Answer to Question #133759 in Chemistry for Dhbrkfkd

Question

An oxide of chlorine has a molar mass of 167g mol-1 and contains 42.5% chlorine by mass. Find it’s molecular formula

Answer 1

Solution:

Since the scale for percentages is 100, it is most convenient to calculate the mass of elements present in a sample weighing 100 g. Taking this into account, the mass percentages provided may be more conveniently expressed as fractions:

42.5% Cl = 42.5 g Cl / 100 g oxide

(100 - 42.5)% = 57.5% O = 57.5 g O / 100 g oxide

The molar amounts of chlorine and oxygen in a 100-g sample are calculated by dividing each element’s mass by its molar mass:

42.5 g Cl × (mol Cl / 35.453 g) = 1.199 mol Cl

57.5 g O × (mol O / 15.999 g) = 3.594 mol O

Coefficients for the tentative empirical formula are derived by dividing each molar amount by the lesser of the two:

1.199 mol Cl / 1.199 = 1

3.594 mol O / 1.199 = 3

Since the resulting ratio is one chlorine to three oxygen atoms, the empirical formula is ClO₃.

Empirical formula mass = Ar(Cl) + 3 × (O) = 35.453 + 3×15.999 = 83.45 (g mol^-1)

Molar mass is 167 g mol^-1.

molar mass / empirical formula mass = n formula units/molecule

(167 g mol^-1) / (83.45 g mol^-1) = 2 formula units/molecule

Finally, derive the molecular formula for oxide from the empirical formula by multiplying each subscript by two: (ClO₃)₂ = Cl₂O₆

Answer: Cl₂O₆ is molecular formula of chlorine oxide.