# The mass ratio of sodium to fluorine in sodium fluoride is 1.21:1. A sample of sodium fluoride produces 34.5 g of sodium upon decomposition. How much fluorine (in grams) forms?

Oct 31, 2015

$\text{28.5 g}$

#### Explanation:

The idea here is that you need to use the given mass ratio to determine how much fluorine will be produced if the decomposition reaction also produces $\text{34.5 g}$ of sodium.

A $1.21 : 1$ mass ratio between sodium and fluorine tells you that can expect to have a mass of sodium that is $1.21$ times bigger than the mass of fluorine, regardless of the mass of the sodium fluoride.

So based on this information, you know that the mass of fluorine must be smaller than that of sodium.

More specifically, the mass of fluorine will come out to be

34.5color(red)(cancel(color(black)("g Na"))) * "1 g F"/(1.21color(red)(cancel(color(black)("g Na")))) = color(green)("28.5 g F")

Alternatively, we can double-check the result by finding the mole ratio that exists between sodium and fluorine in a sample of sodium fluoride.

To do that, use the molar amsses of the two elements. Let's say that you have a sample of sodium fluoride that contains $x$ grams of sodium and $y$ grams of fluoride.

You know that

$\frac{x}{y} = \frac{1.21}{1} = 1.21$

You can say that you have

(x color(white)(x) color(red)(cancel(color(black)("g"))))/(23.0color(red)(cancel(color(black)("g")))/"mol") = x/23.0"moles of Na"

(y color(white)(x) color(red)(cancel(color(black)("g"))))/(19.0color(red)(cancel(color(black)("g")))/"mol") = y/19.0"moles of F"

The mole ratio will thus be

$\frac{x}{23.0} \cdot \frac{19.0}{y} = {\overbrace{\frac{x}{y}}}^{\textcolor{b l u e}{= 1.21}} \cdot \frac{19}{23} = 1.21 \cdot \frac{19}{23} = 0.99956 \approx 1$

This tells you that you get one mole of fluorine for every one mole of sodium.

The decomposition reaction produced

34.5color(red)(cancel(color(black)("g"))) * "1 mole Na"/(23/0color(red)(cancel(color(black)("g")))) = "1.5 moles Na"

Automatically, you know that it also produced $1.5$ moles of fluorine. The mass of fluorine will thus be

1.5color(red)(cancel(color(black)("moles"))) * "19.0 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("28.5 g F")