Download AP Chemistry Ch. 3 Sections 3.5-3.6 Notes Percent Composition of

AP Chemistry
Ch. 3 Sections 3.5-3.6 Notes
Percent Composition of Compounds, Determining the Formula of a Compound
Percent Composition of Compounds
• There are two common ways to describe the composition of a compound: the numbers of its
constituents atoms (chemical formula) or by percentages (by mass) of its elements (percent
composition).
• The mass percents of the elements in a compound can be determined by comparing the mass
of each element present in 1 mole of the compound to the total mass of 1 mole of the
compound.
mass of element in 1 mole of compound
• Mass percent of element =
x 100%
mass of 1 mole of compound
• Example:
• Find the percent composition of the compound N2O5.
• Solution: First determine the mass of each element in the compound.
• We know that 1 mol of N2O5 must contain 2 mol N and 5 mol O. The number of grams of N
and O are found as follows:
14.01 g
Mass of N = 2 mol x
= 28.02 g N
1 mol
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16.00 g
Mass of O = 5 mol x
= 80.00 g O
1 mol
• Total mass of compound =
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28.02 g N + 80.00 g O = 108.02 g N2O5
• The mass percent of N and O in N2O5 can be computed by comparing the mass of N and O in
1 mole of N2O5 to the total mass of 1 mole of N2O5.
Mass percent of N =
28.02 g
x 100% = 25.94% N
108.02 g
Mass percent of O =
80.00 g
x 100% = 74.06% O
108.02 g
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Notice the percents add up to 100.00%; this provides a check that the calculations are correct.
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Example:
A sample of a liquid with a mass of 8.657 g was decomposed into its elements and gave
5.217 g of carbon, 0.9620 g of hydrogen, and 2.478 g of oxygen. What is the percent
composition of this compound?
Solution:
We can apply an equation for the percent by mass of an element to each element. The “mass
of whole sample” here is 8.657 g so we can take each element in turn and perform the
calculations.
mass of element
% by mass of element =
x 100%
mass of whole sample
5.217 g
For C :
x 100% = 60.26% C
8.657 g
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For H :
0.9620 g
x 100% = 11.11 % H
8.657 g
For O :
2.478 g
x 100% = 28.62% O
8.657 g
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The percentages must add up to 100%, allowing for small differences caused by rounding.
Sum of percentages: 99.99%
Determining the Formula of a Compound
• When a new compound is prepared, one of the first items of interest is the formula of the
compound.
• For example, the compound that forms when phosphorus burns in oxygen consists of
molecules with the formula P4O10.
• When a formula gives the composition of one molecule, it is called a molecular formula.
• Molecular formula: the exact formula of a molecule, giving the types of atoms and the
number of each type.
• Notice that in the formula P4O10 that both the subscripts 4 and 10 are divisible by 2, so the
smallest numbers that tell us the ratio of P to O are 2 and 5.
• In a simpler kind of formula, the empirical formula, the subscripts are the smallest whole
numbers that describe the ratios of the atoms in the substance.
• Empirical formula: the simplest whole number ratio of atoms in a compound.
• It is quite common for the empirical and molecular formulas to be different; some examples
are shown in Figure 3.6 in your textbook.
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Example: Calculating an Empirical Formula from Percent Composition
A white powder used in paints, enamels, and ceramics has the following percent
composition: Ba, 69.58%, C, 6.090%, and O, 24.32%. What is its empirical formula?
Solution:
Since mass percentage gives the number of grams of a particular element per 100 grams of
compound, base the calculation on 100 grams of compound. Each percent will then represent
the mass in grams of that element.
If we have 100 g of the above compound, based on the percentages we would have 69.58 g of
Ba, 6.090 g of C, and 24.32 g of O.
Determine the number of moles of each element present in 100 grams of compound using the
atomic masses of the elements present.
1 mol Ba
Ba : 69.58 g Ba x
= 0.5067 mol Ba
137.33 g Ba
C : 6.090 g C x
1 mol C
= 0.5071 mol C
12.01 g C
O : 24.32 g O x
1 mol O
= 1.520 mol O
16.00 g O
Divide each value of the number of moles by the smallest of the values. If each resulting
number is a whole number (after appropriate rounding), these numbers represent the
subscripts of the elements in the empirical formula.
If the numbers obtained in the previous process are not whole numbers, multiply each
number by an integer so that the results are all whole numbers.
0.5067
=1
0.5067
0.5070
C:
= 1.001 = 1
0.5067
1.520
O:
= 3.000 = 3
0.5067
Ba :
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The coefficients are acceptably close to whole numbers, so the empirical formula is
BaCO3.
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Example: One compound of mercury with a molar mass of 519 contains 77.26% Hg, 9.25%
C, and 1.17% H (with the balance being O). Calculate the empirical and molecular formulas,
arranging the symbols in the order HgCHO.
Solution:
First obtain the empirical formula.
Assume a 100 g sample therefore, we have 77.26 g Hg, 9.25 g C, 1.17 g H, and 12.32 g O.
The amount of oxygen was determined by subtracting the total amounts of the other three
elements from the total assumed mass of 100 g.
Convert each of these masses into moles.
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1 mol Hg
= 0.3852 mol Hg
200.59 g Hg
1 mol C
C : 9.25 g C x
= 0.770 mol C
12.01 g C
1 mol H
H : 1.17 g H x
= 1.16 mol H
1.01 g H
1 mol O
O : 12.32 g O x
= 0.7700 mol O
16.00 g O
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Divide each number of moles by the smallest number of moles.
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0.3852
= 1.000 = 1 Hg
0.3852
0.770
C:
= 2.00 = 2 C
0.3852
1.16
H:
= 3.01 = 3 H
0.3852
0.7700
O:
= 1.999 = 2 O
0.3852
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Empirical Formula: HgC2H3O2
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Compute the mass corresponding to the empirical formula.
(1 Hg x 200.59 amu) + (2 C x 12.01 amu) + (3 H x 1.01 amu) + (2 O x 16.00 amu) = 259.64
g/mol
Calculate the ratio:
Hg : 77.26 g Hg x
Hg :
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Molar Mass
519 g / mol
=
= 2.00
Empirical Formula Mass 259.64 g / mol
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The integer from the previous step represents the number of empirical formula units in one
molecule. When the empirical formula subscripts are multiplied by this integer, the
molecular formula results.
Molecular formula = 2(HgC2H3O2) = Hg2C4H6O4