Question

A component whose molecular mass in 90 gm contain 40.0% carbon,6.67% hydrogen, and 53.33% oxygen. What is true formula of the compound?

a)C2H2OH b)CH2O4

b)C3H6O3 c)C3HO3

Answer 1

Answer: b c3h6O3

Explanation: C moles =40/12 =3.3

Moles of hydrogen =6.6/1 =6.6

Moles of oxygen =53.3/16 =3.3

Simple ratio 3.3/3.3=1. C1

6.6/3.3 =2. H2

3.3/3.3=1. O1

Empirical formula CH2O

n = molecular formula mass / Empirical formula mass

n =90/30 =3

3(CH2O) =C3H6O3