Download Amount of Substance

THE MASSES OF ATOMS AND MOLECULES
RELATIVE ATOMIC MASS
Because of the incredibly small size of atoms, it is not possible to weigh them directly.
Instead the masses of atoms are related to an arbitrary standard; for this purpose the mass
of one atom of 12C is taken to be 12.0000. The mass of any other atom compared to that of
the carbon atom is called it’s RELATIVE ATOMIC MASS and is given the symbol A r.
The relative atomic mass of an element is defined as the average mass of one
atom of the element compared with one twelfth the mass of an atom of 12C.
N.B. Relative atomic mass is a ratio, so it is simply a number and has no units.
The 12C atom contains 6 protons, 6 neutrons and 6 electrons. The hydrogen atom contains 1
proton and 1 electron. Since the mass of an electron is negligible compared to that of a
proton or a neutron, the hydrogen atom has only 1/12 the mass of a carbon atom; therefore
the relative atomic mass of hydrogen is 1.
Similarly, the relative mass of an oxygen atom, which contains 8 protons, 8 neutrons and 8
electrons, is 16.
MASS NUMBER AND RELATIVE ATOMIC MASS
The mass number of an isotope is equal to the number of protons plus neutrons in one atom
and is, therefore, a whole number.
In round numbers, mass number and relative atomic mass are equal, but the relative atomic
mass is not an exact whole number, partly because protons and neutrons do not have
exactly the same mass and partly because of the existence of isotopes.
THE COMPLICATION OF ISOTOPES
In their natural form many elements are mixtures of isotopes. The relative atomic mass of
such an element is an average mass for the atom and will depend on the mass of each
isotope and the relative proportion of each. In very many cases, the relative atomic mass
approximates closely to a whole number, but in a few cases the deviation is considerable.
Example: Natural chlorine consists of 75% 35Cl and 25% 37Cl. Calculate the relative
atomic mass of chlorine.
In every 100 atoms of chlorine, there are:
75 of mass 35, total mass 2625
25 of mass 37, total mass 925
The mass of 100 atoms = 2625 + 925 = 3550
The average mass of one chlorine atom = 3550 = 35.5
100
The relative atomic mass of chlorine = 35.5
TOPIC 12.2: AMOUNT OF SUBSTANCE
1
MODA./Mole 12.1.2
RELATIVE MOLECULAR MASS & RELATIVE FORMULA MASS
The average mass of one molecule of an element (diatomic gases) or a simple
covalent compound compared with one twelfth mass of an atom of 12C is called
its RELATIVE MOLECULAR MASS and is given the symbol Mr.
It is calculated by adding together the relative atomic masses of all the atoms present in one
molecule.
For ionic or macromolecular compounds, which have infinite three-dimensional structures,
the term relative molecular mass is replaced by RELATIVE FORMULA MASS. It is also given
the symbol Mr. It is obtained by adding together the relative atomic masses of all the atoms
present in one formula unit.
Like relative atomic mass, relative molecular/formula mass is simply a number and has no
units.
Example: Calculate the relative molecular mass of phosphoric acid, H3PO4.
Mr (H3PO4) = (3x1) + 31 + (4x16)
= 98
The setting out of these calculations is important and should follow the method given in the
example.
TOPIC 12.2: AMOUNT OF SUBSTANCE
2
WORKSHEET 12.1.1
RELATIVE ATOMIC MASS AND RELATIVE
MOLECULAR MASS
1. Natural copper comprises 69% 63Cu and 31% 65Cu.
29
29
Calculate the relative atomic mass of copper.
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
2. Natural potassium comprises 93% 39K and 7% 41K
19
19
Calculate the relative atomic mass of potassium.
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
3. Natural gallium comprises 60.4% 69Ga and 39.6% 71Ga.
31
31
Calculate the relative atomic mass of gallium.
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
TOPIC 12.2: AMOUNT OF SUBSTANCE
3
4. Calculate the relative molecular/formula masses of the following compounds, using the
relative atomic masses given on your periodic table. Make sure you set out the
calculation correctly.
a) H2SO4
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
b) AlCl3
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
c) HNO3
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
d) K2CO3
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
e) Cu(NO3)2
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
TOPIC 12.2: AMOUNT OF SUBSTANCE
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COUNTING ATOMS AND MOLECULES
THE MOLE CONCEPT
Atoms and molecules are such minute particles that they are difficult to count. Nevertheless,
one of the most important things chemists need to know is the NUMBER of particles they are
dealing with.
If 1g of hydrogen is weighed out, it will contain a certain number of atoms (let this no. be x).
Therefore 1g of any other element will contain fewer atoms, because each atom is heavier.
If
then
1g hydrogen contains x atoms
1g carbon contains x/12 atoms
1g oxygen contains x/16 atoms
etc.
