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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. |
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KF |
1 mole K
1 mole F
1 mole KF |
= 39.1 g
= 19.0 g
= 58.1 g |
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| 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% =
58.1 |
67.3% K |
| % F = |
wt. element
x 100% =
wt. compound |
19.0 x 100% =
58.1 |
32.7% F |
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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 Fe2O3 |
= 2 x 55.8
= 3 x 16.0 |
= 111.6 g
= 48.0 g
= 159.6 g |
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| 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
159.6 |
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% O = |
48.0 x 100% = 30.1 % O
159.6 |
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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.)
Answers
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:
Eden Francis
Clackamas Community College
©1998, 2002 Clackamas Community College, Hal Bender
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