Download Teacher Edition Calculations

CALCULATIONS IN
CHEMISTRY
FOR THE
PRELIMINARY AND HSC COURSES
TEACHER EDITION
BRONWEN HEGARTY
1
CALCULATIONS IN CHEM ISTRY
How to Use this Workbook:
 This workbook is designed for use both by students and teachers.
 It provides a guided pathway through the chemical calculations
specified in the Preliminary and HSC courses of the NSW Chemistry
Syllabus.
 Worksheets provide students with multiple practice examples of the
types of calculations found in Year 11 Preliminary examinations and
in the final HSC paper.
 Fully-worked solutions can be found in the Teacher Edition of this
workbook, which is copyright -free and distributed to teachers on CD.
The content of this CD can be copied and distributed to students or
can be placed on a school’s intr anet for use by students.
 A Periodic Table and Data Sheet, including a Table of Reduction
Potentials, are provided.
Please note:
This Student Edition is covered by Australian copyright law and hence may
not be copied as a whole for distribution to st udents.
2
Checklist for students
Syllabus statements relating to chemical calculations – Preliminary course
Completed
√
Gravimetric analysis of a mixture to estimate its percentage composition
Define the mole as the number of atoms in exactly 12 g of carbon-12 (Avogadro’s
Number)
Process information from secondary sources to interpret balanced chemical equations in
terms of mole ratios
Compare mass changes in samples of metals when they combine with oxygen
Perform a first-hand investigation to meas ure and identify the mass ratios of metal to non metal(s) in a common compound and calculate its empirical formula
Describe the contribution of Gay-Lussac to the understanding of gaseous reactions and
apply this to an understanding of the mole concept
Recount Avogadro’s Law and describe its importance in developing the mole concept
Solve problems and analyse information from secondary sources to perform calculations
involving Avogadro’s number and th e equation for calculating the number of moles of a
substance:
n
m
M
Process information from secondary sources to investigate the relationship between the
volumes of gases involved in reactions involving a metal and relate this to an
understanding of the mole
Distinguish between empirical formulae and molecular formulae
Perform an investigation involving calculations of the density of water as a liquid and a
solid using:
density 
mass
volume
Describe the molarity of a solution as the number of moles of solute per litr e of solution
using:
c
n
V
Carry out simple calculations to describe the concentration of given solutions, given
masses of solute and volumes of solution
Express concentrations using different units
Perform a first-hand investigation to make solutions to specified volume-to-volume and massto-volume specifications and dilute them to specified concentrations
(cV = constant)
Calculate mass and concentration relationships in precipitation reactions as they are
encountered
Explain and use the equation
H  mCT
Explain how water’s ability to absorb heat is used to measure energy changes in chemical
reactions
Choose resources and perform a first -hand investigation to measure the change in
temperature when substances dissolve in water and calculate the molar heat of solution
Solve problems and perform a first -hand investigation to measure the change in mass
when a mixture such as wood is burned in an open container
3
Checklist for students
Syllabus statements relating to chemical calculations – HSC course
Completed
√
Solve problems, plan and perform a first -hand investigation to carry out the fermentation
of glucose and monitor mass changes
Identify data sources, choose resources and perform a first -hand investigation to
determine and compare heats of combustion o f at least three liquid alkanols per gram and
per mole
Solve problems and analyse information to calculate the potential E O requirement of
named electrochemical processes using tables of standard potentials and half -equations
Identify data, plan and pe rform a first-hand investigation to decarbonate soft drink and
gather data to measure the mass changes involved and calculate the volume of ga s
released at 25˚C and 100 kPa
Calculate volumes of gases given masses of some substances in reactions, and calculate
masses of substances given gaseous volumes, in reactions involving gases at 0 ˚C and
100 kPa or 25˚C and 100 kPa
Describe the use of the pH scale in comparing acids and bases
Describe acids and their solutions with the appropriate use of the terms strong, weak,
concentrated and dilute
Identify pH as -log10 [H +] and explain that a change in pH of 1 means a ten -fold change
in [H+]
Compare the relative strengths of equal concentrations of citric, acetic and hydrochloric
acids and explain in terms of the degree of ionisation of their molecules
Solve problems and perform a first -hand investigation to use pH meters/probes and
indicators to distinguish between acidic, basic and neutral chemicals
Process information from secondary sources to calculate pH of strong acids, given
appropriate hydrogen ion concentrations
Perform a first-hand investigation and solve problems using titrations and including the
preparation of standard solutions. Use available evidence to quantitatively and
qualitatively describe the reaction between selected acids and bases
Perform a first-hand investigation to determine the concentration of a domestic acidi c
substance
Identify data, plan, select equipment and perform first -hand investigations to measure the
sulfate content of lawn fertiliser and explain the chemistry involved
Gather, process and present information to interpret secondary data from AAS
measurements and evaluate the effectiveness of this in pollution control
Perform first-hand investigations to use qualitative and quantitative tests to analyse and
compare the quality of water samples
4
Atomic Mass
1.
What is meant by “stoichiometry”?
The study of quantitative aspects of formulae and equations.
2.
Which particles in an atom determine its mass?
Protons and neutrons. The sum of the numbers of protons and neutrons is called the mass
number of an atom.
3.
What are isotopes?
Atoms of the same element which differ in their numbers of neutrons.
4.
List the numbers of protons, neutrons and electrons in atoms of the following isotopes.
Isotope
H
Number of
protons
1
Number of
neutrons
0
Number of
electrons
1
1
1
2
1
35
17
H
1
1
1
Cl
17
18
17
37
17
Cl
17
20
17
92
143
92
235
92
U
5.
Define the relative atomic mass of an atom.
The relative atomic mass (or atomic weight) of an element is def ined as the average mass of the
atoms present in the naturally occurring element re lative to the mass of an atom of the carbon12 isotope, taken as 12 exactly.
6.
Chlorine exists as 2 isotopes , chlorine-35 (77.5%) and chlorine-37 (22.5%). What is the
relative atomic mass of naturally occurring chlorine?
(77.5/100 x 35) + (22.5/100 x 37) = 35.45
7.
Calculate the average atomic weight of copper , assuming it consists of 69.0% copper-63 and
31.0% copper-65.
(69.0/100 x 63) + (31.0/100 x 65) = 63.62
5
8.
Calculate the average atomic weight of carbon which consists of 98.89% carbon -12 and 1.11%
carbon-13.
(98.89/100 x 12) + (1.11/100 x 13) = 12.01
9.
Calculate the average atomic mass of magnesium which consists of 78.70% magnesium -24,
10.13% magnesium-25 and 11.17% magnesium -26.
(78.70/100 x 24) + (10.13/100 x 25) + (11.17/100 x 26) = 24.32
10.
The average atomic mass is listed on the Periodic Table for each element. The average atomic
mass for oxygen is listed as 16.00. What assumption coul d you make about the isotopic
composition of oxygen isotopes?
You would assume (incorrectly) that there is only 1 isotope (oxygen -16) or (correctly) that the
proportion of any other isotope is extremely small.
11.
Boron exists as 2 isotopes, boron -10 and boron-11. The Periodic Table lists the relative atomic
mass of boron as 10.81. What is the percentage abundance of each of the boron isotopes?
Let the % abundance of boron -10 be Y. The % abundance of boron -11 is therefore (100 –Y).
(Y/100 x 10) + ((100 –Y)/100 x 11) = 10.81
0.10Y + 11.00 – 0.11Y = 10.81
0.19 = 0.01Y
Y = 19
Hence the % abundance of boron -10 is 19% and the % abundance of boron -11 is 81%.
12.
Bromine exists as 2 isotopes, bromine -79 and bromine-81. The Periodic Table lists the rela tive
atomic mass of bromine as 79.91. What is the percentage abundance of each of the bromine
isotopes?
Let the % abundance of bromine -79 be Y. The % abundance of bromine -81 is (100 –Y).
(Y/100 x 79) + ((100 –Y)/100 x 81) = 79.91
0.79Y + 81.00 – 0.81Y = 79.91
1.09 = 0.02Y
Y = 54.5
Hence the % abundance of bromine -79 is 54.5% and the % abundance of bromine -81 is 45.5%.
6
13.
14.
15.
16.
17.
Silver has a relative mass of 108. How does the mass of a silver atom compare with the mass of
a carbon-12 atom?
Each silver atom is 9 times the mass of a carbon -12 atom.
Copper atoms are 5.3 times the mass of a carbon -12 atom. What is the relative atomic mass of
copper?
The relative atomic mass of copper must be 5.3 x 12.0 = 63.6
A titanium atom is four times the mass of a carbon-12 atom. What is the relative mass of this
titanium atom?
The relative atomic mass of this titanium atom must be 4 x 12.0 = 48.0
What is meant by the term “atomic weight of an atom”?
The weighted average of the mass numbers of the i sotopes of that element or the average mass
of the isotopes of that element compared with the mass of a carbon -12 atom.
What is meant by the “molecular mass of a substance”?
The average mass of a molecule compared with the mass of a carbon -12 atom.
18.
What is the relationship between the molecular mass and the atomic masses of the elements in a
molecule?
The molecular mass is the sum of the atomic masses of the atoms making up the molecule.
19.
Why shouldn’t we use the term molecular mass for ionic compounds?
Ionic compounds do not exist as molecules. Ionic substances exist as 3 -D crystalline structures
not as molecules, which are small discrete units held together by covalent bonds between nonmetal atoms. The term formula weight, being the sum o f the atomic weights of the atoms
represented in the simplest formula (the empirical formula) is used for ionic compounds.
20.
Use the values of atomic weight from the PT, to determine the molecular weights or formula
weights of the following: (Note: Use Atomic weights as per BoS Periodic Table)
CH4
12.01 + 4(1.008) = 16.04
CO2
12.01 + 2(16.00) = 44.01
CuSO4.
5H2O
Al2(CO3)3
H2 O
2(1.008) + 16.00 = 18.02
PbO
NaCl
22.99 + 35.45 = 58.44
BaSO4
Na2CO3
2(22.99) + 12.01 + 3(16.00)
= 105.99
40.08 + 2(19.00) = 78.08
NaHCO3
3(40.08) + 2(30.97
+8(16.00)) = 310.18
C12H22O11
CaF2
Ca3(PO4)2
P 2 O5
63.55 + 32.07 + 4(16.00) +
5(18.02) = 249.72
2(26.98) + 3(12.01 + 48.00)
=233.99
207.2 + 16.00 = 223.2
137.3 + 32.07 + 4(16.00) =
233.37
22.99 + 1.008 + 12.01 +
48.00 = 84.01
2(30.97) +5(16.00) = 141.94
12(12.01) + 22(1.008) +
11(16.00) = 342.30
7
Calculations Involving Moles, Mass and Avogadro’s Number
1.
Atomic weights are measured relative to carbon -12. The actual weight of a single atom is
minute. For example, the mass of an individual carbon -12 atom is 1.99 x 10-23 grams.
Calculate the number of atoms in 12 .0 grams of carbon-12.
Number of atoms = 12 g / 1.99 x 10 -23 g = 6.03 x 1023 (Actual value is 6.0221415 x 10 23)
This number is called Avogadro’s Number or one mole.
2.
If one mole of carbon-12 atoms has a mass of 12.0 g, what is the ma ss of mole of
magnesium-24 atoms?
24.0 g (twice as heavy as 1 mole of carbon -12 atoms)
3.
What is the mass of one mole of magnesium atoms (the natural mixture of the 3 isotopes)?
24.31 g (using the average atomic mass from the PT)
4.
What is the mass of 1 mole of ch lorine atoms (the naturally mixture of isotopes)?
35.45 g (using the average atomic mass from the PT)
5.
What is the mass of 1 mole of water molecules?
18.02 g (using the molecular mass of water from the previous page)
6.
How many molecules in 1 mole of water?
Avogadro’s No. = 6.022 x 10 23
7.
How many atoms of oxygen in one mole of water molecules?
2 moles of oxygen atoms per mole of water or 1.2 04 x 1024 oxygen atoms
8.
The molecular mass, expressed in grams, is the sum of the molar masses of the atoms making
up that molecule.
Complete the following table : (Note all molar masses recorded to 4 sig figures)
Substance
Mass of 1 mole (molar
mass) in grams
Substance
Mass of 1 mole (molar
mass) in grams
O2
32.00 g
Cl2
70.90 g
CO2
44.01 g
NO2
46.01 g
CH4
16.04 g
N2 O4
92.02 g
N2
28.02 g
H2SO4
98.09 g
8
9.
m = mass = number of moles x molar mass = n x M
Calculate the mass of: (Note number of significant f igures in answer determined by number of
significant figures in each question.)
10.

