# TOPIC 7. CHEMICAL CALCULATIONS I

#### Document technical information

Format pdf
Size 496.1 kB
First found May 22, 2018

#### Document content analysis

Category Also themed
Language
English
Type
not defined
Concepts
no text concepts found

#### Transcript

```TOPIC 7.
CHEMICAL CALCULATIONS I - atomic and formula weights.
Atomic structure revisited.
In Topic 2, atoms were described as ranging from the simplest atom, H, containing
a single proton and usually no neutrons in its nucleus with one electron orbiting
outside that nucleus, through to very large atoms such as uranium for example
which contains 92 protons and even more neutrons in its nucleus. The point was
also made that the mass of the atom is almost entirely located in its nucleus, being
attributable to protons and neutrons, while the mass of electrons is negligible.
Amount in chemistry.
Basic principle: atoms form compounds by combining in simple numerical ratios
but not in simple mass ratios.
Because the masses of atoms of different elements are different, in quantitative
calculations it is not possible to say, for example, that if one atom of element A
reacts with one atom of element B, then 1 gram of A reacts exactly with 1 gram of
B. Thus while atoms do actually combine in simple integer ratios (like 1:1 in NaCl
or 2:1 in H 2O), they do not combine in simple mass ratios. In the example of
NaCl, the actual ratio by mass of Na : Cl in this compound is 22.99 g : 35.45 g.
How can this ratio be deduced? To calculate the amounts of substances that are
needed to react exactly with a given amount of other substances, a method of
counting in chemistry is essential. Because a single atom or molecule is such a
small amount of material, we need a more appropriate way of measuring amount
than counting atoms or molecules.
To illustrate this need by analogy, consider our method of counting money. Use of
cents as the basic unit of amount of money would be limited to only very cheap
purchases - indeed the smallest coin now in use is the 5 cent piece. Instead, most
of our transactions are in units of amount of dollars which is the basic unit used in
most cases. The cent is just too small an amount to be the base unit of money. As
another example, when our government engineered the change to metric units for
weights and measures, they decreed the unit of length to be the millimetre. The
result is very large numbers appear on drawn plans and in specifications. The latter
is a good example of how the unit selected as the base unit can be too small to be
convenient. Similarly, as atoms and molecules individually are exceedingly small,
a very large number of atoms or molecules of elements or compounds is required
as the basic unit of amount.
Atomic masses.
The mass of a single atom of hydrogen, 1 H, is 1.66 × 10–24 g while the mass of a
single atom of the most common isotope of uranium, 238U, is 3.95 × 10–22 g. These
masses are not obtained by weighing any individual H or U atoms but are deduced
from the mass of a very large number of these atoms. In calculations, these very
small numbers are inconvenient and, in any case, they refer to invisibly small
amounts of materials. Instead, rather than the actual masses of atoms, a scale of
RELATIVE ATOMIC MASSES is used. The most obvious way to set up a scale
of relative atomic masses is to adopt the mass of the lightest atom, 1H, as exactly 1
and express the masses of all other atoms relative to the 1H atom.
VII - 1
VII - 2
Using this basis, the relative atomic mass of a uranium atom, 238U, would be
mass of 1 238U atom
mass of 1 1H atom
= 3.95 × 10–22 g
1.66 × 10–24 g
=
238
1
=
238
Comparing these two atoms,
Thus a single 238 U atom is 238 times heavier than a single 1H atom. Similarly, the
relative atomic masses or weights of atoms of all elements can be obtained and
tabulated. Note that this number has no units as it is a ratio only.
As another example, consider a helium atom containing 2 protons + 2 neutrons. If
one makes the approximation that a proton has the same mass as a neutron, it could
be predicted that the relative atomic mass of helium would be 4. Similarly, carbon
atoms with 6 protons and 6 neutrons would have a relative atomic mass = 12.
On this basis, the relative atomic weights of the most common isotopes of the first
6 elements would be H = 1, He = 4, Li = 7, Be = 9, B = 11, C = 12.
However, the tables of relative atomic masses, called ATOMIC WEIGHTS or
ATOMIC MASSES for short, (the terms “mass” and “weight” are typically used
interchangeably in this topic) apply to the weighted average for all isotopes for
each element as it occurs in nature, not to just the most common isotope of the
element. Most elements occur naturally as two or more isotopes, and therefore their
relative atomic weights are not always integer quantities. [See the box below.]
VII - 3
CALCULATION OF THE AVERAGE ATOMIC WEIGHT OF SILVER
from isotopic abundance ratios.
The element silver, Ag, atomic number = 47, occurs naturally as only two
isotopes: 107 Ag (abundance = 51.84 %) and 109Ag (abundance = 48.16 %).
The weighted average of the relative atomic masses of these two isotopes is
calculated as follows:
Weighted average = isotopic mass 1 × fractional abundance 1
+ isotopic mass 2 × fractional abundance 2
= 107.0 × 0.5184 + 109.0 × 0.4816
= 55.46 + 52.49
= 107.9
In this calculation, the isotopic masses have been taken as 107.0 and 109.0 on
the basis of the number of protons + neutrons present in each isotope, and
taking the mass of a proton and a neutron as being identical. A more precise
calculation and using the unit of atomic weight scale to be 1/12 the mass of a
12
C atom would give a value of 107.87 for the relative atomic weight of silver.
For various reasons, the basis taken for the atomic weight scale has been changed a
number of times over the past century. The currently used basis is to define the
unit of atomic weight scales as 1/12 of the mass of the atom of the carbon isotope
with 6 protons and 6 neutrons in its nucleus, 12C. The values which result for
atomic weights of the elements using this basis are almost the same as that obtained
by taking the atomic weight of hydrogen as 1. The following table lists the
(relative) atomic weights to 1 decimal place for the first 20 elements.
Element::
H
He
Li
Be
B
Atomic mass:
1.0
4.0
6.9
9.0
10.8 12.0 14.0 16.0 19.0 20.2
Element:
Atomic mass:
C
N
O
F
Ne
Na
Mg Al
Si
P
S
Cl
Ar
23.0 24.3 27.0 28.1 31.0 32.1 35.5 40.0
For a complete table of more precise atomic weights, see the Periodic Table on the
tear-out page at the back of this book. There is no need to rote learn values of
atomic weights as precise tabulations are always available.
