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Butyric acid contains only C, H and O. A 4.24 mg of sample butyric acid is completely burnt. It gives 8.45 mg of carbon dioxide and 3.46 mg of water. What is the mass percentage of each element in butyric acid ? The molecular mass of buryric acid is determined by experiment as 88 amu. What is its molecular formula ? |
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Answer» Solution :Step I : Calculation of mass percentage of different elements Percentage of carbon. Percentage of carbon can be calculated as follows : `CO_(2)-=C` 44 parts 12 parts or 44 mg 12 mg 44 mg of `CO_(2)` CONTAIN C = 12 mg `:. 8.45` g of `CO_(2)` contain `C = (12)/(44)xx8.45 = 2.30` mg Percentage of `C=("Weight of carbon")/("Weight of compound")xx100=((2.30mg))/((4.24mg))xx100=54.3%` Percentage of HYDROGEN. Percentage of hydrogen can be calculated as : `H_(2)O -= 2H` 18 parts 2 parts 18 mg 2 mg 18 mg of `H_(2)O` contain H = 2mg `:. 3.46` of `H_(2)O` contain `H= (2)/(18) xx 3.46 = 0.384` mg Percentage of `H=("Weight of hydrogen")/("Weight of compound")xx100=((0.384mg))/((4.24mg))xx100=9%` The sum of the percentage of C and H `= 54.3 + 9 = 63.3 %` As this comes out to be less than 100, the balance must be oxygen. `:.` Percentage of `O = 100-63.3 = 36.7 %` Step II. Determination of empirical formula of the compound `{:("Elemenet","Percentage","Atomic mass","Gram ATOMS (Moles)","Atomic ratio (Molar ratio)","Simplest WHOLE no. ratio"),("C",54.3,12,(54.3)/(12)=4.52,(4.52)/(2.29)=1.97,2),("H",9.0,1,(9.0)/(1)=9.0,(9.0)/(2.29)=3.93,4),("O",36.7,16,(36.7)/(16)=2.29,(2.29)/(2.29)=1.0,1):}` The empirical formula of the compound `= C_(2)H_(4)O` Step III. Determination of molecular formula of the compound Empirical formula mass `= 2xx 12 + 4 xx1 +16 = 24 + 4+16 = 44` u Molecular mass = 88 u , `n=("Molecular mass")/("Empirical formula mass")=(88)/(44)=2` `:.` Molecular formula `= n xx` Empirical formula `= 2xx C_(2)H_(4)O=C_(4)H_(8)O_(2)` |
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