Question

A compound X contains 32% of A, 20% of B and remaining percentage of C. Then, the empirical formula of X is :(Given atomic masses of A = 64; B = 40; C = 32 u) 

(1) \(A_2BC_2 \)

(2) \(ABC_3 \)

(3) \(AB_2C_2\) 

(4) \(ABC_4\)

Answer 1

Correct option is (2) \(ABC_3 \)

Element	Mass percentage %	No. of moles	No. of moles/ Smallest number	Simplest whole number
A	32%	\(\frac {32}{64} = \frac {1}{2}\)	\(\frac {1}{2} \times 2\)	= 1
B	20%	\(\frac {20}{40} =\frac {1}{2}\)	\(\frac {1}{2}\times 2\)	= 1
C	48%	\(\frac {48}{32} = \frac {3}{2}\)	\(\frac {3}{2}\times 2\)	= 3

So, empirical formula of \(X = \begin{matrix} A&: & B&: & C \\ 1&: & 1&: & 3 \\ \end{matrix}\)

\(\therefore\) The correct empirical formula of compound X is \(ABC_3\).