Question

A carbon compound on analysis gave percentage compositionC=40%,H=6.6%,O=53.4%.Calculate empirical formula of the compound?​

Answer 1

Answer:

Oxygen = 53.34%. Calculate its molecular formula, if molecular mass is 90.

Explanation:

By using element percentage composition - Atomic mass - Atomic ratio

Carbon ⟶40%÷12=

12

40

=3.33

Hydrogen ⟶6.66%÷1=

1

6.66

=6.66

Oxygen ⟶53.34%÷16=

16

40

=3.33

The least atomic ratio is 3.33.

For simplest ratio, we get C:H:O = 1:2:1

Empirical formula = CH

2

O of mass 30 g.

As molar mass = 60, Molecular formula = (CH

2

O)

2

=CH

3

COOH

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Answer 2

Answer:

The correct answer is CH2O

Explanation:

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