From the table above, it is seen for example that each He atom has four times the
mass of one H atom. Consequently, it follows that in 4.0 g of helium atoms there
would be the same number of He atoms present as there are atoms of hydrogen in
1.0 g of H atoms. Similarly, 6.9 g of Li atoms, 9.0 g of Be atoms or 40.0 g of Ar
atoms would all have the same number of atoms present as there are in 1.0 g of H
atoms. By several experimental methods, this number of atoms has been
determined to be 6.022 × 1023 atoms.
Thus,
1.0 g of H atoms contains
6.022 × 1023
H atoms
4.0 g of He atoms contains
6.022 × 1023
He atoms
6.9 g of Li atoms contains
6.022 × 1023
Li atoms
VII - 4
Likewise,
9.0 g of Be atoms contains
.
.
40.0 g of Ar atoms contains
6.022 × 1023
Be atoms
6.022 × 1023
Ar atoms
and for any element, the relative atomic weight expressed in grams (GRAM
ATOMIC WEIGHT) is the mass of that element which contains the same number
of its atoms as there are in 1.0 g of hydrogen atoms, and that number in each case is
6.022 × 1023 atoms of the element.
This number is called the AVOGADRO NUMBER, and the relative atomic mass
expressed in grams for any element contains an Avogadro number of atoms of that
element. The following examples illustrate these concepts.
Example 1. What mass of copper atoms contains 6.022 × 1023 Cu atoms?
As 6.022 × 1023 atoms is the number of Cu atoms in 1 gram atomic weight of Cu,
and the atomic weight tables list Cu = 63.55,
then 63.55 g of copper contains 6.022 × 1023 Cu atoms.
Example 2. How many neon atoms are present in 10.1 g of Ne?
From the atomic weight tables, the atomic weight of neon = 20.2.
Therefore 20.2 g of neon contains 6.022 × 1023 Ne atoms.
By proportion, 10.1 g of neon contains 10.1 × 6.022 × 1023 Ne atoms
20.2
= 3.011 × 1023 Ne atoms.
Example 3. The mass of one atom of an element Y is found to be 2.00 × 10–23 g.
What is the atomic weight of element Y?
If 1 atom of Y weighs 2.00 × 10–23 g,
then an Avogadro number of atoms weighs 6.022 × 1023 × 2.00 × 10–23 g = 12.0 g.
Therefore, the gram atomic weight of Y = 12.0 g and its atomic weight = 12.0.
The (chemical) mole.
It is convenient to have a shorter name for “an Avogadro number of atoms” and the
"relative atomic mass expressed in grams" or “gram atomic weight or mass”. The
name adopted by chemists is the MOLE . When it is necessary to specify the mass
of a mole of atoms in grams, the term used is MOLAR MASS. One mole of atoms
of any element is an amount equal to the atomic weight of that element expressed
in grams and also it is an Avogadro number of atoms of that element.
Thus 1 mole of hydrogen atoms = 6.022 × 1023 H atoms = 1.0 g of H atoms
and
1 mole of helium atoms
= 6.022 × 1023 He atoms = 4.0 g of He atoms
and
1 mole of lithium atoms
= 6.022 × 1023 Li atoms = 6.9 g of Li atoms
VII - 5
and 1 mole of atoms of any element = 6.022 × 1023 atoms of that element
= the atomic weight of that element expressed in grams.
Thus,
moles of atoms
of any element
= mass of element = number of atoms of element
atomic weight
6.022 × 1023
Comparison of one mole of some atoms:
1
H atom
1
4
He atom
4
1 mole of H atoms contains 1 mole of He atoms contains
6.022 × 1023 atoms
6.022 × 1023 atoms
Mass per mole:
1.0 g
4.0 g (as a 4He atom is 4 ×
heavier than an 1H atom)
Mass per atom:
1.7×10–24 g
6.6 × 10–24 g
16
O atom
1 mole of 16O atoms contains
6.022 × 1023 atoms
16 g (as an 16O atom is 16 ×
heavier than an 1H atom)
27×10–24 g
Relative molecular weights.
It has already been seen that most elements occur in larger aggregates rather than as
single atoms. Some occur as molecules, for example hydrogen and oxygen both
normally occur as the diatomic molecules H 2 and O 2 rather than as individual H or
O atoms. These two elements react to form the compound water, H 2O, another
molecule. The H 2 molecule must have twice the mass of a single H atom and
likewise an O 2 molecule must have twice the mass of a single O atom.
The relative masses of any molecules such as these can be calculated by simply
adding all the component atoms' atomic weights.
Thus the H 2 molecule has a RELATIVE MOLECULAR WEIGHT (usually
shortened to “molecular weight”) = 2 × 1.0 = 2.0 (using atomic weights data to
one decimal place.)
The O 2 molecule has a (relative) molecular weight = 2 × 16.0 = 32.0
In the case of the water molecule, each H 2O molecule has a (relative) molecular
weight = 2 × atomic weight of H + 1 × atomic weight of O
= 2 × 1.0 + 16.0 = 18.0
Similarly, the molecular weight of carbon dioxide, CO 2, = 12.0 + (2 × 16.0)
= 44.0
Because molecular weights are based on addition of atomic weights, it necessarily
follows that the molecular weight expressed in grams (gram molecular weight)
for any molecular species would contain 6.022 × 1023 molecules of that species.
This is also an Avogadro number of molecules and the terms "mole" and “molar
mass” are equally applicable to molecules as to atoms. Thus 1 mole of any
molecular species contains 6.022 × 1023 molecules and has a mass equal to the
molecular weight expressed in grams, its molar mass.
VII - 6
For example,
2.0 g of hydrogen contains 6.022 × 1023 H 2 molecules
32.0 g of oxygen contains 6.022 × 1023 O 2 molecules
18.0 g of water
contains 6.022 × 1023 H 2O molecules
i.e. the molar mass of any molecular species contains
molecules and is one mole of that species.
= 1 mole H 2 molecules
= 1 mole O 2 molecules
= 1mole H 2O molecules
Comparison of one mole of some molecules:
It was shown in earlier Topics that ionic species don't form molecules as such, and
that the empirical formula of these compounds is the only formula possible.
