A student found that 1.19 g of chromium ($Cr$) formed 1.74 g f chromium oxide. The molar mass of $Cr$ is 52.00 g/mol. What is the empirical formula of chromium oxide?

Md Ariful Islam

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Md Ariful Islam

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As with all these problems, we calculate the molar quantities of each substituent, and normalize them according to the masses of the atomic constituents:

$"Moles of metal"$ $=$ $(1.19*g)/(52.00*g*mol^-1)=0.0229*mol$

$"Moles of oxygen"$ $=$ $(1.74*g-1.19*g)/(16.00*g*mol^-1)=0.0343*mol$

And we divide thru by the SMALLEST molar quantity, that of chromium, to give:

$Cr, (0.0229*mol)/(0.0229*mol)=1; O, (0.0343*mol)/(0.0229*mol)=1.50$.

But by definition, the empirical formula is the smallest WHOLE number ratio that defines constituent atoms in a species. To get a whole number ratio, clearly we mulitply the empirical ratio by $2$ to $Cr_2O_3$ as required. Capisce?

How did I know there were $0.55*g$ of oxygen present?

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