Since chemical reactions take place between whole numbers of particles (atoms, ions or
molecules), it is more useful to know the number of particles present than the mass. Thus,
continuing the above example,
If
then
1g hydrogen contains x atoms
12g carbon contains x atoms
16g oxygen contains x atoms
etc.
These numbers correspond to the relative atomic masses of these elements expressed in
grammes. Therefore, if elements are measured out in the same proportions as their relative
atomic masses, we are dealing with the same number of atoms, i.e. numbers of atoms can
be controlled by weighing.
The number of atoms contained in 1g hydrogen, 12g carbon, 16g oxygen etc. has been
determined very accurately and has a value of 6.02x1023
This is called the AVOGADRO NUMBER or AVOGADRO CONSTANT and is given the
symbol L. This number is of considerable importance in chemistry and has been given a
special unit called the MOLE.
A MOLE OF A SUBSTANCE IS AN AMOUNT
CONTAINING THE SAME NUMBER OF PARTICLES
AS THERE ARE ATOMS IN 12g OF 12C
(This number of particles is 6.02x1023)
A mole of a substance is its Ar (or Mr) in grams; this is also known as the MOLAR MASS.
Pay attention to the type of particle when calculating molar masses. For example, O=16, so
1mol. of oxygen atoms(O) has a mass of 16 g, but 1mol. of oxygen molecules (O 2) has a
mass of 32g. 1mol. of water molecules (H2O) has a mass of 1+1+16 = 18g
TOPIC 12.2: AMOUNT OF SUBSTANCE
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For most elements:
no. of moles = mass in g
Ar
MASS
Ar
MOLES
For diatomic elements (e.g. O2) and compounds:
no. of moles = mass in g
Mr
MASS
MOLES
TOPIC 12.2: AMOUNT OF SUBSTANCE
Mr
6
THE SPECIAL SITUATION OF GASES
It is usually more important to know the volumes of gases involved in a reaction, rather than
their masses. The volume occupied by a gas depends on:
It’s mass
Its temperature
Its pressure
In 1811, Avogadro, an Italian chemist, put forward the following law:
EQUAL VOLUMES OF ALL GASES UNDER THE SAME
CONDITIONS OF TEMPERATURE AND PRESSURE
CONTAIN EQUAL NUMBERS OF MOLECULES.
Thus it follows that one mole of any gas under a given set of temperature and pressure
conditions occupies the same volume. This is called the MOLAR VOLUME. At 293K and 1
atmosphere pressure (referred to as r.t.p. or room temperature and pressure), the molar
volume is 24 litres (24,000 cm3). This value can be used in calculations.
In general, the relationship between the pressure, volume, temperature and no. of moles for
an ideal gas is given by the ideal gas equation:
pV = nRT
Which can usefully be re-written as:
pV = mRT
Mr
Where
p is the pressure of the gas in N.m-2 (or Pa)
V is the volume occupied by the gas in m3
n is the no. of moles of the gas
T is the temperature of the gas in K
R is the gas constant ( =8.314 J.K-1.mol-1 )
m is the mass of the gas in g
Mr is the relative molecular mass of the gas
Remember always to check that the units you are using are correct before
substituting values into these equations.
Using the above equation to calculate the volume of 1mole of a gas at 293K and atmospheric
pressure ( 101325 N.m-2 )
V = nRT = 1x 8.314 x 293
P
101325
TOPIC 12.2: AMOUNT OF SUBSTANCE
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= 0.024m3 = 24000cm3
WORKSHEET 12.1.2
CALCULATIONS INVOLVING MASSES,
VOLUMES & MOLES
A) MASSES & MOLES
Calculations of this type are based on these simple relationships.
For most elements:
no. of moles = mass in g
Ar
For diatomic elements (e.g. O2) and compounds:
no. of moles = mass in g
Mr
Note that the abbreviation for moles is mol.
1. Calculate the mass of :
a) 3.1 moles of sodium chloride
b) 2.5 moles of MgSO4.7H2O
c) 1.5mol. of chlorine gas (remember that chlorine is diatomic)
d) 0.25mol. of sodium carbonate
2. Calculate the number of moles contained in:
a) 4.98g of potassium iodide
b) 2.37g of potassium manganate (VII)
c) 2.5g of sodium hydroxide
d) 1.00g of magnesium
TOPIC 12.2: AMOUNT OF SUBSTANCE
8
B) VOLUMES AND MOLES
These calculations are based on the relationships:
pV = nRT
pV = mRT
Mr
1. Calculate the volume occupied by 2.1g of chlorine at 150,000Pa and 25 oC.
2. What pressure is exerted by 0.25 mol. of oxygen at 100oC in a container with a capacity
of 1dm3 ?
3. Calculate the relative molecular weight of a gas, 2.2g of which occupy a volume of
1050cm3 at a temperature of 298K and a pressure of 118,000Pa.