3.00 mol copper
3.00 x 63.55 g = 190.65 g = 191 g (to 3 sig figures)

2.6 mol water
2.6 x 18.02 g = 46.85 g = 47 g (to 2 sig figures)

7.00 mol carbon dioxide 7.00 x 44.01 g = 308.1 g = 308 g (to 3 s.f.)

0.0040 mol sodium chloride

0.25 mol water

4.0 mol oxygen atoms

0.40 mol oxygen molecules (O2) 0.40 x 32.00 g = 12.8 g = 13 g (to 2 s.f.)

1.00 mol glucose (C 6H12O6)

0.50 mol glucose 0.50 x 180.156 g = 90.078 g = 90 g (to 2 s.f.)

1.5 mol octane (C 8H18)
0.0040 x 58.44 g = 0.2338 g = 0.23 g (to 2 s.f.)
0.25 x 18.02 g = 4.505 g = 4.5 g (to 2 s.f.)
4.0 x 16.00 g = 64 g (to 2 s.f.)
1.00 x 180.156 g = 180 g (to 3 s.f.)
1.5 x 114.224 g = 171.336 g = 1.7 x 10 2 g (to 2 s.f.)
n = number of moles = mass
=
molar mass
m
M
Calculate the number of moles in:

13 g zinc 13/65.41 mol = 13/65.41 mol = 0.1987 mol = 0.20 mol (to 2 s.f.)

