Y12 Chemistry Objectives
QUANTITATIVE ANALYSIS
Achievement Aims Level 7
Material World

Develop an understanding of and use the
fundamental concepts of chemistry to
interpret observations

Apply knowledge of chemistry to explain
aspects of the natural world and how
chemistry is used in society to meet needs,
resolve issues and develop new
technologies.
Achievement Standards
Chemistry 2.2
By the end of this unit the student should be able to:

Define the mole as the unit of the amount of the substance.

Define molar mass, M.

Relate the amount of a substance to its mass and molar mass.

Define concentration of an aqueous solution as the amount of solute per unit volume of solution.

Relate the amount of a substance to concentration and volume of a solution.

Define empirical and molecular formulae.

Apply the mole concept to the calculation of empirical formulae from % composition.

Record data systematically.

Use significant figures appropriately, based on the least accurate piece of supplied data.

Use correct units.

Recognise qualitative errors in an experiment

Carry out gravimetric analyses to obtain masses of elements in a compound and water of
crystallisation.

Calculate % composition from a given formula.

Calculate the water of crystallisation from data.

Apply the mole concept to chemical equations.

Perform both one step and two step calculations based on masses, concentrations, amounts and
volumes.

Perform three step calculations based on masses, concentrations, amounts, volumes and chemical
equations.

Define concentration of an aqueous solution as the amount of solute per unit volume of solution.

Perform a titration accurately

Prepare a standard solution.

Solve volumetric problems.

