Question

Calculate the empirical formula of a compound having percentage composition:

potassium (K)=26.57, chromium (Cr)=35.36; oxygen (O) 38.07.(Given the atomic

weights of K, Cr and O as 39, 52 and 16 respectively)​

Answer 1

Solution :-

Given :-

Percentage composition of potassium (K) = 26.57%

Percentage composition of chromium (Cr) = 35.36

Percentage composition of oxygen (O) = 38.07%

We know :-

Atomic mass of potassium (K) = 39g

Atomic mass of chromium (Cr) = 52g

Atomic mass of oxygen (O) = 16g

$\bullet \sf \ No \ of \ moles \ of \ potassium \ (K) = \dfrac{26.57}{39}= 0.68 \\\\\bullet \ No \ of \ moles \ of \ chromium \ (Cr) = \dfrac{35.36}{52}=0.68 \\\\\bullet \ No \ of \ moles \ of \ oxygen \ (O) = \dfrac{38.07}{16}= 2.38$

Simplest ratio :-

$\longrightarrow \sf \dfrac{0.68}{0.68} \ : \ \dfrac{0.68}{0.68} \ : \ \dfrac{2.38}{0.68}\\\\\longrightarrow 1 \ : \ 1 : \ 3.5$

Simplest whole number ratio :-

$\longrightarrow \sf 1\times 2 : \ 1 \times 2 : \ 3.5 \times 2\\\\\longrightarrow 2 \ : \ 2 : \ 7$

$\underline{\boxed{\bf Empirical \ formula = K_2Cr_2O_7}}$