TReview Topics [Rererences Use the References to access important values if needed for this question. Furnace O absorber CO, absorber Sample A 10.08 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 14.30 grams of CO, and S.782 grams of H,0 are produced. In a separate experiment, the molar mass is found to be 62.07 g mol. Determine the empirical formula and the molecular formula of the organic compound. Enter the elements in the order C, H. O empirical formula = molecular formula = Previous
Moles of compound combusted = mass of compound / molar mass of compound = 10.08 / 62.07 = 0.1624 approx
Assuming the molecular formula of compound is CaHbOc
Hence moles of C present in 10.08 grams of compound = moles of compound X a = 0.1624a
And moles of H present in 10.08 grams of compound = moles of compound X b = 0.1624b
And moles of O present in 10.08 grams of compound = moles of compound X c = 0.1624c
Since the moles of C present in compound should be equal to the moles of CO2 forming as the moles of carbon should be same
Since moles of CO2 forming = mass / molar mass = 14.3 / 44 = 0.325 = moles of C in the compound
Hence 0.1624 a = 0.325
=> a = 2
And since each mole of H2O has 2 mole of H and also the moles of H in compound should be equal to the moles of H in H2O
hence the moles of H = moles of H2O X 2 = 8.782 X 2 / 18 = 0.976
Hence moles of H in compound = 0.1624 b = 0.976 => b = 6
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