Question

A sample of a mineral contains `26.83%` `"KCl"` and `34.27%` `"MgCl"_2`. Find the percentage of chlorine in this sample?

Answer 1

`38.28%`

Explanation:

Your strategy here will be to use the given percentages to figure out how much chlorine is present in a given sample of this unknown mineral.

To make the calculations easier, let's pick a sample of this mineral that has a mass of `"100.0 g"`. This implies that the sample contains

• `"26.83% KCl in 100.0 g mineral" \ -> \ "26.83 g KCl"`
• `"34.27% MgCl"_2 \ "in 100.0 g mineral" \ -> \ "34.27 g MgCl"_2`

Now, use the molar mass of potassium chloride and the molar mass of atomic chlorine to calculate how many grams of chlorine are present in this sample.

`26.83 color(red)(cancel(color(black)("g KCl"))) * (1color(red)(cancel(color(black)("mole KCl"))))/(74.5513color(red)(cancel(color(black)("g KCl")))) * (1 color(red)(cancel(color(black)("mole Cl"))))/(1color(red)(cancel(color(black)("mole KCl")))) * "35.453 g"/(1color(red)(cancel(color(black)("mole Cl")))) = "12.759 g Cl"`

So, you know that this sample contains `"12.759 g"` of chlorine from the mass of potassium chloride. To find the mass of chlorine coming from the mass of magnesium chloride, use the molar mass of magnesium chloride and the molar mass of atomic chlorine.

`34.27 color(red)(cancel(color(black)("g MgCl"_2))) * (1color(red)(cancel(color(black)("mole MgCl"_2))))/(95.211color(red)(cancel(color(black)("g MgCl"_2)))) * (2color(red)(cancel(color(black)("moles Cl"))))/(1color(red)(cancel(color(black)("mole MgCl"_2)))) * "35.453 g"/(1color(red)(cancel(color(black)("mole Cl")))) = "25.522 g Cl"`

At this point, you know that the sample contains `"25.522 g"` of chlorine from the mass of magnesium chloride.

You can thus say that the total mass of chlorine present in the sample is

`"12.759 g" \ + \ "25.522 g" = "38.281 g"`

So, you have a sample of this unknown mineral that has a mass of `"100.0 g"` and which contains `"38.281 g"` of chlorine, so you can say that the percent composition of chlorine in the sample is

`color(darkgreen)(ul(color(black)("% composition" \ = \ "38.28% Cl")))`

This is the case because, by definition, the mass percent of an element in a substance is given by the mass of said element present in exactly `"100 g"` of the substance.

The answer is rounded to four sig figs, the number of sig figs you have for your data.