# Percent Composition of Compounds

####
**Percent
Composition of Compounds**

**As we have
seen, the formula of a compound tells us the numbers of atoms of each element in
a unit of the compound.

** However, suppose we needed to verify the purity of a compound for use in a laboratory experiment. We could calculate what percent of the total mass of the compound is contributed by each element from the formula. Then, by comparing the result to the percent composition obtained experimentally for our sample, we could determine the purity of the sample.

**

**The percent composition**: is the percent by mass of each element in a compound.
** Percent
composition is obtained by dividing the mass of each element in 1 mole of the compound
by the molar mass of the compound and multiplying by 100 percent.

** Mathematically,
the percent composition of an element in a compound is expressed as:

Where n is the
number of moles of the element in 1 mole of the compound.

** For example,
in 1 mole of hydrogen peroxide (H

The molar masses of H

_{2}O_{2}) there are 2 moles of H atoms and 2 moles of O atoms.The molar masses of H

_{2}O_{2}, H, and O are 34.02 g, 1.008 g, and 16.00 g, respectively. Therefore, the percent composition of H_{2}O_{2}is calculated as follows:
The sum of the
percentages is 5.926 percent + 94.06 percent = 99.99 percent. The small
discrepancy from 100 percent is due to the way we rounded off the molar masses
of the elements. If we had used the empirical formula HO for the calculation,
we would have obtained the same percentages. This is so because both the
molecular formula and empirical formula tell us the percent composition by mass
of the compound.

**Example (1): Phosphoric acid (H**

_{3}PO_{4}) is a colorless, syrupy liquid used in detergents, fertilizers, toothpastes, and in carbonated beverages for a “tangy” flavor. Calculate the percent composition by mass of H, P, and O in this compound.**Strategy:**

** Recall the procedure for calculating a
percentage.

** Assume that
we have 1 mole of H

_{3}PO_{4}. The percent by mass of each element (H, P, and O) is given by the combined molar mass of the atoms of the element in 1 mole of H_{3}PO_{4}divided by the molar mass of H_{3}PO_{4}, then multiplied by 100 percent.**Solution:**

The molar mass of H

_{3}PO_{4}is 97.99 g. The percent by mass of each of the elements in H_{3}PO_{4 }is calculated as follows:**Check:**

Do the percentages add to 100 percent? The sum
of the percentages is:

3.086% + 31.61% + 65.31% = 100.01%.

The small
discrepancy from 100 percent is due to the way we rounded off.

** The procedure used in Example (1) can be reversed if necessary.

** Given the
percent composition by mass of a compound, we can determine the empirical formula
of the compound:

** Because we
are dealing with percentages and the sum of all the percentages is 100 percent,
it is convenient to assume that we started with 100 g of a compound, as Example
(2) shows.

**Example (2): Ascorbic acid (vitamin C) cures scurvy. It is composed of 40.92 percent carbon (C), 4.58 percent hydrogen (H), and 54.50 percent oxygen (O) by mass. Determine its empirical formula.**

**Strategy:**

In a chemical formula, the subscripts
represent the ratio of the number of moles of each element that combine to form
one mole of the compound. How can we convert from mass percent to moles? If we
assume an exactly 100-g sample of the compound, do we know the mass of each
element in the compound? How do we then convert from grams to moles?

**Solution:**

** If we have
100 g of ascorbic acid, then each percentage can be converted directly to
grams.

** In this
sample, there will be 40.92 g of C, 4.58 g of H, and 54.50 g of O. Because the
subscripts in the formula represent a mole ratio, we need to convert the grams
of each element to moles.

** The conversion
factor needed is the molar mass of each element. Let n represent the number of
moles of each element so that:

Thus, we arrive
at the formula C

_{3.407}H_{4.54}O_{3.406 }, which gives the identity and the mole ratios of atoms present. However, chemical formulas are written with whole numbers. Try to convert to whole numbers by dividing all the subscripts by the smallest subscript (3.406):
where the ~ sign
means “approximately equal to.” This gives CH

_{1.33}O as the formula for ascorbic acid. Next, we need to convert 1.33, the subscript for H, into an integer. This can be done by a trial-and-error procedure:
** Because 1.33
× 3 gives us an integer (4), we multiply all the subscripts by 3 and obtain C

_{3}H_{4}O_{3}as the empirical formula for ascorbic acid**Check:**

Are the subscripts in C

_{3}H_{4}O_{3}reduced to the smallest whole numbers?** Chemists often want to know the actual mass of an element in a certain mass of a compound. For example, in the mining industry, this information will tell the scientists about the quality of the ore.

** Because the percent composition by mass of the elements in the substance can be readily calculated, such a problem can be solved in a rather direct way.

**Example (3): Chalcopyrite (CuFeS**

_{2}) is a principal mineral of copper. Calculate the number of kilograms of Cu in 5.93 × 10^{3}kg of chalcopyrite.**Strategy:**

Chalcopyrite is composed of Cu, Fe, and S. The
mass due to Cu is based on its percentage by mass in the compound. How do we
calculate mass percent of an element?

**Solution:**

The molar mass of Cu and CuFeS

_{2}are 63.55 g and 183.5 g, respectively. The mass percent of Cu is therefore:
To calculate
the mass of Cu in a 5.93 × 10

^{3}kg sample of CuFeS_{2}, we need to convert the percentage to a fraction (that is, convert 34.63 percent to 34.63/100, or 0.3463) and write:
We can also
solve the problem by reading the formula as the ratio of moles of chalcopyrite
to moles of copper using the following conversions:

Try it.

**Check:**

As a ballpark
estimate, note that the mass percent of Cu is roughly 33 percent, so that a
third of the mass should be Cu; that is:

This quantity
is quite close to the answer.

**Reference:***General Chemistry: The Essential Concepts / Raymond Chang , Jason Overby. (sixth edition)**.*

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