Composition from Formulas
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Composition from Formulas

Atoms explain why compounds have fixed composition. It's because the atoms combine with one another in specific fixed ratios. Formulas give us the composition of the compounds in terms of the ratio of atoms. In addition, atomic weights give us the relative weights of the atoms. We can combine all that information and use it to calculate the composition in weight percent for a compound, starting with the formula.

Two examples are worked out for you here (and in Example 16 in your workbook). After you have worked through those, try your hand at this skill by doing Exercise 17 in your workbook (the answers are shown below).

First Example

Let's start with a simple case, KF, the formula for potassium fluoride. The first step is to find the atomic weights for potassium and fluorine, which are 39.1 grams for potassium and 19.0 grams for fluorine. Next, you have to figure the formula weight for the compound KF by adding the weights together, and that is 58.1g.
KF 1 mole K
1 mole F
1 mole KF
= 39.1 g
= 19.0 g
= 58.1 g
Now let's take a look at what that really represents. One mole of KF contains one mole of potassium and one mole of fluorine. That means 58.1 grams of KF contains 39.1 grams of potassium and 19.0 grams of fluorine. So we have the individual weights of the elements that are contained in the compound and the total weight of one mole of that compound.
From there you calculate the percentage composition the same way you have done previously. Take the weight of the element and divide by the weight of the compound and convert it into percent. So the percentage of potassium in this compound is calculated out to be 67.3% and the percent of fluorine is calculated to be 32.7%.
% K = wt. element  x 100% =
wt. compound
39.1 x 100% =
67.3% K
% F = wt. element  x 100% =
wt. compound
19.0 x 100% =
32.7% F

Second Example

This second example shows you how to go about doing the same kind of calculation when you have subscripts in the formula.

In this case, one mole of Fe2O3 has a total formula weight of 159.6. That one mole of Fe2O3 contains two moles of iron, which weighs 111.6 grams. It also contains three moles of oxygen, which weighs 48.0 grams.
Fe2O3 2 moles Fe
3 moles O
1 mole
= 2 x 55.8
= 3 x 16.0
= 111.6 g
= 48.0 g
= 159.6 g
So again, by taking into account what those subscripts are, we can figure out the weight of iron and the weight of oxygen in that particular compound and then use those numbers to calculate that there is 69.9% iron and 30.1% oxygen in that compound.
% Fe = 111.6 x 100% = 69.9 % Fe
% O = 48.0 x 100% = 30.1 % O

Practice with Calculating Composition from Formulas

What I would like you to do now is to get some practice calculating compositions starting from the formula by working on Example 17 parts a, b, and c. Get help from the instructor if you need it. Check your answers below and then continue with the lesson.   (If, after completing these, you would like to try some additional problems of this type, you will find several in a "Practice Problems" page in the Wrap-Up for this lesson.  Click here to go there now.)


The composition of CO2 is 27.3% C and 72.7% O. K2CO3 is 56.6% K, 8.68% C (or 8.7% C if you want to keep the decimal place in line with the other percentages) and 34.7% O. Fe(CN)3 contains 41.7% Fe, 26.9% C, and 31.4% N.


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E-mail instructor: Sue Eggling

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