Question

a compound contains 13.2% B and 86.8% Cl by mass. What is the molecular formula of the compound if it has a molar mass of 163g/mol?

Answer 1

Molar mass of B = 10.8 g/mol

Molar mass of Cl = 35.5 g/mol

Given that

B = 13.2 %

Cl = 86.8 %

Dividing each mass percentage by molar mass

For B , 13.2 / 10.8 = 1.22

For Cl, 86.8/35.5 = 2.44

Divide each by smallest among them i.e. 1.22

For B , 1.22/1.22 = 1

For Cl, 2.44/1.22 = 2

Therefore,

B =1

Cl = 2

Hence,

Empirical formula = BCl₂

Therefore,

Molecular formula = (BCl₂)n

n = molar mass / empirical formula mass

= 163 g/mol / 81.8

= 2

Therefore,

molecular formula of the compound =  (BCl₂)₂ = B₂Cl₄