Relate quantitative analysis to practical environmental and commercial situations.
1
SI UNITS IN CHEMISTRY
Basic Units
Mass
Length
Temperature
Amount of Substance
kilogram
metre
(degrees) Kelvin
Centigrade/Celsius
mole
kg
m
K
o
C
mol
cubic metre
litre
joule
m3
L
J
Derived Units
Volume
Energy
Standard Prefixes
10-1
10-2
10-3
103
deci
centi
milli
kilo
Common Units
Mass
-
gram usually used not kg
Volume
-
litre
millilitre
L
mL
Note:
1L
1mL
400mL
1000mL
100g
=
=
=
=
=
1dm3 decimeter3
1cm3
0.4L = 0.4dm3
1L
1kg
Units must always be quoted
2
Moles – A Learning Programme
Chemists frequently deal with reaction equations and are interested in what amount of one substance is
needed to react completely with a given amount of another. Atoms, ions, and molecules are too small to
count or weigh individually; e.g. l g of carbon contains approximately 5 x 1022 atoms.
One molecule of oxygen weighs approximately 5.3 x 10-23g. In order to discuss amounts of particles
which can be handled and weighed easily in the laboratory, a new unit is required.
The basic unit for measuring the amount of a substance is a mole. Just as mass is measured in grams,
distance measured in metres, so amount of substance (with the symbol “n”) is measured in moles.
This unit has been chosen so that one mole of an element will have a mass equal to its molar mass (M)
in grams. This used to be called “relative atomic mass” for elements. Molar masses for elements are
given on your Periodic Table. Their numerical value is similar to that of mass number but is a weighted
mean of all isotopes. Molar masses have the unit g mol-1 (grams per mole)
Definition Learn this
The mole is the amount of substance which contains as many
Elementary entities (atoms, ions, molecules etc) as there are carbon
atoms in 12.000 g of 12C.
Worked Examples:
(a)
What is the mass of one mole of sodium?
Answer : Look up Molar Mass tables and find M (Na) = 23 g mol-1
Thus one mole of Na has a mass of 23 g.
The molar mass of a compound is the sum of the molar masses of the component elements.
(b)
What is the mass of one mole of calcium chloride, CaC12.
Answer : Calculate the molar mass, M.
M (CaC12)
= M (Ca) + 2 x M (C1)
= 40 + 2 x 35.5
= 110 g mol-1
Thus one mole of CaC12 has a mass of 110 g.
(c)
If 1 mole of Helium weighs 4 g (from M values), what is the mass of 2 moles of Helium?
Obviously it is 8 g. Similarly 0.5 (or ½) mole of Helium would have a mass of 2 g.
Now you try some. (Answers are at the end.)
1.
The Molar mass for sulfur is ?
2.
1 mole of sulfur atoms has a mass of ?
3
3.
The Molar mass for silver is?
4.
1 mole of silver atoms has a mass of ?
5.
What is the mass of 1 mole of gold atoms ?
6.
The mass of 1 mole of magnesium atoms is ?
7.
Two moles of mercury have a mass of ?
8.
Ten moles of barium have a mass of?
Thus we have the generalisation:
I
Mass of substance = amount (in moles) x mass of 1 mole
Or, in symbols:
m = n x M
For subsequent problems show your working.
E.g.
What is the mass of 7 moles of magnesium?
m(Mg) = n(Mg) x M(Mg)
= 7 x 24.3
= 170.1 g
Note: it is important that you include the substance in brackets as in the above example. When you learn
how to solve more complex problems this identification is critical. Get in to the habit now.
9.
What is the mass of 15 moles of tin?
10.
What is the mass of 6.3 moles of nickel?
11.
For an experiment we need 0.01 moles of Cu. What mass will we need?
To use this relationship another way.
II
n =
m
M
LEARN THIS
What is the amount (in moles) of 36 g of carbon?
n(C) =
m(C )
M (C )
=
36
3
= 3 moles
12.
What is the amount of zinc in a 392.4 g sample? …….
13.
What is the amount of manganese in a 5.49 g sample? …….
14.
What is the amount of calcium in a 1.27 g sample? …….
4
Suppose we had 7 g of silicon and we knew that this was 0.25 (¼) mole.
Then 1 mole of silicon must weigh 28 g. That is its molar mass is 28.
To get this we use the relationship
III.
Mass of 1 mole, M = mass of material
Amount
OR
M
m
n
15.
If 0.1 mole of boron has a mass of 1.08 g, what is the mass of 1 mole?
16.
2.5 moles of silicon has a mass of 70.25g. What is the molar mass of Si?
CHECK BACK
The most important formula in this group is
Amount (n) = mass of material (m) ;
Mass of 1 mole (M)
n= m
M
So far we have dealt only with elements but exactly the same rules can be applied to compounds. You
might notice also that we have used no gases in our calculations. Let’s start with compounds.
A chemical symbol has 3 meanings.
(a)
It stands for the material for which it is the symbol, e.g. A1 stands for aluminium metal.
CuSO4 stands for copper sulfate;
Na2CO3.10H2O stands for sodium carbonate crystals (washing soda);
CC14 stands for carbon tetrachloride liquid;
NO2 stands for nitrogen dioxide gas.
(b)
It stands for the simplest particle of the substance e.g. A1 stands for 1 atom of aluminium.
