7 Chemical Composition
7.1 Avogadro’s Number
In 1811, the Italian scientist, Amadeo Avogadro proposed that:
Equal volumes of gas at equal temperatures and pressures have the
same number of particles.
This law, known as Avogadro’s Law, is one of the central ideas of chemistry. One consequence of
Avogadro’s Law is that if a sample of an element has a mass in grams that equals the average atomic
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mass of the element, it will have 6.0221415 x 10 atoms of the element. The number, 6.0221415 x 10 is
known as Avogadro’s Number. Avogadro’s Number refers to a package of particles known as a mole (SI
unit: mol)
NA = 6.0221415 x 1023 particles
1 mol
This is a kind of chemical packaging analogous to 1 dozen refers to 12 units. Just as you can refer to 1
dozen eggs or 1 dozen bags of popcorn, one can equally refer to 1 mol of eggs or one mole of bags of
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popcorn. In the latter cases the 1 mole of eggs or 1 mol of bags of popcorn refer to 6.0221415 x 10
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individual eggs or 6.0221415 x 10 individual bags of popcorn. The importance of Avogadro’s Number is
that it relates a unit mass of chemical to the number of particles of that particular chemical. In terms of
atoms, one mole of atoms is obtained when we weigh out exactly the average mass of the atom given in
the Periodic Table, in units of grams. In other words, the number of particles in a sample can be indirectly
counted by weighing out the sample. Thus, for example in the case of hydrogen,
1
H
A 1.008 g sample of H contains 6.0221415 x 10
23
H atoms.
1.008
Example: (the masses of the elements below are obtained from the Periodic Table)
A sample of 12.011 g C contains 6.0221 x 10
A sample of 26.98 g Al contains 6.022 x 10
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Al atom = 1 mol Al atoms.
A sample of 196.97 g Au contains 6.0221 x 10
A sample of 16.00 g O contains 6.022 x 10
23
C atoms = 1 mol C atoms.
23
Au atoms = 1 mol Au atoms.
O atoms = 1 mol O atoms.
Using Avogadro’s Number, we can calculate the number of particles, and moles, contained in any mass of
chemical using the technique of dimensional analysis.
Example:
1. Calculate the number of moles of atoms and the number of atoms in a 3.7514 g sample of carbon.
1 mol C
= 0.31233 mol C
nC = 3.7514 g C
12.011 g C
6.02214 10 23 atoms C
= 1.8809 1023 atoms C
mol C
Of course, if the question simply asked for the number of Carbon atoms, we could simply combine the
steps:
1 mol C
6.02214 1023 atoms C
= 1.8809 10 23 atoms C
N C = 3.7514 g C
12.011 g C
mol C
N C = 0.31233 mol C
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2. Calculate the number of moles of atoms and the number of atoms in a 3.7514 g sample of aluminum.
1 mol Al
= 0.13904 mol Al
nAl = 3.7514 g Al
26.9815 g Al
N Al = 0.13904 mol Al
6.02214 1023 atoms Al
= 8.3729 1022 atoms Al
mol Al
or
N Al = 3.7514 g Al
1 mol Al
26.9815 g Al
6.02214 10 23 atoms Al
= 8.3736 10 22 atoms Al
mol Al
3. Calculate the number of moles of atoms and the number of atoms in a 3.7514 g sample of gold.
1 mol Au
nAu = 3.7514 g Au
= 0.019046 mol Au
196.9665 g Au
N Au = 0.019046 mol Au
6.02214 1023 atoms Au
= 1.1470 10 22 atoms Au
mol Au
or
N Au = 3.7514 g Au
1 mol Au
196.9665 g Au
6.02214 10 23 atoms Au
= 1.1470 10 22 atoms Au
mol Au
The main point with all this is that the same mass of different materials will have different numbers of
moles (and particles). By knowing the mass of the particular substance, we can calculate the number of
particles present.
mass
moles
particles
Through Avogadro’s Number and knowing the unit mass of a substance, can count the number of particles
by weighing. Care must be taken however, to know exactly the type of substance we are dealing with. For
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example, a sample of 16.00 g O contains 6.022 x 10 oxygen-atoms but only 3.011 x 10 oxygenmolecules. This is because the oxygen-molecule is diatomic, that is, it consists of 2 oxygen-atoms.
FG 1 mol O IJ F 6.022 10 atoms O I = 6.022 10 atoms O
H 16.00 g O K GH mol O atoms JK
F 1 mol O IJ F 6.022 10 atoms O I FG 1 molecule O IJ = 3.011 10
= 16.00 g OG
H 16.00 g O K GH mol O atoms JK H 2 atoms O K
N O = 16.00 g O
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23
N O2
2
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molecules O2
The second calculation can be done by using the packaging of moles of atoms in moles of molecules.