Consequently as there are no molecules, the term "molecular weight" strictly has no
meaning when applied to ionic compounds. Instead, for ionic species the term
“molar mass” refers to the GRAM FORMULA WEIGHT where the formula is
the empirical formula of the ionic compound. The gram formula weight is the sum
of the constituent atomic weights of all the atoms in the empirical formula,
expressed in grams. Again, the term “molar mass” is more commonly used than
“gram formula weight” for these species, but the term “gram formula weight” is
used in sections of these notes as it has the advantage of being self-explanatory.
Gram formula weight can, like molar mass, be applied to any species regardless of
whether they be individual atoms or molecules or ionic compounds or aggregates of
atoms as in the case of all metals and most non-metals which are represented by
just their atomic formulas.
For example, using the atomic formula for the metal magnesium, 1 mole of
magnesium represented by the formula Mg, would contain 6.022 × 1023 Mg atoms
and would weigh 24.3 g.
Similarly, 1 mole of the ionic compound sodium chloride, represented by the
empirical formula NaCl, would have a mass = 23.0 + 35.5 = 58.5 g. This mass
of sodium chloride would contain 1 mole of Na+ ions and 1 mole of Cl– ions, i.e.
6.022 × 1023 of each ion.
VII - 7
This familiar container of table salt shown on the
right holds 125 g of sodium chloride.
One mole of sodium chloride has a mass equal
to the sum of the gram atomic weights of Na and Cl.
i.e. the gram formula weight or molar mass of NaCl
= 23.0 + 35.5 = 58.5 g.
Therefore the container holds 125 moles of NaCl
58.5
= 2.1 moles of salt.
The following expression summarises the relationships between moles, mass and
number of molecules.
moles of substance = mass of substance = number of entities
(of specified formula)
molar mass
6.022 × 1023
Note that while 1 mole of hydrogen molecules weighs 2.02 g and it necessarily
contains 6.022 × 1023 H 2 molecules, the same mass of hydrogen if converted to
individual H atoms, would contain 2 × 6.022 × 1023 H atoms = 12.044 × 1023 H
atoms. Thus, 1 mole of H 2 molecules contains 2 moles of H atoms bonded to form
6.022 × 1023 H2 molecules. Similarly, there are 2 moles of O atoms in each mole of
O2 molecules.
This poses the question as to what is meant when referring to a mole of a given
element - is it atoms or molecules of that element?
The answer is that it must be made clear in some way in the question as to which
species is involved. For those few elements that occur normally as individual
atoms, (the noble gases), the mole refers to an Avogadro number of atoms of each
element and would have a mass equal to its gram atomic weight. For those
elements that occur normally as diatomic molecules, (H 2, N 2, O 2, F 2, Cl 2, Br 2 and
I2), the mole refers to an Avogadro number of diatomic molecules and would have
a mass equal to the gram molecular weight. All other elements exist normally as
VII - 8
larger, often indeterminate, aggregates of atoms and the mole is usually taken as
referring to an Avogadro number of atoms of each element even though the element
is not made up of individual free atoms. Thus, a mole of carbon is taken to mean
the gram atomic weight of that element, 12.0 g, even though carbon in all its
naturally occurring forms has enormous numbers of C atoms joined by chemical
bonds. Similarly, all metals consist of extremely large numbers of atoms joined by
metallic bonds at normal conditions, but the mole is taken to mean an Avogadro
number of the atoms and the molar mass of any metal is its gram atomic weight.
Check your understanding of this section.
Why is it that when atoms of elements form compounds, they don’t combine in
simple integer ratios by mass?
Why are relative atomic weights of the elements usually not integers?
Give two definitions for a mole of helium.
Define the term “relative atomic weight”.
How many helium atoms are present in 8.0 g of helium?
What mass of sodium chloride constitutes two moles of this compound?
Define the term “molar mass”.
Calculate the mass of one mole of (i) carbon (ii) oxygen (iii) carbon dioxide.
How many moles of (i) C atoms (ii) O atoms are present in one mole of carbon
dioxide?
The mole in review.
From the above, it is apparent that there are two definitions of the mole:
(i) as the formula weight of any chemical species expressed in grams.
(ii) as simply a number of entities, 6.022 × 1023, atoms or molecules of a given
chemical species.
This second definition can be extended to include things other than specific
chemical species - in particular, electrons. Use will be made of this extended
definition later when moles of electrons transferred in chemical reactions will be
encountered in reactions where oxidation and reduction occur.
The following table summarises how to interconvert mass, moles and number of
constituent formula units (atoms/molecules/empirical formulas).
VII - 9
× NA
÷ molar mass
mass in grams
W
moles
× molar mass
where
W
number of atoms, molecules
÷ NA
or formula units
molar mass is the gram formula weight and
NA is the Avogadro constant, 6.022 × 1023 per mole.
[The “per” associated with any units is represented mathematically by the exponent
–1, so in this case, per mole can be written as “mole–1 ”.]
Abbreviation used for “mole”.
Units attached to numbers are typically abbreviated, such as “g” for grams as in
10.0 g or “L” for litres as in 15.0 L. For moles, the unit is abbreviated to “mol” as
for example in 2.0 g = 1.0 mol of H 2. When first encountered, this might wrongly
be confused as an abbreviation for molecules. There is no shorthand for
“molecules”.
The following examples further illustrate the concepts discussed in this topic.
Example 1. What is the mass of 1.00 mole of carbon dioxide?
The gram formula weight (molar mass) of CO 2 = 12.0 + 2 × 16.0 = 44.0 g mol!1.
Therefore 1.00 mole would be 44.0 g of CO 2 molecules.
Example 2. What is the mass of 1.00 mole of the salt lithium fluoride?
Lithium fluoride, an ionic compound, has the (empirical) formula LiF.
The gram formula weight (molar mass) of LiF = (6.9 + 19.0) g = 25.9 g mol!1.
Therefore 1.00 mole of LiF weighs 25.9 g.
Example 3. What is the mass of 0.500 mole of C 12H 22O 11?
The gram formula weight (molar mass) = (12 × 12.0 + 22 × 1.0 + 11 × 16.0) g
= 342 g mol!1.
Therefore, 0.500 mole weighs 0.500 × 342 g = 171 g.
Example 4. How many molecules are present in 0.500 mole of water?
