Test: Percentage Composition (NCERT) - NEET MCQ

Mass of compound taken = 0.48 g
Mass of boron in sample = 0.192 g
Mass of oxygen in sample = 0.288 g
Percentage of B = 0.192/0.48 x 100 = 40%
Percentage of O = 0.288/0.48 x 100 = 60%

Percentage of element 'X' = 50;
Mole ratio = 50/20 = 2.5
Percentage of element 'Y' = 50, mole ratio = 50/40 = 1.25
For 'X' simple ratio = 2.5/1.25 = 2
For 'Y' simple ratio = 1.25/1.25 = 1
Formula = X₂Y

Since empirical formula is multiplied by n to get molecular formula.
CH₂O₂ will give only C₂H₄O₄ as its molecular formula. (CH₂O₂)_n where n = 1, 2, 3,... etc.

Mol. wt. = 2 x V.D. = 2 x 39 = 78
(CH)_n = (13)_n or 13 x n = 78
n = 78/13 = 6
∴ Molecular formula = C₆H₆

For element Fe, mole of atoms = 69.9/56 = 1.25
For element O, Mole of atoms = 30.1/16 = 1.88
Mole ratio of Fe = 1.25/1.25 = 1
Mole ratio of O = 1.88/1.25 = 1.5
Simplest whole number ratio of Fe and O = 2, 3
Empirical formula of compoimd = Fe₂O₃
Molecular mass of Fe₂O₃ = 160

Molecular formula = Fe₂O₃