# Question d2834

Jun 21, 2015

Percent of $N a B r$: 24.17%.
Percent of $N {a}_{2} S {O}_{4}$: 75.83%.

#### Explanation:

You know that your $N a B r \text{-} N {a}_{2} S {O}_{4}$ mixture contains 29.96 g of sodium by mass. This means that, for every 100 g of mixture, you get 29.96 g of sodium.

So, let's assume that you have 100 g of mixture. The first relationship you can write is

$\text{100 g} = {m}_{N a B r} + {m}_{N {a}_{2} S {O}_{4}}$

In order to find a relationship between the total mass of sodium in the mixture and the mass of sodium each compound contributes to the this valu,e you can use sodium's percent composition by mass in $N a B r$ and $N {a}_{2} S {O}_{4}$.

To do that, use the molar mass of sodium and the molar mass of each compound

(23.0cancel("g/mol"))/(102.9cancel("g/mol")) * 100 = 22.352% $\to$ percent of sodium in $N a B r$

and

(2 * 23.0cancel("g/mol"))/(142.04cancel("g/mol")) * 100 = 32.385% $\to$ percent of sodium in $N {a}_{2} S {O}_{4}$

Since you know that your 100-g sample contains 29.96 g of sodium, you can use $x$ as the mass of $N a B r$ and $\left(100 - x\right)$ as the mass of $N {a}_{2} S {O}_{4}$ to determine how many grams each compound adds to the mixture

"29.96 g" = xcancel("g"NaBr) * ("22.352 g"Na)/(100cancel("g"NaBr)) + (100-x)cancel("g"Na_2SO_4) * ("32.385 g"Na)/(100cancel("g"Na_2SO_4))

$29.96 = 0.22352 \cdot x + 32.385 - 0.32385 x$

$x = \frac{32.385 - 29.96}{0.32385 - 0.22352} = \text{24.17 g}$

Since $x$ represents the mass of $N a B r$, the mass of $N {a}_{2} S {O}_{4}$ will be

${m}_{N {a}_{2} S {O}_{4}} = 100 - 24.2 = \text{75.83 g}$

Since the sample has a mass of 100 g, the mass each compound has also represents its percent by mass in the compound

(24.17cancel("g")NaBr)/(cancel(100)"g mixture") * cancel(100) = color(green)("24.17% "NaBr)

and

(75.83cancel("g")Na_2SO_4)/(cancel(100)"g mixture") * cancel(100) = color(green)("75.83% "Na_2SO_4)#