Question

An organic compound contains C H& O C=52.24% H=9.05% O=36.71% calculate its empirical and molecular formula if its molecular mass is 88
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Answer 1

percentage of C = 52.24 %

percentage of H = 9.05 %

percentage of O = 36.71%

we know, atomic mass of C = 12 amu , atomic mass of = 1 amu and atomic mass of O = 16 amu

Finding percentage-mass ratio

C => 52.24/12 = 4.35

H => 9.05/1 = 9.05

O => 36.71/16 = 2.29

finding simple ratio of above value

C : H : O = 4.35 : 9.05 : 2.29

= 4.35/2.29 : 9.05/2.29 : 2.29/2.29

= 2 : 4 : 1

hence, empirical formula of compound is $C_2H_4O$

we know, molecular formula= n × empirical formula

so, 88 = n × (12 × 2 + 4 × 1 + 16)

or, 88 = n × (24 + 4 + 16)

n = 2

hence, molecular formula is $C_4H_8O_2$