A given sample of pure compound contains 9.81 gm of Zn, 1.8×1023 atoms of chromium and 0.60 moles of oxygen atoms. What is the simplest formula ?

A
${ZnCr}_{2} O_{7}$
B
${ZnCr}_{2} O_{4}$
C
${ZnCrO}_{4}$
D
${ZnCrO}_{6}$

Solution:

$n_{Zn} = \frac{9.81}{65} = 0.15 n_{Cr} = \frac{1.8 \times 10^{23}}{6.02 \times 10^{23}} = 0.3 n_{O} = 0.6 {Zn}_{0.15} {Cr}_{0.3} O_{0.6} {ZnCr}_{2} O_{4}$

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