a compund(60g) on analysis gave 24g of C,4g of H and 32g of O. its empirical formula is
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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
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
subrat1:
why u divide 2/2 or ......
we have to divide all the values by the least value which we got
according the formula
yeah
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