NO2 stands for 1 molecule of nitrogen dioxide.
CuSO4 stands for 1 copper ion paired with 1 sulfate ion.
Na2CO3.10H2O stands for 2 sodium ions combined with 1 carbonate ion and 10 molecules
of water.
(c)
It stands for 1 mole of each substance e.g. A1 stands for 27 g of aluminium, NO2 stands for
1 mole of nitrogen atoms (14 g) combined with 2 moles of oxygen atoms (32 g). That is it
stands for 46 g of the gas nitrogen dioxide.
Na2CO3.10H2O stands for 2 moles of sodium (46 g);
1 mole of carbon (12 g);
3 moles of oxygen atoms (48 g) plus ten moles of water (10x18=180 g).
The formula therefore stands for 286 g of washing soda crystals.
Whenever we consider particles other than atoms the molar mass is the sum of all
component atoms' molar masses.
5
To determine the molar mass of a compound we simply add up the molar masses of all component
atoms in the correct ratio according to the formula of the compound.
N.B.: It is not possible to get the correct answer to these without getting the formula right. This
will be good formula revision too. Use M values to correct to 1 decimal place.
Find the molar masses of the following:
17.
Sodium chloride
18.
Copper (II) fluoride
19.
Lead (II) iodide
20.
Sodium sulfide
21.
Ammonium chloride
22.
Silver nitrate
23.
Copper (II) nitrate
24.
Potassium sulfite
25.
Ammonium sulfate
26.
Carbon dioxide gas
27.
Epsom salts crystals (MgSO4. 7H2O)
28.
Copper sulfate crystals (CuSO4.5H2O)
29.
Oxygen gas
Now consider this problem:
(a)
(b)
What is the amount of aluminium in 3 moles of aluminium nitrate?
What is the amount of oxygen?
First we write the correct formula Al(NO3)3 . This formula stands for 1 mole of aluminium nitrate
containing 1 mole A1; 3 moles of N; and 9 moles of 0.
3Al(NO3)3 stands for 3 moles of aluminium nitrate.
That is for 3 moles of A1; 9 moles of N and 27 moles of O. So
(a)
(b)
n(Al) = n( Al(NO3)3
= 3 moles
n(O) = 9 x n( Al(NO3)3
(from formula)
(from formula)
This figure of 9 is because for every mole of aluminium nitrate there are 9 moles of
oxygen, from the formula
30.
31.
What is the amount of oxygen in 5 mole of copper sulfate?
What is the amount of iron in 2 moles of Iron (III) oxide?
6
We can extend this quite easily to include masses of materials. Work through this problem.
(a) What is the amount of magnesium sulfate in a 30 g sample?
(b) What is the amount of Sulfur?
(c) What is the amount of oxygen?
i.
M(MgSO4)
n(MgSO4)
ii.
= 120 g mol-1
m( MgSO4 )
M ( MgSO4 )
30
=
120
= 0.25 mol
=
Every mole of MgSO4 has 1 mole of sulfur
So amount of S atoms = amount MgSO4;
i.e. n(S) = n (MgSO4)
therefore n(S) = 0.25 mo1
iii.
iv.
Every mole of MgSO4 has 4 moles of O atoms.
n(O) = 4 x n(MgSO4)
n(O = 4 x 0.25
= 1 mol
32.
33.
(a)
What is the amount of magnesium oxide in a 4 g sample?
M(Mg) = 24; M(O) = 16
(b)
What is the amount of Mg in 4 g of MgO?
(a)
What is the amount of silver nitrate in 3.4 g?
[M(Ag) = 108; M(N) = 14; M(O) = 16]
(b)
What amount of silver is this?
(c)
What amount of oxygen is this?
Now let’s have a look at gases. Gases are made up of molecules, so 1 mole of gas will be 1 mole
of its molecules. Thus 1 mole of carbon dioxide will be 1 mole of carbon dioxide molecules.
So M for CO2 = 44 g mol-1 i.e. the mass of 1 mole of C atoms and 2 mole of O atoms.
34.
What is the molar mass of sulfur trioxide?
35.
What is the amount of nitric oxide (NO) in 3 g sample?
36.
What is the mass of 0.01 mole of sulfur trioxide?
These problems are done just the same as all the others.
Now let us consider the gaseous elements oxygen, hydrogen, nitrogen, chlorine etc. As pure gases
these substances exist in the form of diatomic (2 atoms) molecules – we know that the formula for
oxygen gas is O2. Thus while the mole mass of oxygen is 16, the molar mass of oxygen gas is the
mass of 1 mole of molecules
i.e. it is 32 g mol-1.
7
If we are considering the gas oxygen we use its molar mass (32 g mol-1) to match its formula O2.
However, if we are considering compounds containing oxygen we use the molar mass of atomic
oxygen e.g. M for water is 18 g mol-1 (2 mole of H atoms + 1 mole of O atoms.)
37.
(a)
(b)
(c)
(d)
38.
(a)
The molar mass of atomic chlorine is?
The formula for chlorine gas is?
The molar mass for chlorine gas is?
The molar mass for hydrogen chloride is?
The molar mass of atomic fluorine is …….
(b) The molar mass of fluorine gas is …….
(b) What is the mass of 0.02 mol of fluorine atoms?
(c) What is the mass of 0.02 mol of fluorine molecules?
39.
What is the mass of 3 mol of nitrogen gas?
40.
What amount of hydrogen gas is present in a 10 g sample?
41.
What amount of O atoms is present in a 4g sample of water?
42.
What amount of nitrogen atoms is present in a 5 g sample of silver nitrate?
43.
What is the mass of 2.7 mol of sodium bromide?
