Question

a compund(60g) on analysis gave 24g of C,4g of H and 32g of O. its empirical formula is

Answer 1

24g C(1mol/12.01g) = 2mol C

4g H(1mol/1.01g) = 4mol H

32g O(1mol/16g) = 2mol O


we have to divide all the values by the least value which we got
Here the least value is 2
So, C= 2/2; H =4/2; O= 2/2

Therefore empirical formula is CH2O