Question

An organometallic compound on analysis was found to contain C = 64.4% , H = 5.5% and Fe = 29.9%. Determine its empirical formula. ( Atomic mass of Fe = 56u).

Answer 1

If you mean for 100 g then

64 % × 100 = 64 g ----> C

6% ×100 =6 --------> H

30% ×100= 30g-----> Fe

Then no of carbon atoms of C = 64/12 =5.3/0.5735= 9

For hydrogen = 6/1=6/0.5735 =10

Fe =30/56 =0.5375 /0.5735 = 1

Formula is C9H10Fe

Answer 2

If you mean for 100 g then

64 % × 100 = 64 g ----> C

6% ×100 =6 --------> H

30% ×100= 30g-----> Fe

Then no of carbon atoms of C = 64/12 =5.3/0.5735= 9

For hydrogen = 6/1=6/0.5735 =10

Fe =30/56 =0.5375 /0.5735 = 1

Formula is C9H10Fe