4. A sample of gallium chloride weighing 0.1g was vaporised, and it was found to occupy
16cm3 at a temperature of 415oC and a pressure of 101325Pa.
a) Calculate a value for the relative formula mass of gallium chloride under the conditions
of the experiment.
b) Suggest a molecular formula for gallium chloride in the vapour state, indicating how
you arrive at your answer.
(Ga = 69.7, Cl = 35.5)
TOPIC 12.2: AMOUNT OF SUBSTANCE
9
CALCULATIONS OF CONCENTRATION
The concentration of a standard solution is usually expressed as the number of moles of the
solute contained in 1dm3 of the solution. The unit is mol.dm-3, which can be abbreviated to M.
Strictly speaking the concentration given in mol dm-3 should be called the molarity but at
A-level the terms concentration and molarity are used interchangeably.
Thus, 0.100M NaOH (said as “point one molar NaOH”) is a solution containing 0.100 mole of
NaOH dissolved in every dm3 (litre) of the solution.
If no. of moles = mass
Mr
and
concentration =
no. of moles
volume in dm3
These two expressions can be combined to give:
concentration =
mass of solute in g
volume of solution in dm3 x Mr of solute
This expression can be used to determine the value of any one term given enough
information to calculate the other three terms.
Examples:
1. What mass of solute would be needed to prepare the following standard solutions?
a) 250cm3 of 0.1M sodium chloride, NaCl
b) 200cm3 of 0.017M potassium chromate(VI), K2Cr2O7
c) 2dm3 of 0.15M ammonium iron(II) sulphate, (NH4)2SO4.FeSO4.6H2O
d) 750cm3 of 0.100M ethanedioic acid, H2C2O4.2H2O
2. Calculate the concentration of the following standard solutions.
a) 4.0g of sodium hydroxide, NaOH, is dissolved to make 1000cm 3 of solution.
b) 4.25g of silver nitrate, AgNO3 is dissolved to make 250cm3 of solution.
c) 20.75g of potassium iodide, KI, is dissolved to make 100cm3 of solution.
d) 0.79g of potassium manganate(VII), KMnO4 is dissolved to make 500cm3 of
solution.
3. From the data, calculate the relative molecular mass, Mr, of the following substances.
a) 1.325g of substance A is used to make up 250cm3 of a 0.05M solution.
b) 53.5g of substance B is used to make up 500cm3 of a 2M solution.
c) 124.75g of substance C is used to make up 1000cm 3 of a 0.5M solution.
d) 3.15g of substance D is used to make up 500cm3 of a 0.05M solution.
4. To what volume of solution must each of the following masses of solutes be made up to
give a solution of the specified concentration? Give your answer in cm 3.
a) 6.9g of potassium carbonate to give a 0.100M solution
b) 0.117g of sodium chloride to give a 0.02M solution
c) 20.75g of potassium iodide to give a 0.5M solution
d) 9.35g of sodium tetraborate, Na2B4O7.10H2O to give a 0.100M solution
TOPIC 12.2: AMOUNT OF SUBSTANCE
10
VOLUMETRIC ANALYSIS
STANDARD SOLUTIONS AND PRIMARY STANDARDS
A standard solution is a solution of known concentration; i.e. it contains a known mass of
solid in a known volume of solution at a particular temperature. The concentration of a
standard solution is usually expressed in moles per dm 3 (abbreviation mol.dm-3 or M). A
concentration can be determined either by direct weighing, subject to the constraints
described below, or by titration.
A primary standard is a substance which can be used to make up a standard solution by
direct weighing; i.e. a solid is weighed out and made up to a known volume of solution.
However to be a primary standard, a substance must be
a) pure
b) stable
c) non-volatile
d) of fixed composition
e) not hygroscopic, deliquescent or efflorescent
The concentration of an unknown solution can be found (this is called standardising the
solution) by titration. This involves finding what volume of it is equivalent to (i.e. reacts
completely with) a known volume of standard solution; an indicator is used to show when the
reaction is complete. The number of moles of each substance reacting can be deduced from
the balanced equation for the reaction, and the unknown concentration can then be
calculated. Alternatively, the appropriate numbers can be substituted into the formula:
SOURCES OF ERRORS IN TITRATIONS
1. If the standard solution is not shaken thoroughly to ensure that it is homogeneous, very
large errors can arise.
2. There is an error in the titre, because the minimum volume which can be added at one
time from a burette (1 drop) is about 0.05cm 3. This error is reduced by averaging 3 titres.
N.B. burette readings are usually taken to 2 d.p.
3. A graduated flask is calibrated at a certain temperature (usually 20 oC). It must not be
used for hot solutions otherwise large errors arise.
4. Failure to ensure that the volume below the burette tap is full before taking the first
burette reading will lead to an overestimate in the volume of solution added.
