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It has been found that 0.290 g of an organic compound containing C, H and O on complete combustion yielded 0.66 g of CO_2 and 0.27 g of H_2O. The vapour density of the compound is found to be 29.0. Determine the molecular formula of the compound. |
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Answer» Solution :Calculation of percentage of carbon : The gram molecular mass of `CO_(2)` `=12.01 + (2 XX 16.0) = 44.01 g` This means that 44.01 g of `CO_2` contain 12.01 g of carbon. `therefore` The mass of carbon present in 0.66 g of `CO_(2) = 12.01/44.01 xx 0.66 = 0.18 g` This much carbon comes from 0.290 g of the given COMPOUND. `therefore`Percentage of carbon in the given compound `=(0.18)/0.290 xx 100 = 62.1` Calculation of percentage of hydrogen: The gram molecular mass of `H_(2)O` `=(2 xx 1.008) + 16.0 = 18.0 g` This means that 18.0 g of WATER contain = `2 xx 1.008 = 2.016` g of hydrogen. `therefore`The mass of hydrogen present in 0.27 g of `H_(2)O` `=(2.016)/18.0 xx 0.27 = 0.03 g` This much hydrogen comes from 0.290 g of the compound. `therefore`Percentage of hydrogen in the given compound `=0.03/0.290 xx 100 = 10.34` Calculation of percentage of oxygen : Percentage of oxygen = `100 -(62.1 + 10.34) = 27.56` Calculation of empirical FORMULA:  `therefore` The empirical formula of the given compound is `C_3H_6O` Calculation of molecular formula : The empirical formula mass `=(3 xx 12.01) + 6 xx 1.008 + 16.0` = 58.0 amu Molecular mass = `2 xx` VAPOUR density `=2 xx 29.0 = 58.0 amu` `therefore n =("Molecular mass")/("Empirical formula mass") = 58.0/58.0 = 1` `therefore` Molecular formula = `1 xx` Empirical formula `=1 xx C_(3)H_(6)O = C_(3)H_(6)O` Hence, the molecular formula of the given compound is `C_(3)H_(6)O`. |
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