The Mole and Stoichiometry: Question 5

Worked solution

Part (a): Empirical formula from mass data

Mass of oxygen combined with the chromium:

$\text{mass of O} = 15.2-10.4=4.8\ \text{g}$

Convert each mass to moles using $\text{moles}=\dfrac{\text{mass}}{A_r}$:

$\text{moles of Cr} = \frac{10.4}{52} = 0.2\ \text{mol}$

$\text{moles of O} = \frac{4.8}{16} = 0.3\ \text{mol}$

Divide both values by the smaller one ($0.2$) to find the simplest whole-number ratio:

$\text{Cr}:\text{O} = \frac{0.2}{0.2} : \frac{0.3}{0.2} = 1:1.5 = 2:3$

So the empirical formula is $\boxed{\text{Cr}_2\text{O}_3}$.

Part (b): Percentage yield

$\text{percentage yield} = \frac{\text{actual mass obtained}}{\text{target (theoretical) mass}}\times100$

$\text{percentage yield} = \frac{15.2}{16.0}\times100 = 95\%$

Part (c): Formula from ionic charges

For an ionic compound to be electrically neutral, the total positive charge must balance the total negative charge.

Using $2$ ions of $\text{Cr}^{3+}$ and $3$ ions of $\text{O}^{2-}$:

$\text{total positive charge} = 2\times(3+) = 6+$

$\text{total negative charge} = 3\times(2-) = 6-$

These charges balance, so the formula is $\text{Cr}_2\text{O}_3$, the same formula found in part (a), so the two methods agree.

Final answers

• (a) Empirical formula $=\boxed{\text{Cr}_2\text{O}_3}$
• (b) Percentage yield $=\boxed{95\%}$
• (c) Ionic-charge formula $=\text{Cr}_2\text{O}_3$, which agrees with the answer to (a)