15 g sulfur
15/32.07 mol = 0.468 mol = 0.47 mol (to 2 s.f.)

5.0 kg iron
5.0 x 103/55.85 mol = 89.53 mol = 90 mol (to 2 s.f.)

42 g carbon
42/12.01 mol = 3.497 mol = 3.5 mol (to 2 s.f.)

1.6 g oxygen molecules

1.8 g water
1.6/16.00 mol = 0.10 mol (to 2 s.f.)
1.8/18.02 mol = 0.0999 mol = 0.10 mol (to 2 s.f)
9
11.
12.
13.
14.
What is the number of items in 1 mole of anything (the definition of a mole)? 6.022 x 10 23

What is the number of molecules in 2 .0 mole of carbon dioxide?
2 x 6.022 x 1023 = 1.2 x 10 24 (to 2 s.f.)

What is the number of molecules in 0.5 00 mole of oxygen (O 2)?
0.500 x 6.022 x 10 23 = 3.011 x 10 23 = 3.01 x 10 23 ( to 3 s.f.)


How many eggs in a dozen eggs? 12
How many eggs in a mole of eggs? 6.022 x 10 23
Calculate the number of molecules in :
 3.5 mol carbon dioxide
3.5 x 6.022 x 10 23 = 2.1 x 10 24 (to 2 s.f.)

8 mol oxygen (O2)
8 x 6.022 x 10 23 = 4.818 x 1024 = 5 x 1024 (to 1 s.f.)

0.045 mol water
0.045 x 6.022 x 10 23 = 0.27 x 1023 = 2.7 x 1022 (to 2 s.f.)
Calculate the number of formula units in :
 3 mol sodium chloride
3 x 6.022 x 10 23 = 1.8066 x 10 24 = 2 x 1024 (to 1 s.f.)

0.6 mol CuSO4
0.6 x 6.022 x 10 23 = 3.6132 x 10 23 = 4 x 1023 (to 1 s.f.)

2 x 10-3 mol O2
2 x 10-3 x 6.022 x 10 23 = 12.044 x 10 20 = 1.2 x 1021 = 1 x 1021 (to 1 s.f.)
Calculate the number of atoms in:
 12 g carbon
12/12.01 mol = 0.999 mol = 0.999 x 6.022 x 10 23 atoms = 6.0 x 10 23 atoms (to 2 s.f.)

24 g carbon
24/12.01 mol = 1.998 x 10 mol = 1.998 x 6.022 x 10 23 atoms = 1.2 x 10 24 atoms (to 2 s.f.)

1.2 g carbon
1.2/12.01 mol = 0.0999 mol = 0.0999 x 6.022 x 1023 atoms = 6.0 x 1022 atoms (to 2 s.f.)

0.5 g carbon
0.5/12.01 mol = 0.0416 x 6.022 x 10 23 atoms = 0.250 x 10 23 = 3 x 1022 atoms (to 1 s.f.)

10 g of sulfur
10
15.
16.
10/32.07 mol = 0.3118 x 6.022 x 10 23 atoms = 1.9 x 1023 atoms (to 2 s.f.)
Calculate the number of molecules in:

10 g of carbon dioxide
10/44.01 mol = 0.227 x 6.022 x 10 23 molecules = 1.4 x 10 23 molecules (to 2 s.f.)

10 g of chlorine gas (Cl 2)
10/70.90 mol = 0.141 x 6.022 x 10 23 molecules = 8.5 x 10 22 molecules (to 2 s.f.)

10 g of sugar (C 12H22O11)
10/342.3 mol = 0.0292 x 6.022 x 10 23 molecules = 1.8 x 10 22 molecules (to 2 s.f.)
Calculate the mass of:

5.18 mol nitrogen gas
5.18 x 28.02 g = 145.14 g = 145 g (to 3 s.f.)

6 .0 mol hydrogen gas
6.0 x 2.016 g = 12.096 g = 12 g (to 2 s.f.)

8.5 x 10 -3 mol hydrogen gas
8.5 x 10-3 x 2.016 g = 17.136 x 10-3 g = 1.7 x 10-2 g (to 2 s.f.)

0.75 mol argon gas
0.75 x 39.95 g = 29.96 g = 30 g (to 2 s.f.)

6.022 x 1023 atoms of oxygen (O)
16.00 g (1 mole) (to 4 s.f.)

6.022 x 1023 molecules of oxygen gas (O 2)
32.00 g (1 mole) (to 4 s.f.)

5.98 x 1012 molecules of oxygen gas (O 2)
5.98 x 1012/6.022 x 10 23 x 32.00 g = 3.177 x 10 -10 g (to 3 s.f.) = 3.18 x 10 -10 g (to 3 s.f.)

6.022 x 1025 molecules of carbon dioxide (CO 2)
6.022 x 10 25/6.022 x 10 23 x 44.01 g = 4401 g = 4.401 x 10 3 g
11
17.
18.
Calculate the number of molecules in:

7.0 g nitrogen gas (N 2)
7.0/28.02 x 6.022 x 10 23 molecules = 1.5044 x 10 23 molecules = 1.5 x 10 23 molecules (to 2
s.f.)

4.2 g nitrogen dioxide gas (NO 2)
4.2/46.01 x 6.022 x 10 23 molecules = 5.497 x 10 22 molecules = 5.5 x 10 22 molecules (to 2
s.f.)

18.0 g glucose (C6H12O6)
18.0/180.156 x 6.022 x 10 23 molecules = 6.017 x 1022 = 6.02 x 1022 molecules (to 3 s.f.)
Calculate the mass of 1 molecule of:

Chlorine gas (Cl2)
70.90/6.022 x10 23 = 1.7734 x 10-22 g = 1.773 x 10 -22 g (to 4 s.f.)

Ammonia gas (NH3)
17.034/6.022 x 1023 = 2.8286 x 10 -23 g = 2.829 x 10 -23 g (to 4 s.f.)
12
Calculations Involving Percentage Composition by Mass
1.
2.
Calculate the percentage of:
 iron in the compound Fe 2O3
2(55.85)/159.7 x 100 = 69.94%

sulfur in sulfur dioxide SO 2
32.07/64.07 x 100 = 50.05%

sulfur in sodium sulfate Na2SO4
32.07/142.05 x 100 = 22.58%

magnesium in MgCl 2
24.31/95.21 x 100 = 25.53%
Calculate the percentage of each element in:

glucose, C6H12O6
For carbon:
6(12.01)/180.156 x 100 = 40.00%
For hydrogen:
12(1.008)/180.156 x 100 = 6.71%
For oxygen:
6(16.00)/180.156 x 100 = 53.29%

octane, C 8H18
For carbon:
For hydrogen:
8(12.01)/114.224 x 100 = 84.12%
18(1.008)/114.224 x 100 = 15.88%
ethanol, C 2H6O
For carbon:
For hydrogen:
For oxygen:
2(12.01)/46.068 x 100 = 52.14%
6(1.008)/46.068 x 100 = 13.13%
16.00/46.068 x 100 = 34.73%


