35
MOLES AND MOLE CALCULATIONS
INTRODUCTION
The purpose of this section is to present some methods for calculating both how much of
each reactant is used in a chemical reaction, and how much of each product is formed.
In this section theoretical details are kept to the minimum necessary for a mastery of
calculation methods to be achieved. It is desirable, however, that at some stage students
learn of the historical development of the ideas applied here, and of the underlying theory.
It is assumed that students have a good mastery of basic arithmetic and algebra,
especially of calculations involving ratios and proportions.
It is assumed also that students are able to both read and write chemical formulae and
equations, accurately and with fair confidence. Until such knowledge and skills have been
acquired, it is difficult to achieve success with mole calculations.
SCOPE OF CALCULATIONS COVERED IN THIS SECTION
There are three basic requirements to be mastered in this section:
1) to calculate masses of reactants that will be consumed in a reaction, and the masses
of products likely to be formed from a given mass of reactants.
2) to calculate what volume of gas may be either consumed or produced in a reaction
.
3) to calculate what volumes and concentrations of solutions of reactants should be
mixed to produce predicted concentrations of products in the solution.
DEFINITION OF A MOLE AND OF MOLAR MASS
Atoms of different elements have different masses. It is possible both to measure the mass
of individual atoms, and to measure the relative masses of different kinds of atoms. Actual
masses of atoms are very small (an atom of carbon has mass = 2 x 10-23 g approx), so the
masses are given in atomic mass units (amu).
The mass of an atom of carbon has been selected as a standard for comparing the masses
of atoms: an atom of carbon has been assigned a mass of 12.00 amu.1 The relative atomic
mass of an atom of any other element is given in proportion to the mass of an atom of
carbon. For example, the relative atomic mass of an oxygen atom is given as 16.00 amu,
which indicates that an atom of oxygen is
times more massive than an atom of carbon.
The relative atomic mass of hydrogen = 1.00 amu; an atom of hydrogen has
of the
mass of an atom of carbon.
1
The definitions given here do not allow for the existence of different isotopes of
elements. This omission does not affect the validity of the methods described.
Detailed treatment of this topic should consider the presence of isotopes.
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The relative atomic masses (ram) of some common elements are listed below:
Element
ram
Element
ram
Element
ram
Aluminium
27.0
Gold
197.0
Oxygen
16.0
Barium
137.3
Hydrogen
1.0
Phosphorus
31.0
Bromine
79.9
Iodine
126.9
Potassium
39.1
Calcium
40.1
Iron
55.8
Silicon
28.1
Carbon
12.0
Lead
207.2
Silver
107.9
Chlorine
35.5
Magnesium
24.3
Sodium
23.0
Chromium
52.0
Manganese
54.9
Sulfur
32.1
Cobalt
58.9
Mercury
200.6
Tin
118.7
Copper
63.5
Nickel
58.7
Titanium
47.9
Fluorine
19.0
Nitrogen
14.0
Zinc
65.4
The mass of one mole of an element is the relative atomic mass of an
element, stated in grams.
A mole is defined as the number of atoms in 12.00 g of pure carbon. The
value of this number is 6.023 x 10 23. It is called the Avogadro Number, symbol NA.
One mole of any element also contains 6.023 x 1023, or NA, atoms.
WHAT ABOUT COMPOUNDS?
The mass of one mole of any compound can be calculated by adding the relative atomic
masses of all atoms in the formula, and stating the total in grams.
Examples:
Compound
Formula
Sum of masses of atoms in
formula
Mass of
one mole
Water
H2O
(1.0 x 2) + 16.0
18.0 g
Carbon dioxide
CO2
12.0 + (16.0 x 2)
44.0 g
Sodium chloride
NaCl
23.0 + 35.5
58.5 g
Calcium sulfate
CaSO4
40.1 + 32.1 + (16.0 x 4)
136.2 g
Lead nitrate
Pb(NO3)2
207.2 + 2(14.0 + 16.0 x 3)
331.2 g
Ammonium phosphate
(NH4)3PO4
3(14.0 + 1.0 x 4) + 31.0 + (16.0 x
4)
149.0 g
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The mass of one mole of a substance is the MOLAR MASS, for which the symbol is M.
One mole of any substance contains NA, or 6.023 x 1023, of each particle present.
Examples:
One mole of
with
formula
contains the following
numbers of molecules
or ions
and the following
numbers of atoms
Water
H2O
NA molecules of water
2 x NA atoms of hydrogen,
NA atoms of oxygen
Carbon dioxide
CO2
NA molecules of carbon
dioxide
NA atoms of carbon,
2 x NA atoms of oxygen
Sodium chloride
NaCl
NA sodium ions and
NA chloride ions
NA atoms of sodium,
NA atoms of chlorine
Calcium sulfate
CaSO4
NA calcium ions and
NA sulfate ions
NA atoms of calcium,
NA atoms of sulfur,
4 x NA atoms of oxygen
Lead nitrate
Pb(NO3)2
NA lead ions and
2 x NA nitrate ions
NA atoms of lead,
2 x NA atoms of nitrogen,
6 x NA atoms of oxygen.