As 1 mole of water contains 6.022 × 1023 H 2O molecules, then 0.500 mole would
contain 0.500 × 6.022 × 1023 molecules = 3.01 × 1023 molecules.
VII - 10
Example 5. How many molecules are present in 8.0 g of oxygen?
Oxygen is diatomic normally so the molecular formula O 2 is assumed.
The mass of 1 mole of O 2 = 2 × 16.0 = 32.0 g.
Therefore, 8.0 g of O 2 would be 8.0/32.0 mole = 0.25 mole.
As 1 mole contains 6.022 × 1023 molecules of O 2, then 0.25 mole contains
0.25 × 6.022 × 1023 molecules of O 2
= 1.5 × 1023 molecules.
Example 6. How many molecules are present in a 2.00 g cube of table sugar having
the molecular formula C 12H 22O 11.
The mass of 1 mole would be (12 × 12.0 + 22 × 1.0 + 11 × 16.0) g = 342 g
which contains 6.022 × 1023 molecules of C 12H 22O 11.
Therefore, 2.00 g contains (2.00/342) × 6.022 × 1023 molecules of C 12H 22O 11
= 3.5 × 1021 molecules.
Example 7. How many moles of each of the constituent ions are there in one mole
of sodium sulfate?
The formula of sodium sulfate, an ionic compound, is Na 2SO 4. Thus each mole of
Na2SO4 would contain two moles of sodium ions, Na+, plus one mole of sulfate
ions, SO42–.
Example 8. How many moles of each of the constituent ions are there in one mole
of barium chloride and what mass of each species would be present?
The formula of barium chloride, also an ionic compound, is BaCl 2. Thus each mole
of BaCl2 would contain one mole of barium ions, Ba2+, plus two moles of chloride
ions, Cl–.
To obtain the mass of each of these species present, use the atomic weights of Ba
and Cl. [Note that the difference in the mass of an atom such as Ba and its cation,
Ba2+ , is negligible, so atomic weights are used even for cations and anions.]
 mass of one mole of Ba2+ = 137.3 g and
the mass of two moles of Cl– = 2 × 35.5 g = 71.0 g.
VII - 11
Objectives of this Topic.
When you have completed this Topic, including the tutorial questions, you should
have achieved the following goals:
1.
Know why atoms of different elements have different masses.
2.
Understand the terms (relative) atomic mass (weight); (relative) molecular
mass (weight); gram formula weight.
3.
Know how to obtain the atomic weight of any element from the tables.
4.
Know that the atomic weight of any element when expressed in grams
contains 6.022 × 1023 atoms of that element, and that this number is called
5.
Be able to calculate the number of atoms in a given mass of an element, and
be able to calculate the mass of a given number of atoms of an element.
6.
Understand the meaning of the terms "mole" and “molar mass”.
7.
Be able to calculate the molar mass of a compound.
8.
Be able to interconvert a given mass of a compound, moles of the compound
and number of constituent entities.
9.
Recognise when the molar quantity of a compound or element refers to an
empirical formula or to a molecular formula.
10.
Understand that the mole when regarded as an Avogadro number of entities
can apply to species other than elements or compounds, in particular, to
electrons.
VII - 12
SUMMARY
Atoms of each element have their own characteristic masses due to the different
numbers of protons and neutrons in their nuclei. Consequently, while atoms
combine in simple integer ratios when forming compounds, they do not combine in
simple integer ratios by mass.
Masses of individual atoms and molecules are exceedingly small, so calculating
amounts that combine in reactions on the basis of actual masses of each constituent
atom would involve inconveniently small quantities for the basic unit and
extremely large numbers of each atom. Instead, a relative scale of atomic masses
was devised wherein the mass of the atom of each element is expressed relative to
that of the lightest atom, hydrogen. This basis has been altered to replace the mass
of the H atom with 1/12 the mass of the carbon isotope, 12 C, although the resulting
values are almost the same as for the hydrogen-based scale. The atomic mass of
each element is calculated as the weighted average of all the naturally occurring
isotopes of that element and consequently for most elements, the atomic weight is
not an integer.
Because of the relative nature of the atomic weight scale, it follows that the atomic
weight of any element expressed as grams must contain the same number of atoms
as the atomic weight of any other element expressed as grams. This quantity is
called the “gram atomic weight” and the number of constituent atoms is known as
the Avogadro number, determined by experiment to be 6.022 × 1023 atoms.
It is more convenient to use the name “mole” to refer to the gram atomic weight of
an element. Thus one mole of atoms of any element has a mass equal to its atomic
weight expressed as grams and contains an Avogadro number of atoms.
On the same basis, compounds consisting of molecules have a characteristic
relative molecular weight which is the sum of the constituent atoms’ atomic
weights. A mole of such a compound contains an Avogadro number of molecules
and has a mass equal to the relative molecular weight expressed as grams.
A number of synonymous terms are to be found in texts. Relative atomic weight
and relative atomic mass, both usually shortened to atomic weight and atomic mass
are identical. Similarly, relative molecular weight and relative molecular mass,
again shortened to molecular weight and molecular mass, are identical. Gram
atomic weight, gram molecular weight, gram formula weight and molar mass all
refer to the mass of one mole of an element or compound, i.e. the mass containing
an Avogadro number of the constituent entities as defined by its formula.
The formula weight definition of the mole cannot be applied to species other than
those having a defined chemical formula but the Avogadro number definition of the
mole can be applied to any species, for example, electrons.
The mole of the few non-metals that usually occur as diatomic molecules such as
O2, H2 etc refers to their molecular formulas. However, for elements such as metals
and most other non-metals which consist of extremely large aggregates of atoms,
the term mole is applied on the basis of a single atom of that element even though
the element occurs as enormous numbers of bonded atoms. Thus for example, a
mole of carbon is taken as the amount in 12.0 grams of that element and containing
6.022 × 1023 C atoms. For non-molecular compounds such as all ionic compounds,
the mole applies to the empirical formula such as NaCl for sodium chloride.
Interconversion between mass of a substance, moles of substance and number of
constituent entities requires the molar mass (from atomic weight tables) and the
Avogadro number (6.022 × 1023) respectively.
VII - 13
TUTORIAL QUESTIONS - TOPIC 7.
1. Explain the meaning of each of the following terms:
(i) relative atomic mass.
(ii) gram atomic weight.