iron (III) sulfate, Fe2(SO4)3
For iron:
2(55.85)/399.91 x 100 = 27.93%
For sulfur:
3(32.07)/399.91 x 100 = 24.06%
For oxygen:
12(16.00)/399.91 x 100 = 48.01%

sulfuric acid, H2SO4
For hydrogen:
2(1.008)/98.09 x 100 = 2.06%
For sulfur:
32.07/98.09 x 100 = 32.69%
For oxygen:
4(16.00)/98.09 x 100 = 65.25%

urea, CH 4ON2
For carbon:
For hydrogen:
For oxygen:
For nitrogen:
12.01/60.062 x 100 = 20.00%
4(1.08)/60.062 x 100 = 6.71%
16.00/60.062 x 100 = 26.64%
2(14.01)/60.062 x 100 = 46.65%
13
3.
Calculate the percentage of nitrogen, by mass, in TNT, C 7H5N3O6.
Molar mass of TNT = 7(12.01) + 5(1.008) + 3(14.01) + 6(16.00) = 227.14 g
Percentage of nitrogen by mass is:
3(14.01)/227.14 x 100 = 18.50%
4.
Calculate the percentage of copper in the copper ore, chalcopyrite, CuFeS 2.
Molar mass of chalcopyrite = 63.55 + 55.85 + 2(32.07) = 183.54 g
Percentage of copper by mass is:
63.55/183.54 x 100 = 34.62%
5.
6.
Titanium can be extracted from the ores, rutile (TiO 2) and ilmenite (FeTiO 3).

What is the percentage of titanium, by mass, in each of these ores?
For rutile:
% of titanium = 47.87/79.87 x 100 = 59.93%
For ilmenite:
% of titanium = 47.87/151.72 x 100 = 31.55%

What mass of rutile would yield 1.00 tonne of titanium?
Since titanium makes up 59.93% by mass of rutile, then :
100/59.93 x 1.00 = 1.67 tonnes of rutile is needed to produce 1 tonne o f titanium, assuming
100% yield

What mass of ilmenite would yield 1.00 tonne of titanium?
Since titanium makes up 31.55% by mass of ilmenite, then:
100/31.55 x 1.00 = 3.17 tonnes of ilmenite is needed to produce 1 tonne of titanium,
assuming 100% yield

