0.22 g of a hydrocarbon (i.e., a compound containing carbon and hydrogen only) on complete combustion with oxygen gave 0.9 g water and 0.44 g carbon dioxide. Are these results in accordance with the law of conservation of mass (atomic mass of C = 12, H = 1, O = 16).
0.22 g of a hydrocarbon (i.e., a compound containing carbon and hydrogen only) on complete combustion with oxygen gave 0.9 g water and 0.44 g carbon dioxide. Are these results in accordance with the law of conservation of mass (atomic mass of C = 12, H = 1, O = 16).
The correct option is A True
Carbon and hydrogen in the hydrocarbon on combustion form carbon dioxide and water. From the formulae of CO2 and H2O, the weights of carbon and hydrogen contained in 0.44 g CO2 and 0.9 g water, respectively, can be calculated as under:
Molecular weight of CO2 = 12 + 32 = 44
Molecular weight of H2O = 2 + 16 = 18
Weight of carbon in 0.44 g of CO2 = 12×0.4444 = 0.12 g
Weight of carbon in 0.9 g of H2O = 2 × 0.918 = 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 and 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.
True
Carbon and hydrogen in the hydrocarbon on combustion form carbon dioxide and water. From the formulae of CO2 and H2O, the weights of carbon and hydrogen contained in 0.44 g CO2 and 0.9 g water, respectively, can be calculated as under:
Molecular weight of CO2 = 12 + 32 = 44
Molecular weight of H2O = 2 + 16 = 18
Weight of carbon in 0.44 g of CO2 = 12×0.4444 = 0.12 g
Weight of carbon in 0.9 g of H2O = 2 × 0.918 = 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 and 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.
