Question

An organic compound contains 40% C , 6.66% H , and rest oxygen It's molecular mass is 60. Calculate it's E.F & M.F
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Answer 1

★ Given :-

• C = 40 %
• H = 6.66 %
• O = 100 - ( 40 + 6.66 ) = 100 - 46.66 = 54.34

★ Relative no of atoms :-

• Simply divide the % composition of the elements with thier atomic mass

➝ C = 40 / 12 = 3.33

➝ H = 6.66 / 1 = 6.66

➝ O = 53.34 / 16 = 3.33

★ Simple Ratio :-

• Divide all the obtained value with the least number i.e 3.33 ( in this case )

➝ C = 3.33 / 3.33 = 1

➝ H = 6.66 / 3.33 = 2

➝ O = 3.33 / 3.33 = 1

➸ Empirical formula = C H 2 O

➸ Empirical mass = 12 + 2 + 16 = 30 g

➻ Molecular mass = 60 g

we know that ,

➽ Molecular mass = n ( Empirical mass )

On rearranging ,

➽ n = molecular mass / Empirical mass

➽ n = 60 / 30

➽. n = 2

hence ,

➥ Molecular formula = n ( Empirical formula )

➥ Molecular formula = 2 ( C H 2 O )

➥ Molecular formula = C 2 H 4 O 2