Which ore is likely to produ ce a more cost effective source of titanium?
Rutile – as less rutile ore is needed to produce the same mass of titanium.
It was found that 1.364 g of a particular nickel ore produced 0.0345 g nickel. What is the
percentage yield of nickel from this ore?
Percentage yield nickel = 0.0345/1.364 x100 = 2.53%
14
7.
Magnetite (Fe 3O4) and haematite (Fe 2O3) are 2 ores from which iron is extracted. Compare the
theoretical yields of iron from the 2 ores.
Theoretical yield of Fe from magnetite:
3(55.85)/231.55 x 100 = 72.36%
Theoretical yield of Fe from haematite:
2(55.85)/159.70 x 100 = 69.94%
There is a greater theoretical yield from magnetite.
8.
If the theoretical yield of aluminium from a bauxite ore is 15% and if the extraction process is
90% efficient, what mass of bauxite ore ne eds to be processed to produce 1.00 tonn e of
aluminium?
If yield of aluminium from bauxite is theoretically 15%:
then 100/15 x 1.00 = 6.667 tonne of bauxite needs to be processed to give 1.00 tonne of
aluminium.
If the extraction process is 90% efficient:
100/90 x 6.667 = 7.41 tonne of bauxite will need to be processed to give 1.00 tonne aluminium.
9.
What mass of sodium can be produced from the electrolysis of 1.00 tonne of molten sodium
chloride, if the process is 95% efficient?
% of sodium in NaCl = 22.99/58.44 x 100 = 39.34%
Hence 1.00 tonne of sodium chloride theoretically produce s
39.34/100 x 1.00 tonne sodium = 0.3934 tonne
The process is 95% efficient.
Hence mass of sodium produced = 95/100 x 0. 3934 tonne = 0.374 tonne or 374 kg.
 What mass of chlorine would form as a by -product of this electrolysis reaction?
When NaCl is electrolysed, 35.45 g chlorine is produced per 22.99 g sodium.
Hence mass of chlorine produced per 0.374 tonne of sodium = 35.45 x 0.374/22.99 tonne =
0.576 tonne or 576 kg.
10.
What mass of sulfur is present in 2.0 tonne of iron sulphide, FeS?
Percentage of sulfur in FeS = 32.07/87.92 x 100% = 36.48%
Hence in 2.0 tonne FeS, mass of sulfur = 36.48/100 x 2.0 = 0.73 tonne
15
11.
What mass of sulfate ion is present in 6.58 g of sulfuric acid?
Percentage of sulfate ion in sulfuric acid, H 2SO4 = (32.07 + 4(16.00))/98.086 x 100% =
96.07/98.086 x 100 = 97.94%
Hence mass of sulfate ion in 6.58 g sulfuric acid = 97.94 /100 x 6.58 g = 6.44 g (3 s.f.)
12.
What mass of barium sulfate could be formed from 5.0 g sulfate ions?
Percentage, by mass, of sulfate ions in barium sulfate, BaSO 4 = 96.07/233.37 x 100 = 41.1 66%
Hence mass of barium sulfate formed from 5.0 g sul fate ions = 100/41.166 x 5.0 = 12.15 g
= 12 g (2 s.f.)
13.
What mass of silver chloride could be formed from 1.00 g silver?
Percentage Ag in AgCl = 107.9/143.35 x 100 = 75.27%
Mass of AgCl formed from 1.00 g silver = 100/75.27 x 1.00 g = 1.33 g (2 s.f.)
14.
What mass of chlorine could be formed from 56.78 g FeCl 3?
Percentage of chlorine in FeCl 3 = 3(35.45)/162.2 x 100% = 65.57%
Hence mass of chlorine formed from 56.78 g FeCl3 = 65.57/100 x 56.78 g = 37.23 g (4 s.f.)
15.
What mass of calcium is needed to produce 300.7 g of calcium carbonate?
Percentage of calcium in CaCO 3 = 40.08/100.09 x 100 = 40.04%
Hence mass of calcium needed to form 300.7 g calcium carbonate
= 40.04/100 x 300.7 g
= 120.4 g
16
Molecular and Empirical Formulae
The molecular formula tells you the number of each type of atom present in a compound.
The empirical formula of a compound tells you the ratio of the number of the atoms present.
For example, C 2H6 is the molecular formula for ethane. Its empirical formula is CH 3.
1.
What is the empirical formula for each of the following? In each case, the molecular formula is
shown.
 C 6 H6
CH
 C 2 H4
CH2
 C 3 H6
CH2
 C6H12O6
CH2O
 NO2
NO2
 N2 O4
NO2
 N2 O3
N2 O3
 NO
NO
2.
When 10.24 g lithium metal was allowed to react with excess oxygen , it was completely
converted to lithium oxide. 22.10 g of lithium oxide was formed. Calculate the empirical
formula for lithium oxide.
(Hint: work out how many moles of lithium and how many moles of oxygen atoms are in the
lithium oxide. The ratio of the number of moles is the empirical formula.)
Moles lithium = 10.24/6.941 = 1.475 mol
Mass oxygen = 22.10 – 10.24 g = 11.86 g = 11.86/16.00 mol = 0.741 mol
Moles lithium : moles oxygen = 1.475 : 0.741 = 2:1 approx imately
Hence empirical formula lithium oxide = Li 2O
3.
A compound of carbon, fluorine and chlorine only, weighing 2.37 g, was found to contain
1.36 g chlorine, 0.71 g fluorine and 0.30 g carbon.
Calculate the empirical formula of this compound.
(Hint: work out the number of moles of atoms of each element, then find the simplest ratio of
this number of moles.)
Moles chlorine atoms = 1.36/35.45 = 0.038 mol
Moles fluorine atoms = 0.71/19.00 = 0.037 mol
Moles carbon atoms = 0.30/12.01 = 0.025 mol
Simplifying mole ratio (by diving each by the smallest number of moles)
Chlorine
0.038/0.025 = 1.52
Fluorine
0.037/0.025 = 1.48
Carbon
0.025/0.025 = 1.00
Multiplying by 2 to make whole numbers (as there must be whole numbers of atoms in a
molecular formula)
Chlorine
1.52 x 2 = 3
Fluorine
1.48 x 2 = 3
Carbon
1.0 x 2 = 2
Hence empirical formula is C 2F3Cl3
17
4.
A compound of silver contained 69.2% silver, 10.3% sulfur and 20.5% oxygen. Calculate the
empirical formula of the compound.
Assume you have 100 g of the compound.
Moles silver atoms = 69.2 g/107.9 g = 0.64 mol
Moles sulfur atoms = 10.3 g/32.07 g = 0.32 mol
Moles oxygen atoms = 20.5 g/16.00 g = 1.28 mol
Simplifying mole ratio (by diving each by the smallest number of moles)
Silver
0.64/0.32 = 2
Sulfur
0.32/0.32 = 1
Oxygen
1.28/0.32 = 4
Hence the empirical formula is Ag 2SO4
5.
Magnesium combines with nitrogen gas (N 2) to form magnesium nitride. When 2.55 g
magnesium reacted with excess nitrogen until all the magnesium had r eacted, the mass of
magnesium nitride formed was 3.51 g. Calculate the empirical formula of magnesium nitride.
Mass of nitrogen reacting = 3.51 – 2.55 g = 0.96 g
Moles of nitrogen reacting = 0.96/14.01
= 0.069 mol
Moles of magnesium reacting = 2.55 /24.31 = 0.105 mol
Simplifying mole ratio (by diving each by the smallest number of moles)
Nitrogen
0.069/0.069 = 1.00
Magnesium 0.105/0.069 = 1.52
Multiplying by 2 to get whole numbers of atoms
The empirical formula is Mg 3N2
6.
1.00 g of a compound of copper and chlorine contains 0.524 g chlorine. Calculate the empirical
formula of the compound.
Mass of copper reacting
= 1.00 – 0.524 g = 0.476 g
Moles of copper
= 0.476/63.55 = 0.00749 mol
Moles of chlorine reacting
= 0.524/35.45 = 0.0148 m ol
Simplifying mole ratio (by diving each by the smallest number of moles)
Copper
0.00749/0.00749 = 1.00 = 1
Chlorine
0.0148/0.00749 = 1.97 = 2
Hence the empirical formula is CuCl 2
18
7.
Cyclohexane is a compound of carbon and hydrogen only. An alysis shows that its composition
by mass is 85.7% C and 14.3% H, while its molar mass is 84.0 g per mole. Find the molecular
formula of cyclohexane.
Assume 100 g of cyclohexane
Moles of carbon
= 85.7/12.01 = 7.136 mol
Moles of hydrogen = 14.3/1.008 = 14.19 mol
Simplifying mole ratio (by diving each by the smallest number of moles)
Carbon
7.136/7.136 = 1.0
Hydrogen
14.19/7.136 = 2.0
Hence the empirical formula is CH 2.
Empirical mass (sum of atomic masses of atoms) = 12.01 + 2.016 = 14.02
Since the molar mass/empirical mass = 84.0/14.0 = 6
The molecular formula must contain 6 times the number of each atom in the empirical formula.
Hence molecular formula for cyclohexane is C 6H12
8.
The percentage composition, by mass, of ethylene glycol is 38.71% C, 9.68% H, with the
remainder being oxygen. One mole of ethylene glycol has a mass of 62 g. Determine the
molecular formula of ethylene glycol.
Assume 100 g of ethylene glycol
Mass of oxygen must be 100 – (38.71 + 9.68) g = 51.61 g
Moles of oxygen
= 51.61/16.00 = 3.23 mol
Moles of carbon
= 38.71/12.01 = 3.22 mol
Moles of hydrogen = 9.68/1.008 = 9.603 mol
Simplifying mole ratio (by diving each by the smallest number of moles)
Oxygen
3.23/3.22
= 1.00
Carbon
3.22/3.22
= 1.00
Hydrogen
9.68/3.22
= 3.01
Empirical formula of ethylene glycol is CH 3O
Empirical mass = 12.01 + 3.024 + 16.00 = 31.03
Since the molar mass/empirical mass = 62/31 = 2
The molecular formula must be 2 times the empirical formula
Hence the molecular formula is C 2H6O2
19
9.
Glycerol is a compound containin g carbon, hydrogen and oxygen. 12.88 g of glycerol is found
to contain 5.04 g carbon, 1.12 g of hydrogen and oxygen. What is the empirical formula of
glycerol? If 1.8 x 10 22 molecules of glycerol have a mass of 2.7 6 g, what is the molecular
formula of glycerol?
Mass of oxygen must be 12.88 – (5.04 + 1.12) g = 6.72 g
Moles of oxygen
= 6.72/16.00 = 0.420 mol
Moles of carbon
= 5.04/12.01 = 0.420 mol
Moles of hydrogen = 1.12/1.008 = 1.111 mol
Simplifying mole ratio (by diving each by the smallest number of moles)
Oxygen
0.420/0.420 = 1.0
Carbon
0.420/0.420 = 1.0
Hydrogen
1.111/0.420 = 2.65
Multiply each by 3 to get whole number ratio.
Oxygen
3.0
Carbon
3.0
Hydrogen
7.9 (assume this is close enough to 8 )
Empirical formula of glycerol is C3H8O3
Empirical mass = 36.03 + 8.06 + 48.00 = 92.09 = 92 (2 s.f.)
If 1.8 x 10 22 molecules have a mass of 2.76 g
Then 6.022 x 10 23 molecules (1 mole) have a mass of 2.76 x 6.022 x 10 23/1.8 x 1022 = 92 (2 s.f.)
Since the molar mass/empirical mass = 1
Then molecular formula must be the same as the empirical formula
Hence the molecular formula is C3H8O
10.
Analysis of an oxide of nitrogen shows its percentage composition, by mass, to be 30.43%
nitrogen. 1.5 x 10 23 molecules of the oxide have a mass of 23.0 g. Determine the molecular
formula of the nitrogen oxide.
Assume 100 g of the compound.
Mass of oxygen
= 100.0 – 30.43 = 69.57 g
Moles of oxygen
= 69.57/16.00 = 4.3 mol
Moles of nitrogen
= 30.43/14.01 = 2.2 mol
Simplifying mole ratio (by diving each by the smallest number of moles)
Hence empirical formula is NO 2
Since 1.5 x 10 23 molecules have a mass of 23.0 g
Then 1 mole, (6.02 x 10 23 molecules) has a mass of 4 x 23.0 = 92.0 g
Empirical mass = 14.01 + 32.00 = 46.0 g
The molecular mass/empirical mass = 2
Hence the molecular formula must be N 2O4
20
Moles and Chemical Equations
When hydrogen and oxygen reac t together, water is formed:
2H2 (g) + O2 (g)  2H2O (l)
This means:
2 molecules of hydrogen gas (H 2) react with 1 molecule of oxygen gas (O 2) to form 2 molecules of
water.
It also means:
2.00 mol of hydrogen gas (H 2) react with 1.00 mol of oxygen gas (O 2) to form 2.00 mol of water.
It also means:
2 x 6.022 x 1023 molecules of hydrogen gas (H2) react with 1 x 6.022 x 1023 molecules of oxygen gas
(O2) to form 2 x 6.022 x 1023 molecules of water.
It also means:
2 x 2.016 (= 4.032 g) of hydrogen gas (H 2) react with 1 x 32.00 g of oxygen gas (O 2) to form
2 x 18.016 (= 36.03 g) of water.
1.
Use this information to calculate the:

