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

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)`