A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.

Explanation

Assuming 1 gm. Of gas is burnt
The weight % of `C` = `(3.38xx12)/(44)xx100=92.18%`
The weight % of .H. = `(0.69xx2)/(18)=7.67%`

Experical formula of compound = `C_(1)H_(1)`
(ii) Given 10 lit of gas at STP weighs - 11.6 gas
22.44 lit of gas at STP weights = `(22.4xx11.6)/(10)=25.984`
`:.` Molecular weight of given gas = 25.984
(iii) Molecular formula = n (emperical formula)
n= `("Mol wt")/("Emp.wt")=(25.984)/(13)=2` :. Molecular formula = `2(CH)=C_(2)H_(2)`.