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A 1.000 gram sample of an alcohol was burned in oxygen to poduce 1.913g of CO2 and 1.174g of H2O. The alcohol contained only C,H, and O. What is the simplest formula of the alcohol?
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1.913 g CO2/44 = 0.043 moles of CO2*12 = 0.043 moles of C 1.174 g H2O/18 = 0.065 moles of H2O or 0.13 moles of H (because it contains 2 H) O: 1- 0.516 g C - 0.13 g H = 0.354 g O/16 = 0.022 moles of O So we have 0.043:0.13:0.022 as C:H:O Dividing by the smallest no. 0.022 we get 1.95:5.9:1 or 2:6:1 as C:H:O so the formula will be C2H6O. It is ethyl alcohol or ethanol CH3CH2OH
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