mass of water that can be formed when 5 .60 g of hydrogen gas burns completely in oxygen.
If 4.032 g hydrogen gas produces 36.0 3 g water
Then 5.60 g hydrogen gas produces 5.60/4.032 x 36.0 3 = 50.0 g water (to 3 s.f.)

moles of oxygen reacting with 20 .0 g of hydrogen.
If 4.032 g hydrogen reacts with 1.00 mol oxygen
Then 20.0 g hydrogen reacts with 20.0/4.032 x 1.00 = 4.96 mol oxygen (to 3 s.f.)

molecules of hydrogen needed to form 3.6 0 g water.
If 36.032 g water is produced from 2 x 6.022 x 10 23 molecules hydrogen
Then 3.60 g water is produced from 3.60/36.0 32 x 12.044 x 10 23 = 1.20 x 10 23 molecules of
hydrogen (to 3 s.f.)

moles of water formed from 1.2 5 x 1024 molecules of oxygen (O 2).
If 6.022 x 10 23 molecules of oxygen forms 2.00 moles of water
Then 1.25 x 10 24 molecules of oxygen forms 1.25 x 10 24/6.022 x 10 23 x 2.00 = 4.15 mol water
(to 3 s.f.)

mass of oxygen needed to produce 50 .0 moles of water.
If 2.00 mol water is formed from 32.00 g oxygen
Then 50.0 mol water is formed from 50.0/2.00 x 32.00 = 800 g oxygen (to 3 s.f.)
21
2.
Nitrogen (N2) and hydrogen (H2) react to form ammonia (NH 3).
 Write the balanced equation for this reaction.
N2 (g) + 3H2 (g)  2NH3 (g)
 How many moles of hydrogen are needed to form 40 moles of ammonia?
2 mol ammonia form from 3 mol of hydrogen
Hence 40 mol ammonia form from 40/2 x 3 = 60 mol hydrogen
 How many grams of hydrogen are needed?
2 mol of ammonia form from (6 x 1. 008) = 6.048 g hydrogen
40 mol of ammonia form from 40/2 x 6.048 = 120.96 g hydrogen = 1.2 x 10 2 g (to 2 s.f.)
 How many grams of nitrogen are needed?
2 mol ammonia form from 2 x 14.01 g nitrogen
40 mol ammonia form from 40/2 x 2 x 14.01 = 560 g nitrog en = 5.6 x 102 g (2 s.f.)
3.
250 g of limestone (calcium carbonate) is heated strongly to form calcium oxide and carbon
dioxide.
 Write a balanced equation for this reaction.
CaCO3 (s)  CaO (s) + CO2 (g)
 Calculate the number of moles of carbon dioxide formed.
Molar mass of calcium carbonate = 40.08 + 12.01 + 3(16.00) = 100.09
100.09 g calcium carbonate forms 1.00 mol carbon dioxide
Hence 250 g calcium carbonate forms 250/100.09 x 1. 00 mol = 2.50 mol carbon dioxide (3 s.f.)
 Calculate the mass of calcium oxide which is formed.
100.09 g calcium carbonate forms (40.08 + 16.00) = 56.08 g calcium oxide
250 g calcium carbonate forms 250/100.09 x 56.08 = 140 g calcium oxide (3 s.f.)
4.
The metal zinc reacts with hydrochloric acid to produce hydrogen gas.
 Write the balanced equation for this reaction.
Zn (s) + 2HCl (aq)  ZnCl2 (aq) + H2 (g)
 Calculate the mass of zinc required to produce 7.5 g hydrogen.
2 x 1.008 g hydrogen forms from 1 mol Zn (65.41 g Zn)
Hence 7.5 g hydrogen forms from 7.5/2.016 x 65.41 = 243.3 g Zn = 2.4 x 10 2 g (2 s.f.)
22
5.
Sodium reacts with water to form hydrogen and sodium hydroxide solution.
 Write the balanced equation for this reaction.
2Na (s) + 2H2O (l)  2NaOH (aq) + H2 (g)
 What mass of water reacts with 1.32 g sodium?
2 mol Na reacts with 2 mol water
2 x 22.99 g Na reacts with 2 (16.00 + 2.016) g water
45.98 g Na reacts with 36.032 g water
Hence 1.32 g Na reacts with 1.32/45.98 x 36.032 g water = 1.03 g water (to 3 s.f.)
 What mass of hydrogen is formed?
45.98 g Na forms 2.016 g hydrogen
Hence 1.32 g Na forms 1.32/45.98 x 2 .016 g = 0.0579 g hydrogen (3 s.f.)
 How many moles of sodium hydroxide are formed?
45.98 g Na forms 2.00 mol NaOH
Hence 1.32 g Na forms 1.32/45.98 x 2.00 mol NaOH = 0.574 mol NaOH
 How many moles of hydroxide ions are formed?
Since 0.574 mol of NaOH form s, then the same number of moles of hydroxide ions form.
6.
Sulfur dioxide, SO 2, reacts with oxygen to form sulfur trioxide, SO 3.
 Write the balanced equation for this reaction.
2SO2 (g) + O2 (g)  2SO3 (g)
 How many moles of oxygen are needed to f orm 0.60 moles of sulfur trioxide?
2 mol SO 3 form from 1 mol O 2
Hence 0.60 mol SO3 forms from 0.60/2 x 1 mol = 0.30 mol oxygen (2 s.f.)
 How many grams of sulfur dioxide will have reacted?
0.60 mol sulfur trioxide forms from 0.6 0 mol sulfur dioxide
0.60 mol SO2 has a mass of 0.60 x (32.07 + 16.00 + 16.00) = 38.44 g = 38 g (2 s.f.)
 How many molecules of sulfur trioxide are formed?
0.60 mol SO 3 forms.
This contains 0.60 x 6.022 x 10 23 molecules = 3.6 x 10 23 molecules
23
7.
To extract copper from cuprite , Cu2S, the ore was roasted in air. Copper and sulfur dioxide
formed.
What mass of sulfur dioxide wa s produced per tonne of copper by the reaction?
Always write the equation.
Cu2S (s) + O2 (g)  2Cu (s) + SO2 (g)
For 2 mol of copper, 1 mol S O2 is produced
For 2(63.55) = 127.1 g copper, 1(64.07) g SO 2 forms
For 1.000 g copper then 1.000/127.1 x 64.07 g SO 2 = 0.5041 g SO 2 forms (4 s.f.)
For 1.000 tonne copper then 0.5041 tonne sulfur dioxide forms
8.
When a student reacted 2.4 g of magne sium metal in air until no further reaction occurred she
found that the mass of oxide formed was 4.0 g. Determine the:
 number of moles of magnesium atoms that reacted
Moles of magnesium = 2.4/24.31 mol = 0.099 mol (2 s.f.)
 number of grams of oxygen (O 2) that must have reacted with the magnesium
Since mass of oxide = 4.0 g
Then mass of reacting oxygen = 4.0 – 2.4 = 1.6 g
 number of moles of oxygen gas ( O2) that reacted
Moles of oxygen = 1.6/32.00 = 0.050 mol (2 s.f.)
 number of moles of oxygen atoms (O) tha t reacted
Moles of oxygen atoms (O) = 2 x moles of oxygen molecules (O 2)
Hence mol O = 2 x 0.050 = 0.10 mol (2 s.f.)