(iii) gram formula weight.
(iv) the mole.
(v) molar mass
2. Write the chemical formula for each of the following elements or compounds
and calculate the number of moles in the masses specified. Part (i) is an example.
Chemical
formula
Number of moles
(i)
carbon (18.0 g)
(ii)
argon (12.5 g)
(iii) bromine (23.6 g)
(iv) aluminium (36.4 g)
(v) water (36.0 g)
(vi) silicon tetrachloride (200 g)
(vii) sodium hydrogencarbonate (20.0 g)
(viii) iron(II) phosphate (50.0 g)
(ix) sulfur (2.80 g)
(x) chlorine (18.3 g)
(xi) ammonium sulfate (42.8 g)
(xii) potassium dichromate (94.6 g)
(xiii) copper(II) sulfate (20.4 g)
(xiv) sulfur dioxide (15.0 g)
C
18.0 / 12.01 = 1.50
VII - 14
3. Write the chemical formula for each of the following elements or compounds and
calculate the mass of the number of moles specified.
Formula
Mass
(i)
silver (0.150 mol)
(ii)
silicon (2.00 mol)
(iii) hydrogen (2.00 mol)
(iv) iodine (0.100 mol)
(v) water (2.00 mol)
(vi) sodium chloride (3.50 mol)
(vii) barium chloride (0.370 mol)
(viii) sulfur trioxide (1.87 mol)
(ix) iron(II) phosphate (0.550 mol)
(x) potassium dichromate (1.70 mol)
(xi) ammonium sulfate (1.70 mol)
(xii) copper(II) nitrate (0.480 mol)
(xiii) manganese(II) hydroxide (1.50 mol)
(xiv) nickel(II) iodide (2.90 mol)
(xv) lithium carbonate (5.30 mol)
4. Write the formulas for the constituent ions of the compounds below. Calculate
how many moles of each ion is present in the given mass of each compound.
(i)
sodium chloride (65.0 g)
(ii)
potassium sulfide (23.5 g)
(iii)
cobalt(II) chloride (20.0 g)
(iv)
potassium oxide (50.0 g)
(v)
lithium carbonate (32.5 g)
VII - 15
(vi)
iron(III) phosphate (100 g)
(vii)
aluminium sulfate (50.0 g)
5. (a) (i) How many moles of carbon dioxide would be present in a sample
containing 4.53 × 1024 molecules?
(ii) What would be the mass of this amount of carbon dioxide?
(b) (i) How many moles of water are present in a glass containing 7.34 × 1023
molecules?
(ii) What would be the mass of this amount of water?
(c) The mass of a crystal of rock salt, sodium chloride is 5.15 g. Calculate the
number of sodium ions and chloride ions in that crystal.
(d) A sugar cube has a mass = 6.50 g. If the cube consists of pure sucrose of formula
C12H22O11, how many sucrose molecules are present in the sugar cube?
1. (i) Relative atomic mass: The mass of an atom relative to that of, initially, the H
atom, but currently taken as relative to 1/12 of the mass of a carbon-12 atom. Being
a ratio of two masses, it has no units.
(ii) Gram atomic weight: The relative atomic weight of any atom expressed as
grams. Units are grams.
(iii) Gram formula weight: The sum of the atomic weights of all the component
atoms in a compound, expressed as grams. Can be applied to atoms, molecular
formulas or, in the case of ionic compounds, to the empirical formula. Units are
grams.
(iv) Mole: The amount of a substance in an Avogadro number (6.022 x 1023) of its
constituent entities. For an element or compound, the mole is also the amount
present in a mass of that substance equal to its gram formula weight, i.e. the sum of
the atomic weights of all of its constituent atoms expressed as grams. Units are
moles, abbreviated “mol”.
(v) Molar mass: Same as gram formula weight. Can be used to refer to atoms,
molecules or formula units such as the empirical formulas of ionic compounds.
Units are grams.
VII - 16
2.
(i)
(iii)
(v)
(vii)
(ix)
(xi)
(xiii)
C; 1.50 mol
Br2; 1.48 × 10–1 mol
H2O; 2.00 mol
NaHCO 3; 2.38 × 10–1 mol
S; 8.73 × 10–2 mol
(NH 4)2SO 4; 3.24 × 10–1 mol
CuSO 4; 1.28 × 10–1 mol
(ii)
(iv)
(vi)
(viii)
(x)
(xii)
(xiv)
Ar; 3.13 × 10–1 mol
Al; 1.35 mol
SiCl4; 1.18 mol
Fe 3(PO 4) 2; 1.40 × 10–1 mol
Cl2; 2.58 × 10–1 mol
K 2Cr2O 7; 3.22 × 10–1 mol
SO 2; 2.34 × 10–1 mol
Explanations and Worked Solutions.
All parts of the question involve converting mass to moles. Before this can be done,
it is essential that the correct formula for the element or compound must be
used. Then in each part, calculate the molar mass (gram formula weight) of the
element or compound and divide the specified mass of the substance by its molar
mass.
(i) The element carbon is represented by the symbol for its atom, C. The atomic
weight of carbon is 12.01 or, its molar mass = 12.01 g mol–1.
 moles of carbon in 18.0 g = 18.0 ÷ 12.01 = 1.50 mol.
(ii) The element argon is represented by the symbol for its atom, Ar. The atomic
weight of argon is 39.95 or, its molar mass = 39.95 g mol–1.
 moles of argon in 12.5 g = 12.5 ÷ 39.95 = 0.313 mol or 3.13 × 10–1 mol.
(iii) The element bromine consists of diatomic molecules and so it is represented by
the molecular formula, Br2 and not by the symbol for its atom, Br. The atomic
weight of bromine is 79.90 so its molecular weight = 2 × 79.90 = 159.8 or its molar
mass = 159.8 g mol–1.
 moles of bromine in 23.6 g = 23.6 ÷ 159.8 = 0.148 mol or 1.48 × 10–1 mol.
(iv) The element aluminium is represented by the symbol for its atom, Al. The
atomic weight of aluminium is 26.98 or, its molar mass = 26.98 g mol–1.
 moles of aluminium in 36.4 g = 36.4 ÷ 26.98 = 1.35 mol.
(v) The compound water consists of molecules having molecular formula H 2O.
