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

A welding fuel gas contains carbon and hydrogen only. Burning a small

1.	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 weldinggas is found to weight 11.6g. Calculate : (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.
Answer» Solution :`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` Masspercent 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)`