# A Gaseous Hydrocarbon Contains 82.76% Of Carbon. Given That Its Vapour Density Is 29,Find Its Molecular Formula. [C = 12, H =1].

The molecular mass of carbon (C) is 12

The %composition of carbon (C) is 82.76

Total number of carbon atoms is 1 * 2 = 2

Number of carbon atoms present = $$\frac{82.76}{12} = 6.89$$

The simplest ratio =

$$\Rightarrow \frac{6.89}{6.89} = 1$$

The molecular mass of hydrogen (H )is 1

The %composition of carbon (C) is 17.24

Total number of carbon atoms is 1 * 1 = 1

Number of carbon atoms present = $$\frac{17.24}{1} = 17.24$$

The simplest ratio =

$$\Rightarrow \frac{17.24}{6.89} = 2.5$$

Empirical fromula is $$C_{2}H_{5}$$

Empirical formula mass = 2 * 12 + 1 * 5 = 24 + 5 = 29

There, the vapour density is 29

The molecular mass – 2 * vapour density

= 2 * 29 = 58

$$n = \frac{Molecula rmass}{Empirical formula mass}$$ $$\Rightarrow \frac{58}{29} = 2$$

Therefore, the molecular formula is $$C_{4}H_{10}$$

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