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Calculate the percentage of elements in the following compound: Ethanol (C_(2)H_(5)OH) |
Answer» SOLUTION :Molecular MASS of Ethanol `(C_(2)H_(5)OH) = (2 xx "ATOMIC mass of C") + (6 xx "atomic mass of H") + (1 xx "atomic mass of O")` `= (2 xx 12) + (6 xx 1) + (1 xx 16)` = 24 + 6 + 16 = 46 `g//"mole"` `:.` 46 g of ethanol contains 24G carbon, 6g hydrogen and 16 g oxygen. Percentage of carbon =`(24g)/(47g) xx 100 ~~ 52.174` Percentage of hydrogen = `(6 g)/(46 g) xx 100 ~~ 13.043` Percentage of oxygen =`(16 g)/(46 g) xx 100 ~~ 34.783` Carbon: 52.174 %. Hydrogen: 13.043 % and Oxygen : 34.783 % |
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