5. Failure to swirl the flask during the titration can lead to premature indication of the endpoint.
6. Failure to remove the funnel from the top of the burette before the first burette reading is
taken can permit solution to drain down into the burette during the titration.
TOPIC 12.2: AMOUNT OF SUBSTANCE
11
PRESENTATION OF RESULTS
a. Weighings
All weighings taken must be recorded (to 3 d.p.) and should be written down while
still at the balance.
e.g. Mass of weighing bottle + (named solid)
=
g
Mass of weighing bottle
=
g
Mass of (named solid) taken
=
g
b. Titration figures
The burette is read to the bottom of the meniscus to two decimal places; it is necessary
to estimate the second place. The total volume of solution added from the burette is called
the titre. It is usual to obtain three concordant titres (i.e. titres which agree within a range of
0.2cm3). The results are recorded in a table.
An average titre is calculated, ignoring results which are obviously spurious, and is quoted
to two decimal places.
Titration no.
cm3
Final burette reading
1
2
3
4
Initial burette reading
Volume of (named solution) added
Average of titres (specified numbers) = …………………cm3
Before beginning the calculation, the balanced equation for the reaction taking place in the
titration should be written.
TOPIC 12.2: AMOUNT OF SUBSTANCE
12
WORKSHEET 12.1.9
VOLUMETRIC ANALYSIS CALCULATIONS
1. What mass of sodium carbonate dihydrate (Na2CO3.2H2O) is required to react completely
with 50.0cm3 of 0.100M sulphuric acid?
2. What volume of 0.200M potassium hydroxide will react with 50.0cm3 of 0.100M sulphuric
acid?
3. What volume of 0.125M sodium hydroxide solution is needed to titrate 25.0cm 3 of 0.085M
sulphuric acid?
4. Soda lime is a mixture containing 85.0% NaOH and 15.0% CaO. What volume of 0.500M
nitric acid is needed to neutralise 2.50g of soda lime?
5. A 1.00g sample of limestone is allowed to react with 100cm 3 of 0.200M hydrochloric acid.
The excess acid required 24.8cm3 of 0.100M sodium hydroxide for titration. Calculate the
percentage of calcium carbonate in the limestone.
6. 0.500g of impure ammonium chloride is warmed with an excess of sodium hydroxide
solution. The ammonia liberated is absorbed in 25.0cm 3 of 0.200M sulphuric acid. The
excess sulphuric acid requires 5.64cm3 of 0.200M sodium hydroxide solution for titration.
Calculate the percentage ammonium chloride in the original sample.
7. A solution of a metal carbonate, M2CO3, was prepared by dissolving 7.46g of the
anhydrous solid in water to give one litre of solution. 25cm 3 of this solution reacted with
27.0cm3 of 0.100M hydrochloric acid. Calculate the relative formula mass of M2CO3 and
hence the relative atomic mass of the unknown metal M.
8. 10cm3 of concentrated hydrochloric acid were diluted to 100cm 3 with water. 10cm3 of the
diluted acid was found to react exactly with 20cm3 of 1.0M sodium hydroxide solution.
a) Calculate the concentration of the diluted acid solution.
b) Calculate the concentration of the undiluted acid.
9. 1.500g of a sample of limestone were dissolved in 50cm 3 of 1M hydrochloric acid. The
resulting solution was made up to 250cm3 with distilled water. 25cm3 of this solution
required 21.05cm3 of 0.100M sodium hydroxide for neutralisation. Assuming all the basic
material in the rock to be calcium carbonate, calculate the percentage of calcium
carbonate in the limestone.
10. 100cm3 of concentrated hydrochloric acid were diluted to 1dm 3 with distilled water.
26.8cm3 of the diluted acid were needed to neutralise 25.0cm 3 of 0.500M sodium
carbonate solution. Calculate the molarity of the concentrated hydrochloric acid and
hence its concentration in g.dm-3.
11. 8.58g of washing soda were made up to 250cm 3 of aqueous solution. 25cm3 of this
solution required 30.0cm3 of 0.2M HCl for neutralisation with methyl orange as indicator.
Calculate x in the formula for washing soda, Na2CO3.xH2O.
TOPIC 12.2: AMOUNT OF SUBSTANCE
13
THE PERCENTAGE COMPOSITION OF
COMPOUNDS
A pure chemical compound has a fixed chemical composition and is represented by a fixed
formula. This formula conveys information about:
a) the elements present in the substance
b) the mass of 1 mole of the substance (Mr expressed in g)
c) the number of moles of each element present in 1 mole of the substance
The percentage composition of a compound is the % contribution by mass which each
component element makes to the substance.
CALCULATION OF % COMPOSITION
Example: Calculate the % composition by mass of ammonium nitrate.