44.
What amount of oxygen atoms is present in a 10 g sample of copper sulfate?
45.
What is the molar mass of a new element X, if 3 mol of X has a mass of 246g?
8
ANSWER SHEET
1.
32 g mol-1
28.
249.5 g mole-1
2.
32 g
29.
32 g mole-1
3.
108 g mol-1
30.
20 mol
4.
108 g
31.
4 mol
5.
197 g
32a.
0.099 mol
6.
24.3 g
32b
0.099mol
7.
402 g
33a.
0.02 mol
8.
1373 g = 1.373 kg
33b.
0.02 mol
9.
1781 g = 1.78 kg
33c.
0.06 mol
10.
369.8 g
34.
80 g mole-1
11.
0.635 g
35.
0.1 mol
12.
6 mol
36.
0.8 g
13.
0.1 mol
37a.
35.5 g mole-1
14.
0.0316 mol
37b.
Cl2
15.
10.8 g
37c.
71 g mole-1
16.
28.1 g mol-1
37d.
36.5 g mole-1
17.
58.5 g mol-1
38a.
19 g mole-1
18.
101.5 g mol-1
38b.
38 g. mole-1
19.
461 g mol-1
38c.
0.38 g
20.
78 g mol-1
38d
0.76 g
21.
53.5 g mol-1
39.
84 g
22.
170 g mol-1
40.
5 mol
23.
187.5 g mol-1
41.
0.222 mol
24.
158.2 g mol-1
42.
0.294 mol
25.
132 g mol-1
43.
277.8 g
26.
27.
44 g mol-1
246.3 g mol-1
44.
45.
0.251 mol
82 g mol-1
9
MOLE LABORATORY EXERCISE
Station 1
Calculate the mass of 0.1 mole of sodium chloride; weigh this mass and show the teacher your sample.
molar mass of NaCl
=
 mass of 0.1 mol NaC1 =
a)
What amount, in moles, of sodium ions are there in this sample?
b)
What mass of NaCl would you have to weigh to have 2 moles?
M (Na) = 23 ; M (C1) = 35.5
Station 2
Calculate the mass of 0.5 mole of calcium carbonate, weigh this mass and show the teacher your sample.
molar mass of CaCO3
 mass of 0.5 mol CaCO3
=
=
a)
What amount, in moles, of atoms does this contain altogether?
b)
What mass of CaCO3 would you have to weigh to have 10 mol of atoms?
M (Ca) = 40 ; M (C) = 12 ; M (O) = 16
Station 3
Weigh the sample of sulfur in the jar provided to the nearest 0.05g.
DO NOT OPEN THE JAR.
mass of sulfur + jar
mass of jar
 mass of sulfur
=
=
=
a)
What amount, in moles, of sulfur atoms is this?
b)
What amount, in moles, of S8 molecules is this?
M (S) = 32
10
72.50  0.05g
Station 4
Weigh 20 granules of zinc
mass of 20 granules
=
a)
What amount, in moles, of zinc is there in one granule?
b)
How many granules are there in 1 mole of zinc?
M (Zn) = 65
Station 5
Measure out 18mL of water. Weigh this water in a beaker.
mass of water
a)
What amount, in moles, of water is this?
b)
What would the mass of 1 L of water be?
c)
What amount, in moles, of water is in 1L?
=
M (H) = 1 ; M (O) = 16
Station 6
Weigh this 1 metre length of magnesium ribbon
mass of magnesium =
a)
What amount, in moles, of magnesium is this?
b)
What length of ribbon would you need if you required one mole of magnesium?
M (Mg) = 24.3
11
Moles / Molar Mass Problems
SHOW ALL WORKING
1
Find the molar mass of the following ionic compounds:
a) NaHCO3
b) Pb(NO3)2
2
What amount (in moles) is:
a) 20 g of calcium
b) 40.5 g of aluminium
c) 88.75 g of chlorine gas
d) 64 g of oxygen gas
e) 10 g of methane (CH4)
f) 32 g of sodium hydroxide
g) 100 g of potassium dichromate (K2Cr2O7)
3
What mass would the following amounts of compounds have:
a)
b)
c)
4
For the ionic compound calcium carbonate CaCO3, find the:
a)
b)
5
n(Na) = 0.01 mole
n(NaOH) = 0.01 mole
n(Na2CO3) = 0.01 mole
mass of 0.05 moles
amount of CaCO3 in 10 g of CaCO3
H
H
1 mole of NaCl contains:
a)
b)
O
N
H
what amount (in moles) of Na+ ions?
what amount (in moles) of Cl- ions?
NH2
C
N
N
C
C
C
H
6
C
N
3 moles of Na2SO4 contains:
a)
b)
7
What amount, in moles, of Cu2+ ions are in 79.75 g of copper sulfate?
8
What amount, in moles, of OH- are in 161 g of copper hydroxide?
9
Which of the following has three moles of ions in one mole of oxide?
10
b) Al2O3
N
H
what amount (in moles) of ions?
what amount (in moles) of atoms?
a) MgO
c) SO2
d) Na2O
What amount, in moles, of atoms are there in 5 g of ammonium sulfate?
12
H
H
H
PERCENTAGE COMPOSITION AND CHEMICAL FORMULAE
CHEMICAL FORMULAE: HOW THEY ARE WORKED OUT AND WHAT WE CAN LEARN FROM
THEM
A pure chemical compound has a fixed composition and we are therefore justified in representing such a
compound by a fixed formula. In this section we will consider what we can learn from a chemical
formula and the methods by which it can be determined.
Percentage Composition of Compounds
In a compound the mass of each element can be expressed as a percentage of the total mass of the
components and the gravimetric composition is usually given in this way. The method of working is
simple and straightforward, as you will see from the following example:
Example 1
Calculate the percentage composition by mass of ammonium nitrate.
(a)
Write down the formula of the compound: NH4NO3
(b)
Find the molar mass of the compound by adding together the molar masses of all of the atoms
present: 14 + (4 x 1) + 14 + (3 x 16) = 80.
Express each atom’s molar mass as a percentage of the compounds molar mass.
If more than one atom of a particular element is present, then the total mass of these atoms must
