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An organic compound has the following percentage composition , `C = 48 %, H = 8 %, N = 28 %`. Calculate the empirical formula of the compound. |
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Answer» For the available data : Percentage of `C = 48%` , Percentage of `H = 8 %` , Percentage of `N = 28 %` Total percentage of C, H and `N = 48 + 8 + 28 = 84 %` We know that the sum of the percentages must be 100. But in this case, it is only `84 %`. The balance `(100-84)= 16 %` is regarded as the percentage of `"oxygen"^(**)` in the compound Step I. Calculation of simplest whole number ratios of the elements `{:("Element","Percentage","Atomic Mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("C",48.0,12,(47.0)/(12)=4,(4)/(1)=4,4),("H",8.0,1,(8.0)/(1)=8,(8)/(1)=8,8),("N",28.0,14,(28.0)/(14)=2,(2)/(1)=2,2),("O",16.0,16,(16.0)/(16)=1,(1)/(1)=1,1):}` The simplest whole number ratios of different elements are : `C:H:N:O : : 4 : 8 : 2 : 1` Step II. Writing the empirical formula of the compound The empirical formula of compound `= C_(4)H_(8)N_(2)O`. |
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