1. Work out the formula of the compound. The formula of ammonium nitrate is NH 4NO3.
2. Calculate the relative molecular/formula mass of the compound.
Mr (NH4NO3) = 14 + (4x1) + 14 + (3x16) = 80
 80g NH4NO3 contains
28g nitrogen
4g hydrogen
48g oxygen
3. Express the mass of each element as a percentage of the molar mass.
%N = 28 x 100 = 35.0%
80
%H =
4 x 100 = 5.0%
80
%O = 48 x 100 = 60.0%
80
Answers are usually given to 1 d.p.
4. Check that the percentages add up to 100%.
N.B. Errors from rounding up or down may give answers between 99.8 and 100.2%, which
is satisfactory.
TOPIC 12.2: AMOUNT OF SUBSTANCE
14
WORKSHEET 12.1.3
THE PERCENTAGE COMPOSITION OF COMPOUNDS
Calculate the percentage composition by mass of the following compounds, giving your
answers to 1 d.p. Use the relative atomic masses given on your periodic table.
a) Na2SO4
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
b) Ca(NO3)2
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
c) K2CO3
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
d) (NH4)2SO4
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
TOPIC 12.2: AMOUNT OF SUBSTANCE
15
CALCULATION OF EMPIRICAL FORMULAE
The EMPIRICAL FORMULA is the simplest WHOLE number ratio of the atoms of each
element present in a compound.
The MOLECULAR FORMULA is the actual number of atoms of each element present in
one molecule of the compound.
The two are related by the expression:
Molecular formula = Empirical formula x n
where n is a whole number.
For example:
The molecular formulae of methane and ethane are CH4 and C2H6 respectively.
The empirical formula of methane is CH4; in the above expression n = 1.
The empirical formula of ethane is CH3, which is the simplest whole number ratio of carbon
to hydrogen atoms. In the above expression n = 2.
CALCULATION OF AN EMPIRICAL FORMULA
Example: Calculate the empirical formula of a compound having the composition:
lead 8.32g; sulphur 1.28g; oxygen 2.56g
If its molecular mass is 303, what is its molecular formula?
PROCEDURE:
1) Convert the mass of each element into moles: this is done by dividing the mass by the
relative atomic mass.
Moles Pb 8.32 = 0.040 moles
207
Moles S = 1.28 = 0.040 moles
32
Moles O = 2.56 = 0.160 moles
16
This transformation counts the combining atoms; therefore, in the compound,
0.040 moles (i.e. 0.040 x 6 x 1023 atoms ) of Pb
0.040 moles (i.e. 0.040 x 6 x 1023 atoms) of S
0.160 moles (i.e. 0.160 x 6 x 1023 atoms) of O
have combined together.
TOPIC 12.2: AMOUNT OF SUBSTANCE
16
2) This ratio of combining atoms is converted to a whole number by dividing through by the
lowest number:
0.040
0.040
moles Pb
0.040
0.040
moles S
0.160
0.040
moles O
combine together
 1 mole Pb, 1 mole S and 4 moles O combine together.
 The empirical formula is PbSO4
3) To find the molecular formula:
Molecular formula = Empirical formula x n
Molecular mass = Empirical mass x n
The molecular mass (Mr) = 303
The empirical mass (of PbSO4) = 303
 n=1
 Molecular formula is PbSO4
The calculation of an empirical formula can be set out more briefly and more conveniently in
table form.
Element
Mass
Ar
Pb
8.32
207
S
1.28
32
O
2.56
16
Moles
8.32 = 0.040
207
1.28 = 0.040
32
2.56 = 0.16
16
Simplest ratio
0.040 = 1
0.040
0.040 = 1
0.040
0.16 = 4
0.040
The empirical formula is PbSO4
TOPIC 12.2: AMOUNT OF SUBSTANCE
17
Empirical
formula
1
1
4
WORKSHEET 12.1.5
PERCENTAGE COMPOSITION AND
EMPIRICAL FORMULAE
1. Calculate the percentage composition by mass of the following compounds, giving your
answers to 1 d.p. Use the relative atomic masses given on your periodic table.
a) aluminium chloride
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
b) sodium sulphite
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
c) lead (II) nitrate
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
d) iron (III) oxide
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
TOPIC 12.2: AMOUNT OF SUBSTANCE
18
2. Calculate the empirical formulae of the substances having the following percentage
compositions.
a) magnesium
carbon
oxygen
28.57%
14.29%
57.14%
b) potassium
chromium
oxygen
26.53%
35.37%
38.10%
c) sodium
sulphur
oxygen
36.51%
25.40%
38.10%
a) …………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
b) …………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
c) …………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
3. A hydrocarbon has the composition carbon 85.71%; hydrogen 14.29%
a) Calculate its empirical formula
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
b) If its molecular mass is 56, calculate its molecular formula.
