`0.22 g` of a hydrogen (i.e., a compound conatining carbon and hydrogen only) on complete combustion with oxygen gave `0.9 g` water and `0.44 g` carbon dioxide. Show that these results are in accordance with the law of conservation of mass (atomic mass of `C = 12, H = 1, O = 16)`.

Explanation

Carbon and hydrogen in the hydrocarbon on combustion form carbon dioxide and water. From the formula of `CO_(2)` and `H_(2) O`, the weights of carbon and hydrogen contained in `0.44 g CO_(2)` and `0.9 g` water, respectively, can be calculated as under: ltbr Molecular weight of `CO_(2) = 12 + 32 = 44`
Molecular weight of `H_(2)O = 2 + 16 = 18`
Weight of carbon in `0.44 g` of `CO_(2) = (12 xx 0.44)/(44) = 0.12 g`
Weight of hydrogen in `0.9 g` of `H_(2) O = 2 xx (0.9)/(18) = 0.10 g`
Total weight of `C` and `H` in the hydrocarbon after combustion `= 0.12 + 0.10 = 0.22 g`
Since the weight of carbon hydrogen after combustion is the same as the weight of hydrocarbon (containing carbon and hydrogen only) after combustion, the results are in accordance with the law of conservation of mass.