be used, e.g. as two nitrogen atoms appear in the formula, the mass used is 28 and not 14.
28
%N =
x 100 = 35 %
80
(c)
(d)
%H =
4
80
%O =
48
80
x 100 = 5 %
x 100 = 60 %
Check that the percentages add up to 100.
Problems for You
For all calculations use your Table of Molar Masses and round to 1 d.p.
1.
Calculate the percentage composition by mass of the following compounds:
(a)
(b)
(c)
(d)
calcium carbonate
anhydrous copper (II) sulfate (anhydrous means without water)
methane
potassium hydrogen carbonate
Note: If a compound contains water of crystallization, the percentage of water
should be calculated as a separate unit.
See the next worked example.
13
Example 2
Calculate the percentage composition by mass of hydrated magnesium chloride MgC12.6H20.
Molar mass of compound
= 24 + (2 x 35.5) + (6 x 18) = 203
% Mg
=
24
x 100 = 11.82 %.
203
% C1
=
71
x 100 = 34.98 %
203
% H2O
=
108
x 100 = 53.2 %
203
Problems for You
Determine the percentage composition of each of the following hydrates:
2.
(a)
iron (II) sulfate, FeSO4.7H2O
(b)
sodium carbonate Na2CO3.1OH2O
(c)
copper (II) sulfate, CuSO4.5H2O
Empirical Formulae
Questions
3.
What is the meaning of an ‘empirical’ formula?
4. Before you can find the empirical formula of a compound you need to know its composition by
mass. What other information do you need?
Calculating the Empirical Formula of a Compound:
Except in a few simple cases, such as the formation or reduction of metallic oxides, the quantitative
analysis of a compound is too complicated for you to attempt at present. However you will need to know
how to calculate the formula of a compound, given the results of its analysis by mass.
Example 3
Calculate the simplest (empirical) formula for the compound with the following composition:
Lead 8.32 g
sulfur 1.28 g
oxygen 2.56 g
(Molar masses: Pb = 207, S = 32, O = 16)
(a)
(b)
Convert the masses into amounts (in moles) by dividing the mass of the element by the mass of its
m
molar mass. n 
M
8.32
n(Pb) =
= 0.040 moles
207
n(S)
=
1.28
= 0.040 moles
32
n(O)
=
2.56
= 0.16 moles
16
The compound contains lead, sulfur and oxygen atoms combined in the ratio:
14
(c)
0.04 moles of lead: 0.04 moles of sulfur: 0.16 moles of oxygen
Convert these ratios into whole numbers by dividing throughout by the smallest number (0.04 in
this case).
1 mole of lead: 1 mole of sulfur: 4 moles of oxygen
A mole of any substance contains an equal number of (6.02 x 1023) of particles (atoms in this
case).
Therefore (6.02 x 1023) atoms of lead combine with (6.02 x 1023) atoms sulfur and with (4 x
6.02 x 1023) atoms of oxygen.
Therefore 1 atom of lead combines with 1 atom of sulfur and 4 atoms of oxygen.
Therefore Empirical formula is PbSO4.
This type of calculation is conveniently set out in table form.
Element
Lead
Sulfur
Oxygen
Mass (or % Molar
mass)
mass
8.32 g
207 g
1.28 g
32 g
2.56 g
16 g
Mass___
molar mass
8.32/207
1.28/32
2.56/16
Ratio of
Moles
0.040
0.040
0.16
Simplest
ratio
1
1
4
Ratio of atoms 1:1:4
Empirical formula PbSO4
Note: if the composition by mass is expressed as percentages, the method of working is exactly
the same, because then we are considering the mass of each element in 100 g of the
compound.
Problems for You
5. Determine the empirical formula of each of the following compounds for which the composition by
mass is given: (Set out in a table.)
(a)
(b)
(c)
(d)
(e)
magnesium 9.5 g, chlorine 28.4 g
copper 40 %, sulfur 20 %, oxygen 40 % (take Cu=64)
nitrogen 1.40 g, hydrogen 0.40 g, carbon 0.60 g, oxygen 2.40 g
carbon 75 %, hydrogen 25 percent
carbon 40 %, hydrogen 6.67 percent, oxygen 53.3 %.
Note: In the case of hydrated salts, the mass of water of crystallisation is taken as one unit and
divided by the molar mass of water.
Example 4
A hydrate has the following percentage composition:
iron = 20.15 %, sulfur = 11.51 %, oxygen = 23.02 %, water = 45.32 %.
Determine the empirical formula of the hydrated salt.
15
Element
or group
Fe
S
O
H2O
% mass
20.15
11.51
23.02
45.32
Molar
mass
56
32
16
18
% mass
molar mass
20.15/56
11.51/32
23.02/16
45.32/18
Ratio of atoms or groups Fe:S:O: H2O
Formula is FeSO4.7 H2O
Ratio of
moles
0.36
0.36
1.44
2.52
is
Simplest
ratio
1
1
4
7
1:1:4:7
6. Problems
Calculate the empirical formulae of hydrated salts with the following compositions.
(a) Cu = 25.6 %, S = 12.8 %, 0 = 25.6 %, H2O = 36.0 %
(b) Na = 16.09 %, C = 4.20 %, 0 = 16.78 %, H2O = 62.93 %
(c) Mg = 11.82 %, C1 = 34.98 %, H2O = 53.20 %.
Molecular Formulae
The molecular formula gives the number, and not just the ratio of the numbers, of each type of
atom in one molecule of the compound. A molecular formula can be used only for covalent
(molecular) compounds: for ionic substances, the empirical formula is the only one that can be