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
TOPIC 12.2: AMOUNT OF SUBSTANCE
19
WORKSHEET 12.1.6
EMPIRICAL & MOLECULAR FORMULAE
1. Calculate the empirical formulae of the substances having the following percentage
compositions.
a)
potassium 28.2%
b)
sodium
33.3%
chlorine
25.6%
nitrogen
20.3%
oxygen
46.2%
oxygen
46.4%
a) …………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
b) …………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
2. What is the empirical formula of a compound containing:
4.04% hydrogen, 24.24% carbon and 71.72% chlorine
……………………………………………………………………………………………………….
……………………………………………………………………………………………………….
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……………………………………………………………………………………………………….
The compound has a relative molecular mass of 99; calculate its molecular formula
……………………………………………………………………………………………………….
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TOPIC 12.2: AMOUNT OF SUBSTANCE
20
3. a) 5g of an oxide (A) of chromium were found on reduction to give 2.6g of chromium.
Calculate the empirical formula of this oxide.
Ar (Cr) = 52 Ar (O) = 16
………………………………………………………………………………………………………
………………………………………………………………………………………………………
………………………………………………………………………………………………………
………………………………………………………………………………………………………
………………………………………………………………………………………………………
b) Oxide (A) on strong heating forms oxygen and a second oxide of chromium (B), which is
found to contain 24g of oxygen combined with the molar mass of chromium. Calculate the
empirical formula of oxide (B).
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
c) Using the empirical formula for the two oxides, deduce an equation for the formation of
oxide (B) from oxide (A).
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TOPIC 12.2: AMOUNT OF SUBSTANCE
21
WORKSHEET 12.1.7
CALCULATION OF EMPIRICAL FORMULAE
1. Find the empirical formulae of the compounds formed in the reactions described below.
a) 10.800g of magnesium form 18.000g of an oxide
b) 3.400g of calcium from 9.435g of a chloride
c) 3.528g of iron from 10.237g of a chloride
2. A piece of tin of mass 1.527g was placed in a solution of iodine in trichloromethane and
left for several days in a stoppered flask. At the end of that time, the tin remaining in
excess was removed, washed with tridhloromethane and dried; its mass was 1.170g. The
trichloromethane washings were added to the original solution, then the trichloromethane
was evaporated to leave an orange solid, a compound of tin and iodine, of mass 1.881g.
Calculate the empirical formula of the compound of tin and iodine.
3. 0.5g of X on complete combustion gave 0.6875g of CO2 and 0.5625g of H2O as the only
products. Calculate the masses of C and H in the sample of X and hence find the
empirical formula of X.
4. An organic acid contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen; calculate its
empirical formula. If the Mr = 60, deduce its molecular formula.
5. An organic compound has a molar mass of 150g.mol -1 and contains 72.0% carbon,
6.67% hydrogen and 21.33% oxygen. What is its molecular formula?
TOPIC 12.2: AMOUNT OF SUBSTANCE
22
CALCULATION OF REACTING MASSES AND VOLUMES
Since chemical equations are quantitative as well as qualitative statements, it is possible to
calculate the quantities of reactants necessary to produce a given amount of product, or to
calculate the quantity of product(s) given by a specified amount of reactants.
PROCEDURE:
1)
2)
3)
4)
5)
Write the balanced equation for the reaction concerned.
Write the ratio of reacting moles for the chemicals in question.
Convert the masses into moles using moles = mass/(Ar or Mr)
Use the ratio of moles from step 2 to find number of moles of product formed.
Calculate mass of product from the number of moles formed.
Example 1:
What mass of carbon dioxide will be given off when 9g of calcium carbonate is
decomposed by heating?
1) CaCO3(s)
2) 1 mole
3)
9g = 9/100
=0.09mol
4)
CaO(s)
+
CO2(g)
1 mole
Mr (CO2) = 44
Mr (CaCO3) = 100
If 1 mole CaCO3
then 0.09 mol
1 mole CO2
0.09mol
5)
mass = moles x Mr
Mass = 0.09 x 44
=3.96g
9g calcium carbonate will produce 3.96g of carbon dioxide.
Example 2:
If 4g phosphorus is burned in oxygen, what mass of phosphorus(V) oxide will be
formed?
1) 4P(s) + 5O2(g)
2) 4 mole
3)
4g = 4/31
=0.129mol
4)
2P2O5(s)
2 moles
A r (P) = 31
Mr (P2O5) = 142
If 4 mole of P
then 0.129 mol
2 mole P2O5
0.0625mol
5)
mass = moles x Mr
Mass = 0.09 x 142
=9.16g
4g phosphorus will burn to give 9.16g phosphorus(V) oxide.
TOPIC 12.2: AMOUNT OF SUBSTANCE
23
THE SPECIAL SITUATION OF GASES:
It is usually more important to know the volumes of gases involved in a reaction rather than
their masses. Therefore, molar volume is often used in the calculation in place of molar
mass.