The molecular weight of water = sum of the atomic weights of its component atoms
= 2 × 1.008 + 16.00 = 18.02 or its molar mass = 18.02 g mol–1.
 moles of water in 36.0 g = 36.0 ÷ 18.02 = 2.00 mol.
(vi) The compound silicon tetrachloride consists of molecules having molecular
formula SiCl4.
The molecular weight of silicon tetrachloride = sum of the atomic weights of its
component atoms = 28.09 + 4 × 35.45 = 169.9 or its molar mass = 169.9 g mol–1.
 moles of silicon tetrachloride in 200 g = 200 ÷ 169.9 = 1.18 mol.
(vii) The compound sodium hydrogencarbonate is ionic, consisting of Na+ and
HCO3– ions. Thus the formula for sodium hydrogencarbonate is NaHCO 3. For an
ionic compound, its empirical formula (i.e. the simplest whole number ratio of the
constituent ions) is used as the basis for calculating the molar mass.
The molar mass of sodium hydrogencarbonate = sum of the atomic weights of its
VII - 17
component atoms = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.01 g mol–1.
 moles of sodium hydrogencarbonate in 20.0 g = 20.0 ÷ 84.01
= 0.238 mol or 2.38 × 10–1 mol.
(viii) The compound iron(II) phosphate is ionic, consisting of Fe2+ and PO 43– ions.
Thus the formula for iron(II) phosphate is Fe 3(PO 4) 2. For an ionic compound, its
empirical formula is used as the basis for calculating the molar mass.
The molar mass of iron(II) phosphate = sum of the atomic weights of its component
atoms = 3 × 55.85 + 2 × 30.97 + 8 × 16.00 = 357.49 g mol–1.
 moles of iron(II) phosphate in 50.0 g = 50.0 ÷ 357.49
= 0.140 mol or 1.40 × 10–1 mol.
(ix) Sulfur is an element and is represented by the symbol for its atom, S.
The atomic weight of sulfur is 32.07 or, its molar mass = 32.07 g mol–1.
 moles of sulfur in 2.80 g = 2.80 ÷ 32.07 = 0.0873 mol or 8.73 × 10–2 mol.
(x) The element chlorine consists of diatomic molecules and so it is represented by
the molecular formula, Cl2 and not by the symbol for its atom, Cl. The atomic
weight of chlorine is 35.45 so its molecular weight = 2 × 35.45 = 70.90 or its molar
mass = 70.90 g mol–1.
 moles of chlorine in 18.3 g = 18.3 ÷ 70.9 = 0.258 mol or 2.58 × 10–1 mol.
(xi) The compound ammonium sulfate is ionic, consisting of NH 4+ and SO 42– ions.
Thus the formula for ammonium sulphate is (NH 4) 2SO 4. For an ionic compound, its
empirical formula is used as the basis for calculating the molar mass.
The molar mass of ammonium sulphate = sum of the atomic weights of its
component atoms = 2 × 14.01 + 8 × 1.008 + 32.07 + 4 × 16.00 = 132.14 g mol–1.
 moles of ammonium sulfate in 42.8 g = 42.8 ÷ 132.14
= 0.324 mol or 3.24 × 10–1 mol.
(xii) The compound potassium dichromate is ionic, consisting of K+ and Cr 2O 72–
ions. Thus the formula for potassium dichromate is K 2Cr 2O 7. For an ionic
compound, its empirical formula is used as the basis for calculating the molar mass.
The molar mass of potassium dichromate = sum of the atomic weights of its
component atoms = 2 × 39.10 + 2 × 52.00 + 7 × 16.00 = 294.20 g mol–1.
 moles of potassium dichromate in 94.6 g = 94.6 ÷ 294.20
= 0.322 mol or 3.22 × 10–1 mol.
(xiii) The compound copper(II) sulfate is ionic, consisting of Cu2+ and SO 42– ions.
Thus the formula for copper(II) sulphate is CuSO 4. For an ionic compound, its
empirical formula is used as the basis for calculating the molar mass.
The molar mass of copper(II) sulphate = sum of the atomic weights of its component
atoms = 63.55 + 32.07 + 4 × 16.00 = 159.62 g mol–1.
 moles of copper(II) sulfate in 20.4 g = 20.4 ÷ 159.62
= 0.128 mol or 1.28 × 10–1 mol.
(xiv) The compound sulfur dioxide consists of covalent molecules having molecular
formula SO 2. The molecular weight of sulfur dioxide = sum of the atomic weights
of its component atoms = 32.07 + 2 × 16.00 = 64.07 or its molar mass = 64.07 g
mol–1.
 moles of sulfur dioxide in 15.0 g = 15.0 ÷ 64.07 = 0.234 mol or 2.34 × 10–1 mol
VII - 18
3. Explanations and Worked Solutions.
All parts involve calculating the mass of a given number of moles of the specified
element or compound. Again, the starting point is writing the correct formula
for that element or compound. These calculations are all based on the principle
that one mole of any element or compound has a mass equal to its molar mass (gram
formula weight) which is deduced from the sum of the atomic weights of its
constituent atoms.
(i) Silver is an element and its formula is taken as the symbol of its atom, Ag.
The atomic weight of silver is 107.87 so its molar mass = 107.87 g mol–1.
 the mass of 0.150 mol of silver = 0.150 × 107.87 g = 16.2 g.
(ii) Silicon is an element and its formula is taken as the symbol of its atom, Si.
The atomic weight of silicon is 28.09 so its molar mass = 28.09 g mol–1.
 the mass of 2.00 mol of silicon = 2.00 × 28.09 g = 56.2 g.
(iii) Hydrogen is an element which consists of diatomic molecules so its formula is
taken as H 2 rather than as the symbol of the single atom, H.
One mole of hydrogen therefore refers to one mole of H 2 molecules, the molecular
weight of which is 2 × 1.008 = 2.016.
Thus the molar mass of hydrogen = 2.016 g mol–1.
 mass of 2.00 mol of hydrogen = 2 × 2.016 g = 4.03 g.
(iv) Iodine is another element which exists normally as diatomic molecules so its
formula is taken as I2 rather than as the symbol of the single atom, I. One mole of
iodine therefore refers to one mole of I2 molecules, the molecular weight of which is
2 × 126.90 = 253.80
Thus the molar mass of iodine = 253.8 g mol–1.
 mass of 0.100 mol of iodine = 0.100 × 253.8 g = 25.4 g.