Ammonium phosphate
(NH4)3PO4
3 x NA ammonium ions
and NA phosphate ions
3 x NA atoms of nitrogen,
12 x NA atoms of hydrogen,
NA atoms of phosphorus,
4 x NA atoms of oxygen
PERCENTAGE COMPOSITION
The proportion by mass of an element in a compound can be calculated easily by using
molar masses.
For example, the proportion of sodium present in sodium chloride, NaCl, is the ratio of the
mass of one mole of sodium to the mass of one mole of sodium chloride, expressed as a
percentage.
Proportion of sodium in sodium chloride =
= 0.393
Percentage composition of sodium in sodium chloride = 39.3%.
What is the percentage composition of nitrogen in ammonium nitrate, NH4NO3?
Molar mass of ammonium nitrate = 80.0 g
Mass of nitrogen in one mole of ammonium nitrate = 2 x 14.0 g = 28.0
Percentage composiition of nitrogen in ammonium nitrate =
= 35.0%.
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USE OF MOLAR MASS TO PREDICT REACTING MASSES
A balanced chemical equation is a way of describing the relative quantities of reactants and
products that are involved in a reaction. The coefficients may be read to indicate the relative
numbers of atoms, ions, or molecules involved in the reaction.
It is more useful to read the coefficients as indicating the numbers of moles of each
substance involved.
The following equation, for magnesium metal burning in oxygen, can be
read "two atoms of magnesium combine with one molecule of oxygen to
form two 'molecules' (actually ions Mg2+ and O2-) of magnesium oxide."
A better way to read it is "two moles of magnesium metal combine with one
mole of oxygen gas to form two moles of magnesium oxide".
From this, the relative masses of the reactants and products can be predicted:
2 moles
1 mole
2 moles
2 x 24.3 g
32.0 g
2 x 40.3 g
If the relative masses of reactants and products in a reaction can be predicted from a
balanced equation and knowledge of molar masses, then the actual mass of any reactant or
product can be calculated by the use of ratios.
Example one:
What mass of sodium carbonate will be obtained if 3.36 g of pure sodium hydrogencarbonate is heated? (The other products of the reaction are carbon dioxide and water.)
1) Write a balanced equation:
2) Write mole ratios
underneath equation:
3) Calculate molar masses
and write them under:
4) Write in known value,
and "x" for unknown:
Use ratio to calculate x:
2 moles
1 mole
2 x 84.0 g
106.0 g
3.36 g
=
xg
x =
= 2.12 g
The mass of sodium carbonate formed by heating 3.36 g of sodium hydrogencarbonate
= 2.12 g.
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Example two:
What mass of carbon will be converted to carbon monoxide in reducing 1000 g of iron(III)
oxide to iron metal? What masses of iron and carbon monoxide should be formed?
Note that with three "unknowns" in this problem, three algbraic symbols, x, y, z, are used.
The values of x, y, and z are calculated by simple ratio:
=
=
=
Mass of carbon converted = x = 226 g
Mass of iron formed = y = 700 g
Mass of carbon monoxide formed = z = 526 g
The total mass of reactants should equal the total mass of products:
1000 g + 226 g = 700 g + 526 g
Example three:
What mass of lead can be extracted by heating 120 g of solid lead sulfide in air, forming
lead oxide and sulfur dioxide, then heating the lead oxide with carbon, to form metallic lead
and carbon monoxide?
This problem can be solved as above, by writing the equations and carrying out all ratio
calculations. An alternative method uses percentage composition: the problem can be
summed up as "how much lead can be separated from 120 g of lead sulfide?"
Molar mass of PbS = (207.2 + 32.1) g = 239.3 g.
Percentage of lead in lead sulfide =
= 86.6%
86.6% of 120 g = 104 g = mass of lead that can be extracted from 120 g of lead sulfide.
Exercises:
1. What mass of copper can be extracted from 5.0 g of copper(II) sulfate by dissolving the
copper sulfate in water and adding zinc metal? (The other product is zinc sulfate).
2. What mass of potassium iodide is needed to react exactly with 8.0 g of lead nitrate, to
form lead iodide? (The other product is potassium nitrate).
3. When calcium carbonate is heated strongly, it forms calcium oxide and carbon dioxide.
What mass of calcium carbonate is needed to make 50.0 g of calcium oxide?
4. Sodium carbonate reacts with hydrochloric acid to form sodium chloride, water, and
carbon dioxide. Some hydrochloric acid was added to some sodium carbonate: 6.0 g of
sodium chloride were formed. What mass of carbon dioxide was produced?
5. What mass of lead oxide would need to be reacted with nitric acid to produce 10.0g of
lead nitrate?