written, as ionic substances have giant structures in which there are no free units such as
molecules. Covalent substances can have both empirical and molecular formulae, e.g. the
empirical formula for ethane is CH3, its molecular formula is C2H6, but the formula CH4 for
methane is both its empirical and its molecular formula.
Determination of Molecular Formulae
To find the molecular formula of a compound, we need to know both its empirical formula and its
molecular mass. The molecular mass of a gas or a volatile liquid can be found experimentally by
using the fact that a mole of any gas occupies 22.4 litres at S.T.P. (standard conditions of
temperature and pressure).
A large container such as a glass globe is weighed when evacuated, and then reweighed when
filled with the gas in question. If the volume, temperature, and pressure of the gas are known, its
volume at S.T.P. can be calculated and from this the mass of a mole of that gas, i.e. its molecular
mass.
16
Example 5
Determine the molecular formula of a compound which has the following percentage composition:
carbon = 40 %, hydrogen = 6.66 %,
Oxygen = 53.33 %. Molecular mass = 180.
(a)
Determine the empirical formula.
Consider 100g. 40% C will thus be 40g and so on.
mass
Element
Carbon
Hydrogen
Oxygen
40
6.66
53.33
Molar
mass
12
1
16
n
mass
molar mass
40/12
6.66/1
53.33/16
Ratio of
moles
Simplest
ratio
3.33
6.66
3.33
1
2
1
Ratio of atoms C:H:O = 1:2:1
Empirical formula CH2O
(b)
Use the empirical formula and molecular mass to determine the molecular formula.
Molecular mass = 180
Total mass of atoms in empirical formula = (12 + 2 + 16) = 30
Molecular mass = (30)n where n is the number of empirical formula units in the molecular
formula.
Therefore molecular mass = (30)n = 180
Therefore n = 6
Molecular formula = (6 x CH2O) = C6H12O6
7. Problems
Determine the molecular formulae of compounds having the following percentage composition by
mass.
(a) Carbon = 80 %, hydrogen = 20 %, molecular mass = 30
(b) Hydrogen 5.9 %, oxygen 94.1 %, molecular mass = 34
(c) Carbon 38.75 %, hydrogen 16.1 %, nitrogen 45.2 %, molecular mass = 31.
17
ANSWERS
1. (a) 40% Ca, 12% C, 48% O
(b) 39.8% Cu, 20.1% S, 40.1% O
2.
(c)
75% C, 25% H
(d)
39.1% K, 1% H, 12% C, 48% O
(a)
20.1 Fe, 11.5% S, 23% O, 45.3% H2O
(b) 16.1% Na, 4.2% C, 16.8% O, 62.9% H2O
(c) 25.5% Cu, 12.8% S, 25.7% O, 36.1% H2O
3.
The simplest ratio of the number of atoms in a compound.
4.
Molar mass of the compound.
5.
(a)
MgC12
(b)
CuSO4
(c)
N2H8CO3
(d) CH4
(e) CH2O
6.
7.
(a)
CuSO4.5H2O
(d)
Na2CO3.10H2O
(e)
MgC12.6H2O
(a)
C2H6
(d)
H2O2
(c)
CH5N
18
Percentage Composition Problems
1
Calculate the percentage of magnesium in magnesium oxide.
2
What is the percentage composition of each element in potassium nitrate (KNO3)?
3
Find the percentage of carbon in acetylene, C2H2.
4
Find the empirical formula of the substances which have the following percentage compositions:
a) 80% copper, 20% oxygen
b) 53% aluminium, 47% oxygen
c) 1.6% hydrogen, 22.2% nitrogen, 76.2% oxygen
5
Calculate the empirical formula of the oxide of sulfur which is 40% sulfur by weight.
6
A hydrocarbon contains 90% carbon. The empirical formula of the hydrocarbon is:
a) CH
7
b) CH2
d) CH4
e) C3H4
An analysis gave the following data for three unknown substances. Find the molecular formula of
each substance.
a) 85.7% C and 14.3% H,
b) 30.4% N and 69.6% O,
c) 2% H, 33% S and 64% O,
8
c) CH3
Molar mass = 28 g mol-1
Molar mass = 92 g mol-1
Molar mass = 98 g mol-1
The empirical formula of a substance is HO. Its molar mass is 34 g. the molecular formula of the
substance is:
a) OH
b) H2O
c) H2O2
d) HO2
e) H3O4
9
A weak acid has a molar mass of 60 g mol-1 and the following percentage composition:
40 % C
6.67% H
53.33% O
Find the molecular formula.
10
A sweet smelling substance has a molar mass of 88 g mol-1 and the following percentage
composition:
54.5 % C
9.1% H
36.4% O
Find the molecular formula.
Answers:
1. 60.3 %
4a. CuO
6. (e)
8. (c )
2. K: 38.7%; N: 13.8%; O: 47.5%
4b. Al2O3
4c. HNO3
7a. C2H4
7b. N2O4
9. C2H4O2
10. C4H8O2
19
3. 92.3%
5. SO3
7c. H2SO4
Copper Sulfate - Water of Crystallisation Problem
A student recorded the following results:
Mass crucible + lid
Mass crucible, lid + hydrated crystals
Mass crucible, lid + anhydrous crystals
m1 = 32.24 + 0.01 g
m2 = 35.26 + 0.01 g
m3 = 34.17 + 0.01 g
M (CuSO4) = 159.5 g mol-1.
Calculations (SHOW ALL WORKING!)
M (H2O) = 18 g mol-1.
B1
What is the mass of anhydrous copper sulfate?
B2
Calculate the amount (in moles) of anhydrous copper sulfate remaining.
B3
What is the mass of water removed on heating to form the anhydrous salt?
B4
Calculate the amount (in moles) of water removed.
B5
Calculate the mole ratio of CuSO4 to H2O and hence determine the formula of the hydrated copper
salt. In this formula give the value of x as a whole number. Show all your working.
