A compound made up of two elements A and B has A = 70%, B = 30%. Their

A compound made up of two elements A and B has A = 70%, B = 30%. Their relative number of moles in the compound is 12.5 and 1.88, calculate: (i) Atomic masses of the elements A and B (ii) Molecular mass is found to be 160.

Answer»

Atomic mass of elements A

= \(\frac{\%\,of\,elements\,A}{Relative\,number\,of\,moles}\)

= \(\frac{70}{1.25}\)

Atomic mass of element B

= \(\frac{\%\,of\,elements\,B}{Relative\,number\,of\,moles}\)

=\(\frac{30}{1.88}\)

= 15.957 ≈ 16

(ii)

\(\therefore\) Empirical formula of compound = \(A_2B_3\)

Molecular mass = 160

Empirical formula mass = 2(56) + 3(16)

= 112 + 48 = 160

n = \(\frac{Molecular\,mass}{Empirical\,formula\,mass}\) = \(\frac{160}{160}\) = 1

\(\therefore\) Molecular formula = n x Empirical formula

= 1 x A₂B₃

= A₂B₃