# 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?

##### 1 Answer

#### 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

A **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

You know that

#x/y = 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

#x/23.0 * 19.0/y = overbrace(x/y)^(color(blue)(=1.21)) * 19/23 = 1.21 * 19/23 = 0.99956 ~~ 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

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