ratio of the no. of moles of magnesium atoms to the number of moles of oxygen atoms in
the oxide
Mg : O = 0.099 mol : 0.10 mol = 1:1

the empirical formula of the oxide
MgO
24
Calculations Involving Limiting Reagents
1.
Water is formed when 6.0 g of hydrogen and 40.0 g of oxygen are mixed and the reaction
initiated by an electric spark. Determine which of hydr ogen and oxygen is the limiting reagent
and hence calculate the mass of water produced.
2H2 (g) + O2 (g)  2H2O (l)
2 mol H2 reacts exactly with 1 mol O 2
Hence 4.032 g hydrogen reacts exactly with 32.00 g O 2
6.0 g hydrogen would require 6.0/ 4.032 x 32.00 g oxygen = 47.62 g oxygen
Only 40.0 g oxygen is present, so oxygen is the limiting reagent in that i t determines how much
water can form.
Since 32.00 g oxygen forms 36.032 g water
40.0 g oxygen forms 40.0/32.00 x 36.032 g water = 45.0 g (to 3 s.f.)
2.
A mixture of 60 g of calcium oxide and 100 g of ammonium chloride is warmed and the
following reaction occurs.
CaO + 2NH 4Cl  CaCl2 + 2NH3 + H2O
Calculate the mass of ammonia formed.
1 mol CaO reacts exactly with 2 mol NH 4Cl
Hence 56.08 g Ca reacts exactly with 2(14.01 + 4.032 + 35.45) = 106.98 g NH 4Cl
60 g CaO would require 60/56.08 x 106.98 g NH 4Cl = 114 g NH 4Cl
Only 100 g NH 4Cl is present, so NH 4Cl is the limiting reagent in that it determines how much
ammonia can form.
106.98 g NH4Cl produce 2(14.01 + 3.024) = 34.07 g NH 3
Hence 100 g NH4Cl produces 100/106.98 x 34.07 g NH 3 = 31.8 g NH3 (to 3 s.f.)
25
3.
When 24.8 kg of hydrogen sul fide gas is mixed with 16.6 kg of sulfur d ioxide gas, the
following reaction occurs.
SO2 + 2H2S  2H2O + 3S
What mass of sulfur does the reaction produce?
1 mol SO 2 reacts exactly with 2 mol H 2S
Hence (32.07 + 32.00) g SO 2 reacts exactly with 2(2.016 + 32.07) g H 2S
64.07 g SO 2 reacts exactly with 68.17 g H 2S
16.6 kg SO2 would react with 16.6/64.07 x 68.17 = 17.66 kg H2S
In this reaction hydrogen sul fide is in excess as 24.8 kg is present, so sulfur dioxide is the
limiting reagent and determines what mass of sulfur will form.
1 mole SO 2 produces 3 moles sulfur
64.07 g SO 2 produces 3(32.07) = 96.21 g sulfur
16.6 g SO 2 produces 16.6/64.07 x 96.21 = 24.93 g sulfur
Hence 16.6 kg SO 2 produces 24.93 kg sulfur.
4.
Consider the reaction
Al2(SO4)3 + 3BaCl 2  3BaSO4 + 2AlCl 3
For a reaction involving 15.0 g of aluminium sulfate and 30.0 g of barium chloride, calculate:
 the mass of aluminium chloride produced
Molar masses:
Al2(SO4)3
= 2(26.98) + 3(32.07 + 64.00) = 342.17 g
BaCl2
= 137.3 + 2(35.45) = 208.2 g
AlCl3
= 26.98 + 3(35.45) = 133.33 g
In the reaction,
1 mol Al 2(SO4)3 reacts with 3 mol BaCl 2
342.17 g Al 2(SO4)3 reacts with 3 x 208.2 = 624.6 g BaCl 2
15.0 g Al 2(SO4)3 would react exactly with 15.0/342.17 x 624.6 g BaCl 2 = 27.38 g BaCl 2
Hence there is excess BaCl 2 available and Al 2(SO4)3 is the limiting reagent.
1 mol Al 2(SO4)3 forms 2 mol AlCl 3
342.17 g Al 2(SO4)3 forms 2(133.33) g AlCl 3 = 266.7 g AlCl 3
15.0 g Al 2(SO4)3 would form 15.0/342.17 x 26 6.7 g AlCl 3 = 11.7 g AlCl3 (3 s.f.)

