How many grams of potassium permanganate contains 27.5 grams of oxygen?

Answer 1

This is more of a math question that requires a bit of knowledge of chemistry. So it helps to know the steps of this answer mathematically. Additionally it's worth noting that there are a number of ways to answer this question. The method I provide may take an extra step, but it allows for a better understanding of the process.

First we need to know some basic information about potassium permanganate, KMnO4. This basic information can be found on a Periodic Table, like the one in the link below. The first step is finding the weight of oxygen in one mole of potassium permanganate as a percent. For this you need to know the atomic weights of the elements involved.

K: 39.1 grams

Mn: 54.9 grams

O: 16.0 grams × 4 atoms = 64.0 grams

KMnO4: 39.1 + 54.9 + 64.0 = 158.0 grams/mol

So now we know the weight of one mole of potassium permanganate (158.0 grams). Because we also know the weight of oxygen, we can find the percent of oxygen in the compound by mass.

64.0 grams O ÷ 158.0 grams KMnO4 = 0.405 = 40.5%

In one mole of potassium permanganate, 64.0 grams of it is oxygen, meaning 40.5% of it is oxygen. Because of the Law of Definite Proportions, we know that in any amount of potassium permanganate, 40.5% of it is oxygen.

Then you can set up an equation.

40.5% of (some amount of KMnO4) = 27.5 grams oxygen

Let's set the amount of KMnO4 as the variable "x".

0.405x = 27.5

x = 67.9 grams KMnO4