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 litre (Measured at STP) of this welding gas is found weigh `11.6 g`. Calculate
(i) empirical formula,
(ii) molar mass of the gas, and
(iii) molecular formula.

Explanation

Amount of carbon in `3.38 g CO_(2)`
`=12/44xx3.38 g=0.9218 g`
Amount of hydrogen in `0.690 g H_(2)O`
`=2/18xx0.690 g=0.0767 g`
As the compound contains only `C` and `H`, therefore, total mass of the compound `=0.9218+0.0767 g=0.9985 g`
`%` of `C` in the compound`=0.9218/0.9985xx100=92.32`
`%` of `H` in the compound`=0.0767/0.9985xx100=7.68`
`|{:("Element",% "by mass","Atomic mass","Moles of the element","Simplest molar ratio","Simplest whole number molar rato"),(C,92.32,12,92.32/12=7.69,1,1),(H,7.68,1,7.68/1=7.68 1,1,):}|`
`:.` Empirical formula `=CH`
`10.0 L` of the gas at `STP` weigh`=11.6 g`
`:. 22.4 L` of the gas at `STP` will weigh `=11.6/10.0xx22.4`
`=25.984 g ~~26 g`
:. Molar mass `=26 g mol^(-1)`