Example 3:
What volume of hydrogen at r.t.p. will be produced when 3.25g zinc is dissolved in dilute
sulphuric acid?
1) Zn(s) + H2SO4(aq)
2) 1 mole
3)
ZnSO4(aq) + H2(g)
1 mole
Ar (Zn) = 65
Molar volume (H2) = 24000cm3
24000cm3
65g
4)
24000 cm3
65
1g
3.25g
24000 x 3.25 cm3
65
3.25g
1200cm3
3.25g zinc reacts to produce 1200cm3 hydrogen.
TOPIC 12.2: AMOUNT OF SUBSTANCE
24
WORKSHEET 12.1.8
CALCULATION OF REACTING MASSES AND VOLUMES
Use the relative atomic masses given on your periodic table. The molar volume of a gas at
r.t.p. is 24000cm3.
1. If 4.5g of calcium is burned in oxygen according to the equation:
2Ca + O2
2CaO
a) What mass of oxygen is needed?
b) What mass of calcium oxide is produced?
2. Iron(III) oxide reacts with carbon monoxide to give iron and carbon dioxide according to
the equation:
Fe2O3 + 3CO
2Fe + 3CO2
If 160g of iron(III) oxide reacts, what volume of carbon dioxide (at r.t.p.) will be given off?
3. Limestone decomposes on heating according to the following equation:
CaCO3
CaO + CO2
What mass of 94% pure limestone is needed to make 100 tonnes of quicklime (calcium
oxide)?
4. Write an equation for the fermentation of glucose to give ethanol and carbon dioxide.
What mass of ethanol could be produced by the fermentation of 1kg of glucose? What
volume of carbon dioxide, measured at r.t.p., would be produced in this fermentation?
5. Iron reacts with sulphur to give iron(II)sulphide according to the equation:
Fe + S
FeS
What mass of sulphur must react to give 6.2g iron(II)sulphide?
6. Potassium reacts with water according to the equation:
2K + 2H2O
2KOH + H2
What mass of potassium will give 150cm3 of hydrogen at r.t.p.?
7. Zinc dust burns brightly in air to form zinc oxide, according to the following equation:
2Zn + O2
2ZnO
What mass of zinc oxide is formed when 1.626g of zinc reacts?
8. Methane burns in air to form water and carbon dioxide, according to the following
equation:
CH4(g) + O2(g)
CO2(g) + 2H2O(l)
What mass of water and what volume of carbon dioxide (measured at r.t.p.) will be given off
when 15000cm3 of methane burns?
TOPIC 12.2: AMOUNT OF SUBSTANCE
25
9. Nickel carbonate decomposes when heated according to the following equation:
NiCO3
NiO + CO2
Calculate the mass of nickel oxide produced when 2.5g of nickel carbonate decomposes.
10. Magnesium reacts with dilute sulphuric acid to form magnesium sulphate and hydrogen,
according to the equation:
Mg + H2SO4
MgSO4 + H2
a) Calculate the mass of magnesium needed to produce 22.3g of magnesium sulphate
b) Calculate the volume of hydrogen (measured at r.t.p.) given off when 22.3g of
magnesium sulphate are formed.
11. Write a balanced chemical equation for the reaction described in the following word
equation:
copper(II) sulphate + sodium hydroxide
sodium sulphate + copper(II) hydroxide
Calculate the masses of copper(II) sulphate and sodium hydroxide needed to produce
12.4g of copper(II) hydroxide.
12. Iron reacts when heated with sulphur to form iron(II) sulphide according to the following
equation:
Fe + S
FeS
In one experiment, 5.2g of iron was heated with 4.5g of sulphur; one of the reactants was
present in excess.
Calculate
a) which reactant is present in excess
b) how much of this reactant remains unreacted at the end of the experiment
c) the mass of iron(II) sulphide which is formed
13. A mixture of calcium and magnesium carbonates weighing 10.0000g was heated until it
reached a constant mass of 5.0960g. Calculate the percentage composition of the mixture by
mass.
14. In the thermit reaction:
2Al(s) + Cr2O3(s)
2Cr(s) + Al2O3(s)
Calculate the percentage yield when 180g of chromium are obtained from a reaction
between 100g of aluminium and 400g of chromium(III) oxide.
TOPIC 12.2: AMOUNT OF SUBSTANCE
26
How efficient are reactions?
We can gain an understanding of the efficiency of a reaction by considering the percentage
yield or by looking at the atom economy of a reaction.
Percentage yield
The amount of product obtained, usually measured in grams or kilograms, is known as the
yield. Often, when we carry out a reaction, the starting materials do not fully react to form
products. We say the reaction does not go to completion. This results in many reactions
producing a lower yield than expected.
