An organic compound on analysis gave the following data - C = 57.82%, H = 3.6% and 38.58% is oxygen. Find its empirical formula.
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C = 57.82%
H = 3.6%
O = 38.58%
For C = 57.82(1 mol C÷12.01 g) = 1.99/2
For H = 3.6(1 mol H÷1.01 g) = 3.56
For O = 38.58(1 mol O÷16.00 g) = 2.41
The compound has the emperical formula = C2HO...
Hope this will help!!!
H = 3.6%
O = 38.58%
For C = 57.82(1 mol C÷12.01 g) = 1.99/2
For H = 3.6(1 mol H÷1.01 g) = 3.56
For O = 38.58(1 mol O÷16.00 g) = 2.41
The compound has the emperical formula = C2HO...
Hope this will help!!!
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