(v) The compound water consists of molecules having molecular formula H 2O.
The molecular weight of water = sum of the atomic weights of its component atoms
= 2 × 1.008 + 16.00 = 18.02 or its molar mass = 18.02 g mol–1.
 mass of 2.00 mol of water = 2.00 × 18.02 g = 36.0 g
(vi) The compound sodium chloride is ionic, consisting of Na+ and Cl– ions. Thus
the formula for sodium chloride is NaCl.
For an ionic compound, its empirical
formula is used as the basis for calculating the molar mass.
The molar mass of sodium chloride = sum of the atomic weights of its component
atoms = 22.99 + 35.45 = 58.44 g mol–1.
 mass of 3.50 mol of sodium chloride = 3.50 × 58.44 g = 205 g
(vii) The compound barium chloride is ionic, consisting of Ba2+ and Cl– ions. Thus
the formula for barium chloride is BaCl 2. For an ionic compound, its empirical
formula is used as the basis for calculating the molar mass.
The molar mass of barium chloride = sum of the atomic weights of its component
atoms = 137.34 + 2 × 35.45 = 208.24 g mol–1.
 mass of 0.370 mol of barium chloride = 0.370 × 208.24 g = 77.0 g
(viii) The compound sulfur trioxide consists of covalent molecules having molecular
formula SO 3.
VII - 19
The molecular weight of sulfur trioxide = sum of the atomic weights of its
component atoms = 32.07 + 3 × 16.00 = 80.07 or its molar mass = 80.07 g mol–1.
 mass of 1.87 mol of sulfur trioxide = 1.87 × 80.07 g = 150 g.
(ix) The compound iron(II) phosphate is ionic, consisting of Fe2+ and PO 43– ions.
Thus the formula for iron(II) phosphate is Fe 3(PO 4) 2. For an ionic compound, its
empirical formula is used as the basis for calculating the molar mass.
The molar mass of iron(II) phosphate = sum of the atomic weights of its component
atoms = 3 × 55.85 + 2 × 30.97 + 8 × 16.00 = 357.49 g mol–1.
 mass of 0.550 mol of iron(II) phosphate = 0.550 × 357.49 g = 197 g.
(x) The compound potassium dichromate is ionic, consisting of K+ and Cr 2O 72– ions.
Thus the formula for potassium dichromate is K 2Cr 2O 7. For an ionic compound, its
empirical formula is used as the basis for calculating the molar mass. The molar
mass of potassium dichromate = sum of the atomic weights of its component atoms
= 2 × 39.10 + 2 × 52.00 + 7 × 16.00 = 294.20 g mol!1.
 1.70 mol of potassium dichromate has a mass = 1.70 × 294.20 g = 500 g.
(xi) The compound ammonium sulfate is ionic, consisting of NH 4+ and SO 42! ions.
Thus the formula for ammonium sulfate is (NH 4)2SO 4. For an ionic compound, its
empirical formula is used as the basis for calculating the molar mass.
The molar mass of ammonium sulfate = sum of the atomic weights of its component
atoms = 2 × (14.01 + 4 × 1.008) + 32.07 + 4 × 16.00 = 132.15 g mol–1.
 1.70 mol of ammonium sulfate has a mass = 1.70 × 132.15 g = 225 g.
(xii) The compound copper(II) nitrate is ionic, consisting of Cu2+ and NO 3– ions.
Thus the formula for copper(II) nitrate is Cu(NO 3) 2. For an ionic compound, its
empirical formula is used as the basis for calculating the molar mass.
The molar mass of copper(II) nitrate = sum of the atomic weights of its component
atoms = 63.55 + 2 × (14.01 + 3 × 16.00) = 187.56 g mol–1.
 0.480 mol of copper(II) nitrate has a mass = 0.480 × 187.56 g = 90.0 g.
(xiii) The compound manganese(II) hydroxide is ionic, consisting of Mn2+ and OH–
ions. Thus the formula for manganese(II) hydroxide is Mn(OH)2. For an ionic
compound, its empirical formula is used as the basis for calculating the molar mass.
The molar mass of manganese(II) hydroxide = sum of the atomic weights of its
component atoms = 54.94 + 2 × (16.00 + 1.008) = 88.96 g mol–1.
 1.50 mol of Manganese(II) hydroxide has a mass = 1.50 × 88.96 g = 133 g.
(xiv) The compound nickel(II) iodide is ionic, consisting of Ni2+ and I– ions. Thus
the formula for nickel(II) iodide is NiI 2. For an ionic compound, its empirical
formula is used as the basis for calculating the molar mass.
The molar mass of nickel(II) iodide = sum of the atomic weights of its component
atoms = 58.69 + 2 × 126.90 = 312.49 g mol–1.
 2.90 mol of nickel(II) iodide has a mass = 2.90 × 312.49 g = 906 g.
(xv) The compound lithium carbonate is ionic, consisting of Li+ and CO 32– ions.
Thus the formula for lithium carbonate is Li 2CO 3. For an ionic compound, its
empirical formula is used as the basis for calculating the molar mass.
The molar mass of lithium carbonate = sum of the atomic weights of its component
atoms = 2 × 6.94 + 12.01 + 3 × 16.00 = 73.89 g mol–1.
 5.30 mol of lithium carbonate has a mass = 5.30 × 73.89 g = 392 g.
VII - 20
4. Explanations and Worked Solutions.
(i) Sodium chloride has the formula NaCl. Thus one mole of NaCl contains one
mole of Na+ ions and one mole of Cl– ions.
i.e. moles of Na+ = moles of Cl! = moles of NaCl.
Molar mass of NaCl = sum of the atomic weights of the component atoms
= 22.99 + 35.45 = 58.44 g mol–1.
Moles of NaCl in 65.0 g = 65.0 ÷ 58.44 = 1.11 mol
 moles of Na + ions = 1.11 mol and moles of Cl– ions = 1.11 mol.
(ii) Potassium sulfide has the formula K 2S. Thus one mole of K 2S contains two
moles of K+ and one mole of S2– ions.
i.e. moles of K+ = 2 × moles of K 2S and
moles of S2! = moles of K 2S.
