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

No. of moles of `CO_(2)=3.38/44=0.07682`
No. of moles of `H_(2)O=0.69/18=0.03833`
Ratio of the moles of `CO_(2)=H_(2)O`
= 0.07682 : 0.03833 = 2 : 1
`therefore` The ratio of carbon and hydrogen atoms is 1 : 1 (because 1 `CO_(2)=H_(2)O`)
Empirical formula = CH
10.0 L at STP weigh 11.6 g
22.4 L at STP weigh ?
`(22.4xx11.6)/10=26`
`therefore` Molecular wt. of compound = 26
Empirical formula weight = 13
`=26/13=2`
Molecular formula = `(EF)xx2=C_(2)H_(2)`