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combustion of 6.51 mg of a compound gave 20.47 mg of carbon dioxide and 8.3 mg of water. The molecular weight was found to be 84 gm/mole. Calculate molecular formula of the compound (Atomic Weight of C=12.01, H=1.008, O=16.0 gm/mole)
moles CO2 = mass / molar mass = 20.47 mg/44.01 g/mol = 0.4651 mmol moles C = moles CO2 = 0.4651 mmol mass C = 0.4651 mmol x 12.01 g/mol = 5.586 mg % mass C = 5.586 mg / 6.51 mg x 100/1 = 85.8 % moles H2O = 8.36 mg / 18.016 g/mol = 0.4640 mmol moles H = 2 x moles H2O = 0.9281 mmol % mass H = 100% - 85.8 % = 14.2 % % composition 85.8% C, 14.2 % H ratio moles C : moles H = 0.4651 : 0.9281 divide each number in the ratio by the smallest number 0.4651 / 0.4651 : 0.9281 / 0.9281 = 1 : 2 empirical formula CH2 to determine the molecular formula divide the molar mass by the mass of the empirical formula = 84 g/mol / 14 g/mol = 6 there are 6 empirical units in the molecular formula molecular formula = 6 x (CH2) = C6C12
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