## Chemistry (7th Edition)

There are 1.27 moles of anions in 35.6 g of $AlF_3$.
1. Calculate the molar mass $(AlF_3)$: 26.98* 1 + 19.00 * 3 = 83.98g/mol 2. Calculate the number of moles $(AlF_3)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 35.6}{ 83.98}$ $n(moles) = 0.424$ 3. Find the amount of anions: - Each $AlF_3$ has 1 $Al^{3+}$ (Cation) and 3 $F^-$ (Anion) Therefore, each mole of $AlF_3$ has 3 moles of anions. $0.424 mol(AlF_3) \times \frac{3 mol(Anion)}{1 mol (AlF_3)} = 1.27 mol (Anion)$