Question

A compound ( 60g) on analysis gave C = 24 g, H = 4 g and O = 32 g. Its empirical

formula is​

Answer 1

Answer:

The ratio of number of gram atoms gives,

C

12

24

2

:

:

:

H

2

4

4

:

:

:

O

16

32

2

Therefore, the empirical formula is =C

2

H

4

O

2

=CH

2

O.

Answer 2

Answer:

C$H_{2}$O

Explanation:

Total mass of compound = 60 g

Mass of C = 24g

Mass of H = 4g

Mass of O= 32 g

%ge of C = 24/60× 100 = 40 %

%ge of H = 4/60 × 100 = 6.66 %

%ge of O = 32/60 × 100 = 53.33 %

Gram atom = %ge / atomic mass

So, no. of gram atoms of C = 40/12= 3.33

     no. of gram atoms of H =  6.66/1 = 6.66

    no. of gram atoms of O = 53.33/16 = 3.33

So, smallest gram atom = 3.33

Dividing each element with the smallest gram atoms-

C= 3.33/3.33 =1  

H = 6.66/3.33 = 2

O = 3.33/3.33 = 1

So, C:H:O = 1:2:1

Therefore, emperical formula - C$H_{2}$O

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