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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