Molar mass of K 2S = sum of the atomic weights of the component atoms
= 2 × 39.10 + 32.07 = 110.27 g mol–1.
Moles of K 2S in 23.5 g = 23.5 ÷ 110.27 = 0.213 mol
 moles of K+ ions = 2 × 0.213 = 0.426 mol or 4.26 × 10–1 mol
and moles of S2– ions = 0.213 mol or 2.13 × 10–1 mol.
(iii) Cobalt(II) chloride has the formula CoCl2. Thus one mole of CoCl 2 contains
one mole of Co2+ and two moles of Cl– ions.
i.e. moles of Co2+ = moles of CoCl 2 and
moles of Cl– = 2 × moles of CoCl2.
Molar mass of CoCl2 = sum of the atomic weights of the component atoms
= 58.93 + 2 × 35.45 = 129.83 g mol–1.
Moles of CoCl2 in 20.0 g = 20.0 ÷ 129.83 = 0.154 mol
 moles of Co2+ ions = 0.154 mol or 1.54 × 10–1 mol
and moles of Cl– ions = 2 × 0.154 = 0.308 mol or 3.08 × 10–1 mol.
(iv) Potassium oxide has the formula K 2O. Thus one mole of K 2O contains two
moles of K+ and one mole of O2– ions.
i.e. moles of K+ = 2 × moles of K 2O and
moles of O2– = moles of K 2O.
Molar mass of K 2O = sum of the atomic weights of the component atoms
= 2 × 39.10 + 16.00 = 94.20 g mol–1.
Moles of K 2O in 50.0 g = 50.0 ÷ 94.20 = 0.531 mol
 moles of K+ ions = 2 × 0.531 = 1.06 mol
and moles of O2– ions = 0.531 mol or 5.31 × 10–1 mol.
(v) Lithium carbonate has the formula Li 2CO 3. Thus one mole of Li 2CO 3 contains
two mole of Li+ and one mole of CO 32– ions.
i.e. moles of Li+ = 2 × moles of Li 2CO 3 and
moles of CO32– = moles of Li2CO 3.
Molar mass of Li2CO 3 = sum of the atomic weights of the component atoms
= 2 × 6.94 + 12.01 + 3 × 16.00 = 73.89 g mol–1.
Moles of Li2CO 3 in 32.5 g = 32.5 ÷ 73.89 = 0.440 mol
 moles of Li+ ions = 2 × 0.440 = 0.880 mol or 8.80 × 10–1 mol
and moles of CO 32– ions = 0.440 mol or 4.40 × 10–1 mol.
(vi) Iron(III) phosphate has the formula FePO 4. Thus one mole of FePO 4 contains
one mole of Fe3+ and one mole of PO 43– ions.
i.e. moles of Fe3+ = moles of PO 43– = moles of FePO 4.
Molar mass of = sum of the atomic weights of the component atoms
VII - 21
= 55.85 + 30.97 + 4 × 16.00 = 150.82 g mol–1.
Moles of FePO 4. in 100 g = 100 ÷ 150.82 = 0.663 mol
 moles of Fe3+ ions = 0.663 mol or 6.63 × 10–1 mol and
moles of PO 43– ions = 0.663 mol or 6.63 × 10–1 mol
(vii) Aluminium sulfate has the formula Al 2(SO 4) 3. Thus one mole of Al 2(SO 4) 3
contains 2 moles of Al3+ and 3 moles of SO 42– ions.
i.e. moles of Al3+ = 2 × moles of Al 2(SO 4) 3 and
moles of SO 42– = 3 × moles of Al2(SO 4)3
Molar mass of Al2(SO 4) 3 = sum of the atomic weights of the component atoms
= 2 × 26.98 + 3 × (32.07 + 4 × 16.00) = 342.17 g mol–1.
Moles of Al2(SO 4)3 in 50.0 g = 50.0 ÷ 342.17 = 0.146 mol
 moles of Al3+ ions = 2 × 0.146 = 0.292 mol or 2.92 × 10–1 mol
and moles of SO42– ions = 3 × 0.146 mol = 0.438 mol or 4.38 × 10–1 mol.
5. Explanations and Worked Solutions.
(a) (i) One mole of any compound contains an Avogadro number of molecules.
i.e. 6.022 × 1023 molecules is one mole.
 4.53 × 1024 molecules is (4.53 × 1024 ) ÷ ( 6.022 × 1023 ) moles of CO 2 = 7.52 mol
(ii) One mole of CO 2 has a molar mass = sum of the atomic weights of the
constituent atoms = 12.01 + 2 × 16.00 = 44.01 g mol–1.
 7.52 mol has a mass = 7.52 × 44.01 g = 331 g.
(b) (i) One mole of any compound contains an Avogadro number of molecules.
i.e. 6.022 × 1023 molecules of water is one mole.
 7.34 × 1023 molecules is (7.34 × 1023 ) ÷ ( 6.022 × 1023) moles of H 2O = 1.22 mol
(ii) One mole of H 2O has a molar mass = sum of the atomic weights of the
constituent atoms = 2 × 1.008 + 16.00 = 18.016 g mol–1.
 1.22 mol has a mass = 1.22 × 18.016 g = 22.0 g.
(c) One mole of NaCl contains one mole of Na+ ions and one mole of Cl– ions.
i.e. 6.022 × 1023 Na+ ions and 6.022 × 1023 Cl– ions.
Molar mass of NaCl = 22.99 + 35.45 g = 58.44 g mol–1
 moles of NaCl in 5.15 g = 5.15 ÷ 58.44 = 8.812 × 10–2 mol
Number of Na+ ions present = number of Cl– ions
= number of moles of NaCl × Avogadro number
= (8.812 × 10–2 ) × (6.022 × 1023) = 5.31 × 1022 of each.
(d) Moles of sucrose in a sugar cube containing 6.50 g of sucrose, C 12H 22O 11
= 6.50 ÷ molar mass of sucrose
= 6.50 ÷ (12 × 12.01 + 22 × 1.008 + 11 × 16.00) mol
= 6.50 ÷ 342.30 mol = 1.90 × 10–2 mol.
One mole of sucrose contains an Avogadro number of C 12H 22O 11 molecules.
 1.90 × 10–2 moles contains (1.90 ×10–2) × 6.022 × 1023 sucrose molecules
= 1.14 × 1022 molecules
```