Question for Ashish Siva

Chemistry Level 2

A hydrocarbon , $\ce{C}_\ce{x}\ce{H}_\ce{y}$ contains 10.5 grams of Carbon per gram of Hydrogen. One litre vapours of the hydrocarbon at 127 degrees centigrade and 1 atm pressure weighs 2.8 grams. Find the value of $\dfrac{x}{y}$ .

The answer is 0.875.

1 solution

Ashish Menon
Apr 28, 2016

First lets find the empirical formula.

Element	No. of moles	Simplest ratio	Simplest whole number ratio
Carbon	10.5 ÷ 12 = 0.875	0.875 ÷ 0.875 = 1	1 × 7 = 7
Hydrogen	1 ÷ 1 = 1	1 ÷ 0.875 = 1.142	1.142 × 7 = 8

So, the empirical formula of the given compound is $C_7H_8$

Number of moles of the hydrocarbon = $\dfrac{\text{Given weight}}{\text{Molecular weight}}$
Let the molecular weight be $z × \text{Empirical weight}$
Now applying the formula:-
$PV = nRt\\ 1 × 1 = \dfrac{2.8 × 0.082 × (127 + 273)}{z × ((7 × 12) + (8 × 1))}\\ z = \dfrac{2.8 × 0.082 × 400}{92}\\ z = 0.998\\ z \approx 1$

This proves that the Empirical formula is equal to the molecular formula. So, the molecular formula of the compound is $C_7H_8$ .

$\therefore$ $\dfrac{x}{y} = \dfrac{7}{8} = \boxed{0.875}$

Thope! You are