An organic compound has C=40%, H=6.67% and O=53.3% What is the empirical formula of the compound?

MCQS

Sindh MDCAT 2023

Q-124

Jump to Deck

<p>An organic compound has C=40%, H=6.67% and O=53.3% What is the empirical formula of the compound?</p>

Chemistry

Miscellaneous

Question 124 of 200

Multiple Choice

CH₃O

16.24%

Explanation:

C₂H₂O

13.26%

Explanation:

CH₂O

56.52%

Explanation:

Emprical formula can be calculated as follows;

Number of atoms of C = 40/12 =3.33

Number of atoms of H = 6.68/1 = 6.67

Number of atoms of O = 53.3/16 =3.33

Taking ratio;

C:H:O

3.33:6.67:3.33

1:2:1

Thus, Emprical forumla will be CH2O.

Hence, option C is correct.

C₂H₆O

13.97%

Explanation:

Explanation

Emprical formula can be calculated as follows;

Number of atoms of C = 40/12 =3.33

Number of atoms of H = 6.68/1 = 6.67

Number of atoms of O = 53.3/16 =3.33

Taking ratio;

C:H:O

3.33:6.67:3.33

1:2:1

Thus, Emprical forumla will be CH2O.

Hence, option C is correct.