Question

Calculate the empirical formula for a compound that contains 26.6% potassium,35.4% chromium and 38.1% oxygen.(Given :K=39.1,Cr=52,O=16)

Answer 1

empirical formula is K2Cr2O7

Answer 2

Given,

$\qquad \qquad \sf K \: \: \: \: Cr \: \: \: \:O$

To Calculate the empirical formula :

Step-1 :

Weight percentage of elements taken as weight

$\qquad\qquad \sf 26.6 \: \: \: \: 35.4 \: \: \: \:38$

Step-2 :

Weight of element divided with atomic weight

$\displaystyle \qquad\qquad \sf \frac{26.6}{39.1} \: \: \: \: \frac{35.4}{52} \: \: \: \: \frac{38}{16}$

$\displaystyle \qquad\qquad \sf 0.68\: \: \: \: 0.68 \: \: \: \: 2.37$

Step-3 :

Find a simple ratio of atoms of elements

(This can be done by divided with smallest number)

Divide with 0.68

$\displaystyle \qquad\qquad \sf \frac{0.68}{0.68} \: \: \: \: \frac{0.68}{0.68} \: \: \: \: \frac{2.37}{0.68}$

Divide with 2

$\displaystyle \qquad\qquad \sf 1\times 2\: \: \: \: 1 \times 2 \: \: \: \: 3.5 \times 2$

$\displaystyle \qquad\qquad \sf 2\: \: \: \: 2 \: \: \: \: 7$

Step-4 :

Write simple ratio of atom aganist of each element as prefix

$\qquad \qquad \sf K_{2} \: \: \: \: Cr_{2}\: \: \: \:O_{7}$

The empirical formula is

$\qquad \qquad \longmapsto \: \: \underline{\boxed{ \: \bf K_{2}Cr_{2}O_{7}} }$

$\:$