Question 3.5.3: A 2.64-g sample of Cr is heated in the presence of excess ox......
A 2.64-g sample of Cr is heated in the presence of excess oxygen. A metal oxide (Cr_{x}O_{y}) is formed with a mass of 3.86 g. Determine the empirical formula of the metal oxide.
You are asked to determine the empirical formula of a binary compound.
You are given the mass of a reactant and the mass of the binary compound formed in the reaction.
Step 1. Use the mass of the metal oxide and the mass of the metal to determine the amount of oxygen present in the metal oxide sample.
3.86 g Cr_{x}O_{y} – 2.64 g Cr = 1.22 g O
Step 2. Use the mass of Cr and mass of O to determine moles of each element present in the compound.
2.64 g Cr \times \frac{1\text{ mol Cr}}{52.00 \text{ g}} = 0.0508 mol Cr
1.22 g O \times \frac{1\text{ mol O}}{16.00 \text{ g}} = 0.0763 mol O
Step 3. To determine the simplest whole-number ratio of elements, divide each by the smallest value.
\frac{0.0508\text{ mol Cr}}{0.0508} = 1.00 mol Cr = 1 mol Cr
\frac{0.0763\text{ mol O}}{0.0508} = 1.50 mol O
In this case, the ratio is not made up of two integers. Writing the relative amounts as a ratio and rewriting the ratio as a fraction results in a whole-number ratio of elements and the correct empirical formula.
\frac{1.50 \text{ mol O}}{1\text{ mot Cr}}=\frac{\frac{3}{2}\text{ mol O} }{1 \text{ mol Cr}}=\frac{3 \text{ mol O}}{2\text{ mol Cr}}The empirical formula is Cr_{2}O_{3}.