Samples: Free-Response Questions

1. An unknown hydrocarbon is burned in the presence of oxygen in order to determine its
   empirical formula. Another sample of the hydrocarbon is subjected to colligative
   property tests in order to determine its molecular mass.

(a) Calculate the empirical formula of the hydrocarbon, if upon combustion at

STP, 9.01 grams of liquid H 2 O and 11.2 liters of CO 2 gas are produced.

(b) Determine the mass of the oxygen gas that is used.

(c) The hydrocarbon dissolves readily in CCl 4. A solution prepared by mixing 135

grams of CCl 4 and 4.36 grams of the hydrocarbon has a boiling point of 78.7°C.

The molal boiling-point-elevation constant of CCl 4 is 5.02°C/molal, and its normal

boiling point is 76.8°C. Calculate the molecular weight of the hydrocarbon.

(d) Determine the molecular formula of the hydrocarbon.

Answer

1. Given: Unknown hydrocarbon is combusted in O 2 and subjected to colligative tests.

(a) Given: 9.01 g H 2 O + 11.2 liters CO 2 produced.

Restatement: Find empirical formula of hydrocarbon.

.

.

/mole.

HO

18 02 mole H O

901

g 0 500

g 2

= 2

.

. mole H O
mole H O

mole H mole H

11

0 500 (^22100)
2

=

.
.
liters mole/.
liters CO
22 4 mole CO
11 2
0 500
2
= 2
.

. mole CO
mole CO

mole C mole C

11

0 500 (^21) 0 500
2

=

empirical formula = C0.5H 1 →CH 2
(b) Restatement: Calculate mass of O 2 required for complete combustion.
2 CH 2 (g) + 3O 2 (g) →2 CO 2 (g) + 2 H 2 O(,)

..
.

mole H O

moles H O

moles O

mole O

gO

1 gO

0 500

2

3

1

32 00

240

2

2

2

2

2

##= 2

(c) Given:

• The unknown hydrocarbon dissolves in CCl4.


• 135 grams of CCl 4 + 4.36 grams of hydrocarbon.


• B.P. of CCl 4 = 76.8°C


• New B.P. = 78.7°C

Part II: Specific Topics