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 weight 11.6g. Calculate : (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.

Explanation

`44g CO_(2) = 12 g` carbon
`3.38 g CO_(2) = (?)`
`=(12)/(44) xx 3.38g = 0.9218` g carbon
`18 g H_(2)O = 2g H_(2)`
`0.69 g H_(2)O = (2)/(18) xx 0.69`
`= 0.0767` g hydrogen
Total mass of compound `= 0.9218+0.0767`
`= 0.9985g`
Mass percent of `C = (0.9218)/(0.9985) xx 100 = 92.32%`
Mass percent of `H=(0.0767)/(0.9985) xx 100 = 7.68%`

Empirical formula `= CH`
mass of 10 L gas at STP `= 11.6g`
mass of 22.4 L at STP `=(11.6xx22.4)/(10) = 25.984`
Molar mass `=25.984 g/mol ~=26 g/mol`
Empirical formula mass `= 12 +1= 13`
`:.n = ("Molar mass")/("Empirical formula mass")`
`=(26)/(13)=2`
Molecular formula `=n xx` Empirical formula
`=2xxCH=C_(2)H_(2)`