N O2 = 16.00 g O
FG 1 mol O IJ FG 1 mol O IJ F 6.022 10 molecules O I = 3.011 10
JK
mol O
H 16.00 g O K H 2 mol O K GH
2
23
2
23
molecules O2
2
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7.2 Molecular Mass
We have seen that we can count the number of atoms and molecules present in an element by weighing
out a sample of the element and using Avogadro’s Number to convert the mass to the number of atoms
and molecules. The question is can we do the same when we have a molecule rather than an element?
The answer is yes and this is demonstrated in the following examples.
Examples:
1. What is the unit mass of carbon monoxide, CO?
Carbon monoxide consists of one atom of carbon and one atom of oxygen. Thus its mass is the sum of the
masses of its component atoms.
Unit mass of CO = 12.011 amu C + 16.00 amu O = 28.01 amu CO
Molecular Mass = sum of the masses of the individual atoms
This unit mass of CO is known as the molecular mass (or molar mass) of CO.
2. What is the molecular mass of methane, CH4?
From the Periodic Table, the atomic mass of C is 12.011 and of H is 1.008.
The Molecular Mass of CH4 is:
1 atom C
amu H I
.
FG 12.011 amu C IJ + 4 atoms H FG 1008
H atom C K
H atom C JK = 16.043 amu CH
4
3. What is the molecular mass of aluminum oxide, Al2O3?
From the Periodic Table, the atomic mass of Al is 26.98 and of H is 16.00.
The Molecular Mass of Al2O3 is:
2 atom Al
FG 26.98 amu Al IJ + 3 atoms OFG 16.00 amu O IJ = 10196
H atom Al K
H atom O K . amu Al O
2
3
4. What is the molecular mass of iron (II) sulphate, FeSO4?
From the Periodic Table, the atomic mass of Fe is 55.85, S is 32.07 and O is 16.00.
1 atom Fe
. amu Al I
FG 5585
F 32.07 amu S IJ + 4 atom OFG 16.00 amu O IJ = 15192
+ 1 atom S G
J
H atom Al K
H atom S K
H atom O K . amu FeSO
4
5. What is the molecular mass of ammonium phosphate, (NH4)3PO4?
From the Periodic Table, the atomic mass of:
H = 1.008
N = 14.01
O = 16.00
P = 30.97
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There are two main ways to go about calculating the molar mass of ionic compounds such as ammonium
phosphate.
i)
Ammonium phosphate consists of 3 atoms N, 12 atoms H, 1 atom P and 4 atoms O.
14.007 amu N
1.008 amu H
30.97 amu P
16.00 amu O
+ 12 atoms H
+ 1 atom P
+ 4 atoms O
= 149.09 amu ( NH 4 )3 PO4
atom N
atom H
atom P
atom O
3 atoms N
ii)
An alternative is to calculate the mass of the individual ions and put them together in the molecule.
Mass of NH 4+ = 1 atom N
Mass of PO43 = 1 atom P
14.007 amu N
1.008 amu H
+ 4 atoms H
= 18.04 amu NH 4+
atom N
atom H
30.97 amu P
16.00 amu O
+ 4 atoms O
= 94.97 amu PO43
atom P
atom O
Molecular Mass ( NH 4 )3 PO4 =3 ions NH 4+
18.04 amu NH 4+
+ 1 ion PO43
ion NH 4+
94.97 amu PO43
ion PO43
= 149.09 amu ( NH 4 )3 PO4
Once we have calculated the molecular mass (sometimes referred to as the molar mass) of a compound
we may calculate the number of molecules in a given sample of the compound.
Examples:
1. Calculate the number of molecules of FeSO4 in a 14.327 g sample of FeSO4 . The molar mass of
FeSO4 is 151.91.
N FeSO = 14.327 g FeSO4
4
1 mol FeSO4
151.91 g FeSO4
6.0221 1023 molecules
= 5.6795 1022 molecules FeSO4
mol
2. Calculate the number of molecules of (NH4)3PO4 in a 212.447 g sample of (NH4)3PO4.
N (NH
4 )3PO 4
= 212.447 g (NH 4 )3 PO4
1 mol (NH 4 )3 PO4
149.09 g (NH 4 )3 PO4
6.02214 10 23 molecules
= 8.5813 10 23 molecules (NH 4 )3 PO4
mol
We can also calculate the number of atoms and molecules of the elements that made up the compound.
