Question

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?

• A. $ZnCr_2O_7$
• B. $ZnCr_2O_4$
• C. $ZnCrO_4$
• D. $ZnCrO_6$

Answer 1

Answer: (B)

Moles of $Zn= \frac{9.81}{65.39}=0.15$
Moles of $Cr=\frac{1.8\times10^{23}}{6.023\times10^{23}}=0.3$
Moles of $O = 0.6$
So, whole molar ratio
Hence, formula $= ZnCr_2O_4$