# Question 7071f

Nov 12, 2016

$\text{1.27 g Na}$

#### Explanation:

The trick here is to realize the every sample of sodium bromide, regardless of its mass, will contain 22.34% sodium by mass.

This is true because sodium and bromide always combine in definite proportions, i.e. stable ratios by mass, to form sodium bromide.

Now, a 22.34% percent concentration by mass means that every $\text{100 g}$ of sodium bromide will contain $\text{22.34 g}$ of sodium and

$\text{100 g " - " 22.34 g" = "77.66 g Br}$

Therefore, you can say that the mass ratio that exists between sodium and bromine in any sample of sodium bromide will be

$\text{22.34 g Na"/"77.66 g Br} = \frac{22.34}{77.66}$

You can thus use this definite proportion to say that a $\text{5.69 g}$ sample of sodium bromide will contain

5.69 color(red)(cancel(color(black)("g NaBr"))) * "22.34 g Na"/(100color(red)(cancel(color(black)("g NaBr")))) = "1.271 g Na" = color(green)(bar(ul(|color(white)(a/a)color(black)("1.27 g Na")color(white)(a/a)|)))

The answer is rounded to three sig figs.

You can use the known definite proportion to find the mass of bromide present in the sample

1.271 color(red)(cancel(color(black)("g NaBr"))) * "77.66 g Br"/(22.34 color(red)(cancel(color(black)("g NaBr")))) = "4.42 g Br"

This checks out because

overbrace("5.69 g")^(color(blue)("mass of NaBr sample")) - overbrace("1.27 g")^(color(purple)("mass of Na")) = overbrace("4.42 g")^(color(darkgreen)("mass of Br"))#