Question

A compound contains 54.55 % carbon, 9.09% hydrogen , 36.36% oxygen. The empirical formula of this compound is :
(a) $C_3H_5O$
(b) $C_4H_8O_2$
(c) $C_2H_4O_2$
(d) $C_2H_4O$

Answer 1

First we calculate the molar ratios of each element:

Carbon: 54.55 ÷ 12 = 4.55

Hydrogen: 9.09 ÷ 1 = 9.09

Oxygen: (100 – 54.55 – 9.09) ÷ 16 = 2.27

The ratio of oxygen is 2.27 ÷ 2.27 = 1 (it's the smallest ratio so using oxygen as the benchmark gives a simpler ratio).

The ratio of carbon is 4.55 ÷ 2.27 = 2

The ratio of hydrogen is 9.09 ÷ 2.27 = 4

Hence the empirical formula is C₂H₄O.

. it can be the butirric acid

Answer 2

Answer:

First we calculate the molar ratio of each element:-

carbon = 54.55÷12 = 4.55

hydrogen = 9.09 ÷1 = 9.09

oxygen = (100-54.55-9.09) ÷ 16 =2.27

the ratio of oxygen is 2.27 ÷2.27= 1

the ratio of carbon = 4.55÷2.27 = 2

the ratio of hydrogen = 9.09÷ 2.27 = 4

hence, the empirical formula is C2H4O

Explanation:

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