It is useful, therefore, to know the percentage yield of a reaction. This compares the amount
of product that the reaction really produces with the maximum amount it could possibly
produce if it went to completion.
Percentage yield = amount of product produced by the reaction x 100
maximum amount of product possible
Examples:
1. It was calculated that a reaction could produce 24g of iron. When it was carried out, the
mass of iron produced was 15.6g. Calculate a percentage yield for this reaction.
2. 12g of magnesium were burned in oxygen. 15.8g of magnesium oxide were produced.
a) write a word equation and then a balanced symbol equation for the reaction
b) calculate the number of moles of magnesium that were used in the reaction
c) using the equation, state the number of moles of magnesium oxide that would be
formed
d) calculate the mass of magnesium oxide that would be formed if the reaction went to
completion
e) thus calculate the percentage yield of magnesium oxide for this reaction.
TOPIC 12.2: AMOUNT OF SUBSTANCE
27
Atom Economy
Some reactions could have very high percentage yields of product – the reaction could go to
completion – but if a lot of unwanted materials are produced the process may still not be
considered to be efficient.
If a reaction produces 3 different products but only one is useful, even a reaction with a
100% yield could produce a large amount of waste. A way to measure the amount of starting
materials that end up as useful products is to calculate the atom economy (or atom
utilisation) of the reaction. It is important for sustainable development and for economical
reasons to use large scale reactions with a high atom economy.
% atom economy = (Mr of useful product) x its balancing number in the equation x 100
(Mr of ALL reactants) x their balancing numbers
e.g. Iron oxide is converted into iron by reduction with carbon monoxide:
Fe2O3 + 3CO  2Fe + 3CO2
In this reaction, the carbon dioxide is a waste product. The iron is the useful product.
Percentage atom economy is calculated by:
Mr of Fe x balancing number
x 100
[(Mr of Fe2O3 x balancing number) + (Mr of CO x balancing number)]
So:
(56 x 2)
(160) + (28 x 3)
=
x 100
45.9%
Example 1:
Calcium carbonate decomposes on heating to form quicklime (calcium oxide), a useful
product, according to the following equation:
CaCO3  CaO + CO2
Calculate the atom economy for this reaction.
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TOPIC 12.2: AMOUNT OF SUBSTANCE
28
Example 2:
Ethanol can be manufactured by fermentation according to the following equation:
C6H12O6  2C2H5OH + 2CO2
Equation 1
It can also be manufactured by direct hydration of ethene:
C2H4 + H2O  C2H5OH
Equation 2
(a) Calculate the atom economy for the production of ethanol by fermentation.
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(b) What is the atom economy for the direct hydration of ethene?
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(c) Suggest one advantage and one disadvantage of each method of manufacture.
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TOPIC 12.2: AMOUNT OF SUBSTANCE
29
TOPIC 12.1: AMOUNT OF SUBSTANCE
LOWER SIXTH TEST: FORMULAE & MOLE CALCULTIONS
1. Write down the formulae of:
a) calcium chloride
………………………….
b) potassium sulphide
………………………….
c) iron (II) hydroxide
………………………….
d) magnesium nitrate
………………………….
e) aluminium oxide
………………………….
(5)
2. Calculate the relative molecular masses of :
a) Na2CO3 .……………………………………………………………………………….……
……………………………………………………………………………………………………
b) Cu(NO3)2…………………………………………………………………………………….
……………………………………………………………………………………………… (4)
3. Natural magnesium comprises 78.6% 24Mg, 10.1% 25Mg and 11.3% 26Mg. Calculate the
relative atomic mass of magnesium.
……………………………………………………………………………………………………………
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……………………………………………………………………………………………………… (3)
4. Calculate the percentage composition ( to 1 d.p.) of MgCO3.
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………………………………………………………………………………………………………. (4)
TOPIC 12.2: AMOUNT OF SUBSTANCE
30
5. A solid is found to contain 31.8% K, 29.0% Cl, 39.2% O. Calculate its empirical formula.
(4)
6. Calculate the number of moles contained in:
a) 2.8g of sodium hydroxide
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
b) 22.5cm3 of 0.055M sodium carbonate
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………………………………………………………………………………………………………. (4)
7. Calculate the relative molecular mass of a gas, 4.4g of which occupy a volume of
2100cm3 at a temperature of 25oC and a pressure of 118,000Pa.
……………………………………………………………………………………………………………
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……………………………………………………………………………………………………… (3)
8. Potassium chlorate(V) decomposes on heating according to the following equation:
2KClO3(s)
2KCl(s) + 3O2(g)
What mass of potassium chlorate(V) must be heated to give 1dm 3 of oxygen at r.t.p.?
(molar volume at r.t.p. = 24000cm3)
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
………………………………………………………………………………………………………. (3)
TOPIC 12.2: AMOUNT OF SUBSTANCE
31