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?

1 Answer
anor277
Nov 28, 2016

Cr2O3

Explanation:

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⋅g52.00⋅g⋅mol−1=0.0229⋅mol

Moles of oxygen = 1.74⋅g−1.19⋅g16.00⋅g⋅mol−1=0.0343⋅mol

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

Cr,0.0229⋅mol0.0229⋅mol=1;O,0.0343⋅mol0.0229⋅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 Cr2O3 as required. Capisce?

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

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