Examples:
1. Calculate the number of each type of atom in a 28.115 g sample of FeSO4 .
nFeSO4 = 28.115 g FeSO4
mol FeSO4
= 0.18508 mol FeSO4
151.91 g FeSO4
N Fe = 0.18508 mol FeSO4
1 mol Fe
1 mol FeSO4
6.0221 1023 atoms
= 1.1146 1023 atoms Fe
mol
N S = 0.18508 mol FeSO4
1 mol S
1 mol FeSO4
6.0221 10 23 atoms
= 1.1146 10 23 atoms S
mol
N O = 0.18508 mol FeSO4
4 mol O
1 mol FeSO4
6.0221 1023 atoms
= 4.4582 10 23 atoms O
mol
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2. Calculate the number of each type of molecule of the element in a 28.115 g sample of FeSO4 .
nFeSO4 = 28.115 g FeSO4
mol FeSO4
= 0.18508 mol FeSO4
151.91 g FeSO4
N Fe = 0.18508 mol FeSO4
1 mol Fe
1 mol FeSO4
6.0221 1023 molecule
= 1.1146 1023 molecule Fe
mol
N S = 0.18508 mol FeSO4
1 mol S
1 mol FeSO4
6.0221 1023 molecule
= 1.1146 1023 molecule S
mol
N O2 = 0.18508 mol FeSO4
4 mol O
1 mol FeSO4
1 mol O2
2 mol O
6.0221 1023 molecule
= 2.2291 10 23 molecule O2
mol
Note that the elemental form of sulphur is S8 but usually the monatomic form is used for simplicity.
We can also calculate the mass of atoms and molecules of the elements that made up the compound.
Example:
Calculate the mass of each type of atom in a 28.115 g sample of FeSO4 .
nFeSO4 = 28.115 g FeSO4
mol FeSO4
= 0.18508 mol FeSO4
151.91 g FeSO4
mFe = 0.18508 mol FeSO4
1 mol Fe
1 mol FeSO4
55.85 g Fe
= 10.336 g Fe
mol Fe
mS = 0.18508 mol FeSO4
1 mol S
1 mol FeSO4
32.066 g S
= 5.9348 g S
mol S
mO = 0.18508 mol FeSO4
4 mol O
1 mol FeSO4
16.00 g O
= 11.845 g O
mol O
7.3 Percent Composition of Compounds
So far the formulas of compounds have been expressed in terms of counting the atoms present in the
molecule, i.e., the compounds sodium hydrogen phosphate, Na2HPO4 , consists of 2 atoms sodium, 1 atom
each hydrogen and phosphorus and 4 atoms of oxygen.
There is an older method of expressing the composition of compounds, one that predates the discovery of
atoms. This method expresses the composition of compounds through the relative mass of each of the
elements in the compound. The traditional way is to express the composition as a mass percentage of
each component.
% mass =
mass of element in the compound
100%
total mass of the compound
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Example:
Calculate the % mass of each element in sodium hydrogen phosphate, Na2HPO4 .
Mass of Na
=
2 x 22.99 =
45.98
Mass of H
=
1 x 1.008 =
1.008
Mass of P
=
1 x 30.974 =
30.974
Mass of O
=
4 x 16.00 =
64.00
Molar mass of Na2HPO4
=
141.96
45.98
100% = 32.39%
141.96
1.008
% mass H =
100% = 0.7100%
141.96
30.974
% mass P =
100% = 21.819%
141.96
64.00
% mass O =
100% = 45.08%
141.96
% mass Na =
Note that
= % mass Na + % mass H + % mass P + % mass O
% mass total
= 32.39%+0.7100%+21.819%+45.08%
=100.0%
Example:
Calculate the % mass of each element in glucose, C6H12O6.
Mass of C
=
6 x 12.011 =
72.066
Mass of H
=
12 x 1.008 =
12.10
Mass of O
=
6 x 16.00 =
96.00
Molar mass of C6H12O6
=
180.17
72.066
100% = 40.000%
180.17
12.10
% mass H =
100% = 6.716%
180.17
96.00
% mass O =
100% = 53.28%
180.17
% mass C =
To two decimal places, the total % mass sums to 100.00% as it should.
Note that the mass percent of each element in a given compound is always constant, regardless of the
actual mass of the sample. Thus, for example, a 20.000 g sample of glucose will still consist of 40.000%C,
6.716%H and 53.28%O. This fact can be used (if the % composition is known) to calculate the mass of
elements that make up a particular compound of any sample size.
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Example: A sample of an organic compound has a mass of 47.29 mg. If the compound has a composition
of 37.0%C, 2.21%H, 42.4%O and 18.4%N, calculate the mass of each element in the sample.
mC = 47.29 mg compound
37.0 g C
= 17.5 mg C
100.00 g compound
mH = 47.29 mg compound
2.21 g H
= 1.05 mg H
100.00 g compound
mO = 47.29 mg compound
42.4 g O
= 20.1 mg O
100.00 g compound
mN = 47.29 mg compound
18.4 g N
= 8.70 mg N
100.00 g compound
Check: 17.5 mg C + 1.04 mg H + 20.0 mg O + 8.70 mg N = 47.3 mg compound , which is correct within the
limits of significant figures.
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