Question

A given sample of pure compound contains $32.69g$ of $Zn$, $1mol$ of chromium and $12.04\times {10}^{23}$ number of oxygen atoms. What is the simplest formula of the compound?
(Given molar mass of $Zn=65.38g/mol$)

• A. $ZnC{r}_2{O}_7$
  
• B. $ZnC{r}_2{O}_4$
  
• C. $ZnCr{O}_4$
  
• D. $ZnCr{O}_6$
  

Answer 1

The correct option is B $ZnC{r}_2{O}_4$


moles of $Zn=\frac{32.69}{65.38}=0.5$
moles of $Cr=1$
moles of $O=\frac{12.04\times {10}^{23}}{6.02\times {10}^{23}}=2$
So, whole molar ratio
$=0.5:1:2=1:2:4$
Hence the simplest formula of the compound is $=ZnC{r}_2{O}_4$