20
Formula = _____________________
INTERPRETING GRAVIMETRIC DATA – ZINC IODIDE
The empirical formula of zinc iodide can be determined by carrying out the following procedure.
1
2
3
4
5
6
Weigh a clean dry test tube on a balance to 0.01 g
Add about 0.5 g of zinc powder and weigh accurately.
Add excess iodine solid (about 1g but this need not be accurate) and ethanol to dissolve the iodine.
Shake well and allow the zinc to mix with the dissolved iodine (a lot of heat will be generated)
Allow the solid zinc iodide to settle and discard the solvent. Dry the solid in the test tube
thoroughly.
Weigh the dried zinc iodide and the test tube.
Results:
Mass of test tube
Mass of test tube plus zinc
Mass of test tube plus dried zinc iodide
= 20.04 + 0.01g
= 20.54 + 0.01g
= 22.48 + 0.01g
Calculations
M(Zn) = 65.4 g mol –1
M(I) = 126.9 g mol –1
1
Calculate the mass of zinc added.
2
Calculate the amount of zinc added.
3
Calculate the mass of iodine incorporated into the zinc iodide.
4
Calculate the amount of iodine in the zinc iodide.
5
Calculate the empirical formula of zinc iodide.
21
Concentration of Solutions
The concentration of a solution is usually given as:
n
c
with the units mol L-1
V
However, concentration can also be expressed as mass per volume i.e.
m
c
with the units g L-1
V
We need to be able to interconvert these quantities. This is done by using the relationship
m=nxM
Worked example 1
A solution of NaOH is given as 4 g L-1. Express this in mol L-1.
(4 g L-1 means that in 1L there are 4g of NaOH). Thus:
Answer
In 1L
m(NaOH)
Hence in 1L
n(NaOH)
Thus
c(NaOH)
= 4g
m( NaOH )
=
M ( NaOH )
4
=
40
=
0.1 mol
=
0.1 mol L-1
---------------------------------------------------------In a similar manner concentrations can be converted from mol L-1 to g L-1
Concentrations as % solutions
Solutions can also be expressed as being say 5% solution. This is the mass of the substance are dissolved
to make 100 mL of solution.
Worked example 2
What mass of starch is required to make 50 mL of 5% solution?
This solved by using simple proportion:
Answer
A 5% starch solution contains 5 g starch in 100 mL solution.
Hence 50 mL of 5% starch solution contains 50 X 5 g = 2.5g
100
Problems:
1
What mass of glucose is needed to make 500 mL of 3% solution?
2
What is the % concentration of a salt solution made from dissolving 3g to make 200 mL of
solution?
3
What mass of sugar is need to make 3L of 4.5% solution?
4
Convert the concentration of 5 g L-1 KCl solution to mol L-1
5
Convert the concentration of 0.2 mol L-1 NH3 solution to g L-1, then express as a % solution.
ANSWERS: 1) 15g
2) 1.5%
3) 135g
22
4) 0.067 mol L-1
5) 3.4 g L-1; 0.34%
Volumetric Analysis Techniques
Pipette (Syrette)
- Rinse with distilled water
- Rinse with solution going into pipette
- Suck up solution to the desired mark
- Wipe the outside of the pipette with
lab paper
- Release the solution down to the
desired mark.
Burette
-Rinse with distilled water
- Rinse with the solution going into the
burette
- Fill using a clean funnel which has
been rinsed with the solution passing
through it
- Wipe the outside of the burette with
lab paper
- Note the starting volume and release
solution into the conical flask until
colour change occurs
Conical Flask
- Rinse with distilled water
- Continually swirl conical flask
- Add solution from burette dropwise
in latter part of titration
- Use wash bottle to wash lingering
drops at the base of the burette into the
conical flask
- Stop titration when indicator changes
- Note the final volume.
23
Quantitative Chemistry Problem Technique
Step 1 Write all appropriate equations, omitting spectator ions.
e.g.
H  (aq)  OH  (aq)  H 2O (l )
Step 2 Deduce the relationship between amounts of chemical species of interest in the particular problem.
i.e. n(H+) = n(OH-) for above example.
Step 3 Inspect the problem again and relate to parent compound if necessary.
Step 4 Decide which relationship you can replace each n(substance) with. This may be
concentration/volume information or mass/molar mass information. Remember:
concentration =
amount(in moles)
volume(in L)
and amount =
mass
molar mass
Step 5 Rearrange to have unknown as subject then substitute data.
Step 6 Inspect the problem again and decide whether the units are correct.
Worked Example 1
25.0 mL of HCl was titrated against 0.082 mol.L-1 NaOH and the endpoint detected at 22.5 mL.
This can be reworded as: 25.0 mL of HCl was neutralised by 22.5 mL of 0.082 mol.L-1 NaOH.
Calculate the c(HCl).
Step 1.
H  (aq)  OH  (aq)  H 2O (l )
Step 2.
n(H+) = n(OH-) from equation
n(HCl) = n(NaOH) from formulae
Step 3 -6.
c(HCl) x V(HCl)
c(HCl)
= c(NaOH) x V(NaOH)
= 0.082 x 0.0225
0.025
= 0.0738 mol L-1
Worked Example 2
What mass of CaCO3 is required to neutralise 150 mL of 0.2 mol L-1?
