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Question

1)6.42g of a compound of carbon,hydrogen and oxygen gave on complete combustion 16.2g of carbon dioxide and 2.84g of water.Calculate the empirical formula of the original compound.
2)A volatile liquid X is a compound of carbon,hydrogen and chlorine.0.40 mole of X contains 9.6g of carbon,1.6g of hydrogen and 28.4g of chlorine. Calculate the relative molecular mass and formula of compound X.

Answer 1

1 )
no. of mole of 16.42 g CO2 = 16.42/44 = 0.3732
mass of 0.3732 mole carbon = 0.3732X12 = 4.478 g
no. of mole of 2.84 g water = 2.84/18 = 0.1578
mass hydrogen = 0.1578X2 = 0.3156 g
mass of oxygen = 6.42 - 4.478 - 0.3156 = 1.6242 g

no. of mole of carbon = 0.3732
no. of mole of hydrogen = 0.3156
no. of mole of oxygen = 1.6242/16 = 0.1017

mole ratio of C : H : O = 0.3732 : 0.3156 : 0.1017 = 3.67 : 3.1032 : 1 = 7 : 6 : 2
empirical formula = C7H6O2

2)
no. of mole of carbon = 9.6/12 = 0.8
no. of mole of hydrogen = 1.6/1 = 1.6
no. of mole of chlorine = 28.4/35.5 = 0.8
empirical formula = CH2Cl

There are not enough data for further calculation. But the compound can be C2H4Cl2.