composition of acetic acid is found to be 39.9% C, 6.7% H, and
53.4% O. Determine the empirical formula of acetic acid.
39.9
6.7
53.4
= 3.32 mol C
= 6.6 mol H
= 3.34 mol O
12.011
1.0079
15.999
3.32
6.6
3.34
=1
=2
=1
3.32
3.32
3.32
CH2O
4. The molar mass for question #3 was determined by experiment to be 60.0 g/mol.
What is the molecular formula?
EF = 12.011 + 2(1.0079) + 15.999 = 30.0258 g/mol
60.0 / 30.0258 = 2
C2H4O2
5. Calculate the mass percent of carbon, nitrogen and oxygen in acetamide,
C2H5NO.
24.022
5.0395
(100) = 40.669 % C
(100) = 8.5318 % H
59.0675
59.0675
14.007
15.999
(100) = 23.714 % N
(100) = 27.086 % O
59.0675
59.0675
6. A compound with an empirical formula of C4H4O and a molar mass of 136 grams
per mole.
136
EF = 68.0746
=2
C8H8O2
68.0746
7. A 50.51 g sample of a compound made from phosphorus and chlorine is
decomposed. Analysis of the products showed that 11.39 g of phosphorus atoms
were produced. What is the empirical formula of the compound? PCl3
50.51 – 11.39 = 39.12 g Cl
11.39
39.12
= 0.3677 mol P
= 1.103 mol Cl
30.974
35.453
1.103

PCl3
= 2.9997
0.3677
8. When 2.5000 g of an oxide of mercury, (HgxOy) is decomposed into the elements
by heating, 2.405 g of mercury are produced. Calculate the empirical formula.
2.5000 – 2.405 = 0.095 g O
2.405
0.095
= 0.0119896 mol Hg
= 0.00593787 mol O
200.59
15.999
0.0119896

Hg2O
=2
0.00593787
9. The compound benzamide has the following percent composition. What is the
empirical formula?
C = 69.40 % H= 5.825 % O = 13.21 % N= 11.57 % C7H7NO
10. A component of protein called serine has an approximate molar mass of 100
g/mole. If the percent composition is as follows, what is the empirical and
molecular formula of serine?
C = 34.95 % H= 6.844 % O = 46.56 % N= 13.59 %
C3H7NO3 empirical formula
C3H7NO3 molecular formula