Ans:
CaCO3 ( s)  2 H  (aq)  Ca 2 (aq)  CO2 ( g )  H 2 O(l )
n(CaCO3)
=
n( H  )
2
from equation
n(CaCO3)
=
n(HCl )
2
from formula
m(CaCO3 )
M (CaCO3 )
=
c( HCl ) x V ( HCl )
2
m(CaCO3)
=
=
0.2 x 0.150 x 100
2
1.5 g
24
Year 12 - MODEL CALCULATION FOR HCl CONCENTRATION
Av. titre = 20.1 + 0.1 mL
c(Na2CO3) = 0.10980 + 0.00004 mol L-1
10mL pipette used.
CO32 (aq)  2 H  (aq)  H 2 O (l )  CO2 ( g )
n(CO32-)
= n(H+)
2
n(CO32-)
= n(Na2CO3) from formula
(from eqn)
- (A)
= c(Na2CO3) x v(Na2CO3)
= 0.10980 x 0.01 mol
= 0.10980 x 0.01 mol
= 0.0010980 + 0.04% mol
n(H+) =
0.0021960 mol
n(HCl)=
n(H+)
c(HCl) =
n =
V
0.0201
(This degree of accuracy can be calculated)
(from A)
(from formula)
0.0021960
=
0.109253731 +0.54% mol L-1
=
0.1093 + 0.0006 mol L-1
(This degree of accuracy can be calculated)
*
Obviously your figures will vary depending upon your c(Na2CO3) and your average titre.
*
Each titre has absolute error of + 0.1 mL. For convenience, give the average titre the same error.
*
Remember to convert volumes to Litres prior to calculating amounts.
*
Always write net ionic equations.
25
Volumetric Analysis Problems
Use molar masses from your Periodic Table, as required.
Set A
1
Calculate the concentration of NaOH solution, if 2g are dissolved to make 100 mL of solution.
2
What is the concentration of OH- ions in a solution made from dissolving 4.5 g of NaOH to make
250 mL of solution?
3
What mass of KOH is needed to make 250 mL of 0.2 mol L-1 solution?
4
What is the amount of H+ ions in 25 mL of 0.109 mol L-1 HCl solution?
5
What is the concentration of H+ ions in a solution made from dissolving 0.98 g of H2SO4 to make
500 mL of solution?
6
Express the concentration of 0.2 mol L-1 NaOH as a % solution. [Remember this is the mass of
the substance dissolved to make 100 mL of solution.]
7
Express the concentration of 0.35 mol L-1 HCl as a % solution.
8
Express the concentration of a 2% solution of HCl as mol L-1.
9
What amount of NaOH is required to prepare a 4% solution of NaOH?
10
What amount of HCl is required to prepare a 3.6% solution of HCl?
Set B
1
A solution of 0.1 mol L-1 HCl was titrated against 0.05 mol L-1 NaOH. What volume of NaOH
was needed to neutralise 20 mL of the acid?
2
A 25 mL aliquot (sample) of NaOH required 22.4 mL of 0.100 mol L-1 HCl solution to reach the
endpoint. Calculate the concentration of the NaOH solution.
3
20 mL of 0.150 mol L-1 HCl required 18.6 mL of NaOH to reach the endpoint. Calculate the
concentration of the NaOH solution.
4
10 mL of sodium carbonate solution required 9.7 mL of 0.16 mol L-1 HCl to reach the endpoint.
Calculate the concentration of the Na2CO3 solution.
5
25 mL of H2SO4 solution required 21.4 mL of 0.13 mol L-1 NaOH to reach the endpoint.
Calculate the concentration of the H2SO4 solution.
6
What volume of 0.25 mol L-1 HCl solution is needed to neutralise 10 mL of 0.08 mol L-1 Na2CO3
solution?
7
What volume of 0.2 mol L-1 H2SO4 solution is needed to neutralise 20 mL of 0.15 mol L-1 NaOH
solution?
26
8
A solution of 0.16 mol L-1 HCl was titrated against 0.43 mol L-1 KOH. What volume of KOH was
needed to neutralise 20 mL of the acid?
Set C
1
100 mL of nitric acid solution reacted with 0.85 g of calcium carbonate solid. Calculate the
concentration of the HNO3 solution.
2
It takes 50 mL of 0.2 mol L-1 HCl to react completely with a piece of limestone (CaCO3). What
was the mass of the limestone?
3
What mass of zinc will react completely with 25 mL of 1.5 mol L-1 H2SO4 solution?
4
If 200 mL of caustic soda (NaOH) solution are neutralised by 100 mL of HCl solution which
contains 7.3g of HCl per litre, what is the concentration of the caustic soda solution?
5
What mass of KOH would be needed to exactly neutralise 30 mL of 2 mol L-1 HCl?
6
0.1g of Mg is requires 25 mL of HCl for complete reaction. What is the concentration of the acid?
7
What volume of 0.1 mol L-1 HCl is required to neutralise a 20 mL of a 3% NaOH solution?
8
What mass of Na2CO3 will exactly neutralise 40 mL of 5% HCl solution?
9
A 0.15 g impure sample of Mg is dissolved in 100 mL of 0.1 mol L-1 HCl. The excess (unreacted)
HCl takes 8.9 mL of 0.1 mol L-1NaOH to reach the endpoint. What is the % Mg in the original
sample?
10
A 0.5 g impure sample of Zn is dissolved in 150 mL of 0.1 mol L-1 HCl. The excess (unreacted)
HCl takes 12.5 mL of 0.1 mol L-1NaOH to reach the endpoint. What is the % Zn in the original
sample?
Answers
Set A
1. 0.5 mol L-1
2. 0.45 mol L-1
3. 2.81 g
4. 2.73 X 10-3 mol
5. 0.04 mol L-1
6. 0.8%
7. 1.28 %
8. 0.548 mol L-1
9. 1 mol
10. 0.986 mol
Set B
1. 40 mL
2. 0.0896 mol L-1
3. 0.161 mol L-1
4. 0.0776 mol L-1
5. 0.0556 mol L-1
6. 6.4 mL
7. 7.5 mL
8. 7.44 mL
27
Set C
1. 0.17 mol L-1
2. 0.5g
3. 2.45 g
4. 0.1 mol L-1
5. 3.37 g
6. 0.329 mol L-1
7. 150 mL
8. 0.666 g
9. 73.8 %
10. 89.9%