Question

measurements show that unknown compound x has the following composition: element mass % chromium 68.4% oxygen 31.5% write the empirical chemical formula of x.

Explanation:

Step1: Assume 100g of the compound

If we assume 100g of compound X, then the mass of chromium ($Cr$) is $m_{Cr}=68.4g$ and the mass of oxygen ($O$) is $m_O = 31.5g$.

Step2: Calculate the number of moles of each element

The molar - mass of $Cr$ is $M_{Cr}=52g/mol$, and the number of moles of $Cr$, $n_{Cr}=\frac{m_{Cr}}{M_{Cr}}=\frac{68.4g}{52g/mol}=1.315mol$. The molar - mass of $O$ is $M_O = 16g/mol$, and the number of moles of $O$, $n_O=\frac{m_O}{M_O}=\frac{31.5g}{16g/mol}=1.969mol$.

Step3: Find the mole - ratio of the elements

Divide each number of moles by the smaller number of moles. The smaller number of moles is $n_{Cr}=1.315mol$. The mole - ratio of $Cr$ to $O$ is: For $Cr$, $\frac{n_{Cr}}{n_{Cr}} = 1$, and for $O$, $\frac{n_O}{n_{Cr}}=\frac{1.969mol}{1.315mol}\approx1.5$. Multiply both by 2 to get whole - number ratios. So the ratio of $Cr$ to $O$ is $Cr:O = 2:3$.

Answer:

$Cr_2O_3$