A given sample of pure compound contains 9.81 g of Zn, 1.8 × 1023 atoms of chromium, and 0.60 mol 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:

$\begin{array}{l} Moles of Zn = \frac{Wt .}{Mol . wt .} = \frac{9 .81}{65 .39} = 0 .15 \\ Moles of Cr = \frac{given no . of atoms}{avogadro' s no .} = \frac{1 .8 \times 10^{23}}{6 .023 \times 10^{23}} = 0 .3 \end{array}$

Moles of O = 0.6

So, whole molar ratio

= 0.15 : 0.3 : 0.6 = 1 : 2 : 4

Hence, formula = ${ZnCr}_{2} O_{4}$

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A given sample of pure compound contains 9.81 g of Zn, $1.8 \times 10^{23}$ atoms of chromium, and 0.60 mol of oxygen atoms. What is the simplest formula ?

Solution:

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