Question

An unknown gaseous hydrocarbon consists of 85.63% carbon by mass. A 0.100-g sample of the gas occupies a volume of 0.0533 L at STP. What is the identity of the gas? A. C4H8 B. C3H6 C. C2H4 D. C5H10 E. CH2

Answer 1

% of carbon = 85.63%

% of Hydrogen = 100 - 85.63 = 14.37%

Calculate atomic ratio by dividing % composition by atomic mass.

Mass ratio of carbon = 85.63/12 = 7.13

Mass ratio of hydrogen = 14.37/1 = 14.37

Now calculate simplest ratio by dividing atomic ratio by smallest atomic ratio

Simplest ratio of carbon = 7.13/7.13 = 1

Simplest ratio of hydrogen = 14.37/7.13 = 2.015 = 2

Empirical formula of compound = CH₂

According to avogadro's law 1 mole occupy volume 22.414 L

then 0.0533 L = 0.0533/22.414 = 0.002377978 mole

0.002377978 mole = 0.100 gm then 1 mole = 1 $\times$ 0.100 / 0.002377978 =42.05 gm

molar mass of gas = 42.05 gm

Empirical formula mass = (12$\times$1) + (1$\times$2)  = 14

n = molecular formula mass / Empirical formula mass = 42.05/14 = 3

Molecular formula = n $\times$ Empirical formula = 3 $\times$ Empirical formula

Molecular formula of compound = C₃H₆

Answer - B. C₃H₆