# Question 1ff7e

May 24, 2017

A 4.50-mole sample of $\text{SnF"_2}$ will contain 171 g of fluoride ions.

#### Explanation:

The formula unit for the ionic compound stannous fluoride $\left({\text{SnF}}_{2}\right)$ indicates that there are two moles of fluoride ions $\left({\text{F}}^{-}\right)$ in one mole of stannous fluoride.

4.50color(red)cancel(color(black)("mol SnF"_2))xx(2"mol F"^(-))/(1color(red)cancel(color(black)("mol SnF"_2)))="9.00 mol F"^(-)

Multiply the mol ${\text{F}}^{-}$ by its molar mass: $\text{18.998 g/mol}$. (The atomic weight of F to five sig figs on the periodic table in g/mol.)

9.00color(red)cancel(color(black)("mol F"^(-)))xx(18.998"g F"^(-))/(1color(red)cancel(color(black)("mol F"^(-))))="171 g F"^(-)# (rounded to three sig figs)