the mass of unreacted chemical remaining on completion of the reaction
The unreacted chemical is BaCl 2.
Excess BaCl 2 = 30.0 g – 27.4 g = 2.6 g
26
Volumes of Gases in Reactions
1.
Describe Dalton’s Atomic Theory of Matter .
John Dalton’s Atomic Theory of Matter (1803-1808) stated:
 Matter is composed of tiny indivisible particles called atoms.
 All atoms of one element are identical, but differ from the atoms of all other elements.
 Chemical reactions consist of combining, separating or rearranging atoms in simple whole
number ratios.
His theory led to use of symbols for atoms of elements, formulae and equations.
The theory had limitations as it assumed that gases such as hydrogen and oxygen were
monatomic. As a result, the calculations of atomic weights were incorrect .
2.
State Gay-Lussac’s Law. Use an example to illustrate this law.
Joseph Gay-Lussac’s Law of Combining Gas Volumes stated:
When measured at constant temperature and pressure, the volumes of gases combining or being
formed in a chemical reaction are in a simple whole number ratio .
1 litre of hydrogen gas reacts completely with 1 litre of chlorine gas to form 2 litres of hydrogen
chloride gas. The volume ratio is 1 : 1 : 2.
3.
State Avogadro’s Hypothesis. Use an example to illustrate this hypothes is.
Avogadro put forward an hypothesis that “equal volumes of gases (at the same T and P) contain
the same number of molecules”.
1 molecule of hydrogen reacts with one molecule of chlorine to form 2 molecules of
hydrogen chloride.
4.
Consider the reactions:
 between hydrogen and chlorine to form hydrogen chloride
 between hydrogen and nitrogen to form ammonia gas
Explain what Gay-Lussac would have observed in carrying out these experiments.
1 litre of hydrogen gas reacts completely with 1 litre of chlorine gas to form 2 litres of hydrogen
chloride gas. The volume ratio is 1 : 1 : 2.
3 litres of hydrogen gas reacts with 1 litre of nitrogen to form 2 litres of ammonia gas. The
volume ratio is 3 : 1 : 2.
Explain how Avogadro would have interpreted these observations.
1 molecule of hydrogen reacts with one molecule of chlorine to form 2 molecules of
hydrogen chloride. Hence molecules of hydrogen and chlorine must contain 2 (or multiples of
2) atoms.
1 molecule of nitrogen reacts with 3 molecul es of hydrogen to form 2 molecules of ammonia.
Hence ammonia must contain 1 atom of nitrogen for every 3 atoms of hydrogen.
27
5.
When a gaseous reaction at con stant temperature and pressure occurs, 300 mL of carbon
monoxide reacts with 100 mL of ozone to form 300 mL of carbon dioxide. Assume you do not
know the formulae of carbon monoxide, ozone or carbon dioxide, deduce the number of atoms
of oxygen in ozone.
Volume ratio:
carbon monoxide :
300
:
3
:
Ratio of molecules:
3
:
ozone :
100 :
1
:
1
:
carbon dioxide =
300
3
3
Since the reaction involves 3 molecules incorpor ating 1 molecule of ozone, there must be 3
atoms (or multiples of 3 atoms) of oxygen in the ozone molecule.
6.
What volume of oxygen is required to react with 10.0 L of pentane, C 5H10, both gas volumes
being measured at the same temperature and pressure?
2C5H10 (g) + 15O2 (g)  10CO2 (g) + 10H2O (l)
Since the gases react in molar ratios 2 mol pentane : 15 mol oxygen in the balanced equation,
the volumes will be in the same ratio.
Hence 10.0 L pentane will react with 75.0 L oxygen, at the same T and P.
7.
200 mL of carbon monoxide and 120 mL of oxygen react, producing carbon dioxide. Calculate
the total gas volume at the end of reaction, given that all gas volumes are measured at the same
temperature and pressure.
2CO (g) + O2 (g)  2CO2 (g)
2 mol
1 mol
2 mol
2 vol
1 vol
2 vol
Hence reacting volumes should be:
200 ml
100 mL
200 mL
Since 120 mL of oxygen was present, oxygen was in excess and carbon monoxide is the
limiting reagent.
After all the carbon monoxide reacts, there will be 200 mL of carbon dioxide formed and 20 mL
of oxygen remaining.
Hence a total of 220 mL of gas remaining at the end of the reaction.
28
8.
What volume of chlorine is required to react with 5 mL of methane, assuming all gases are
measured at the same temperature and pressure?
CH4 (g) + 2Cl2 (g)  CH2Cl2 (g) + 2HCl (g)
Methane and chlorine react in 1:2 mole ratio, so the volume ratio will also be 1:2.
Hence 5 mL of methane will react with 10 mL of chlorine, at the same T and P.
9.
What volume of sulfur dioxide is formed when 300 mL of hydrogen sulfide and 500 mL of
oxygen react? All volumes are measured at the same temperature and pressure.
2H2S (g) + 3O2 (g)  2H2O (l) + 2SO2 (g)
Mole ratio:
2
3
2
Volume ratio will also be
2
3
2
300 mL
450 mL
300 mL
Oxygen will be in excess, with hydrogen sulfide the limiting reagent.
300 mL of sulfur dioxide will for m (same number of moles as of hydrogen sul fide, so same
volume).
10.
80 mL of ethyne, C 2H2, and 240 mL of oxygen react to produce carbon dioxide and water.
Calculate the total volume of gas at the completion of the reaction. Assume all measurements
are made at the same temperature and pressure.
2C2H2 (g) + 5O2 (g)  4CO2 (g) + 2H2O (l)
Mole ratio:
2
5
4
2
Volume ratio of gases only:
2
5
4
80 mL
200 mL
160 mL
Hence oxygen is in excess, by 40 mL, and 160 mL of carbon dioxide forms
The total gas volume at the end of reaction = 40 mL + 160 mL = 200 mL
11.
Calculate the total volume of gas remaini ng when 120 mL of propene gas, C 3H6, and 500 mL of
oxygen react to produce carbon dioxide gas and liquid water. (Assume all volumes at 25° C and
1atm pressure).
2C3H6 (g) + 9O2 (g)  6CO2 (g) + 6H2O (l)
Volume ratio of gases only (ignore water):
2
9
6
120 mL 540 mL
360 mL
BUT not enough oxygen is available, so oxygen is the limiting reagent.
111.1 mL 500 mL 333.3 mL
So, at the end, 333.3 mL of carbon dioxide has formed and (120 – 111.1) = 8.9 mL of propene
remain. So total gas at the end = 333.3 + 8.9 = 342 mL
29
Molar Volumes of Gases, Part 1. (Part 2 is covered in Year 12 w ork)
As a result of Avogadro’s work, it was understood that one mo le of any gas would occupy the same
volume at standard conditions of temperature and pre ssure. We now know this value (known as molar
volume of ideal gas) to be:
 24.79 L per mole at 25°C and 1 atm pressure (100 kPa) and
 22.71 L at 0°C and 1 atm pressure (100 kPa).
We can use this information to calculate the volume of gas formed in reactions.
1.
Calculate the volume of hydrogen gas which can be formed from the reaction of 6 .54 g zinc
with hydrochloric acid at 25°C and 1 atm pressure.
 Write the equation for this reaction .
Zn (s) + 2HCl (aq)  ZnCl2 (aq) + H2 (g)

Calculate the number of moles of hydrogen gas formed .
Molar mass zinc = 65.41 g.
Hence 6.54/65.41 = 0.100 mol hydrogen is formed (3 s.f.)

Calculate the volume that this number of moles of hydrogen gas would occupy at 25°C and
1 atm pressure.
0.100 mol of hydrogen (or any ideal gas) occupies 0.100 x 24.79 L = 2.48 L, at 25°C and 1 atm
pressure.
2.
When magnesium reacts completely with sulfuric acid, hydrogen gas is formed. Calculate the
mass of magnesium needed to produce 1.00 L of hydrogen at 25°C and 1 atm pressure.
Mg (s) + H2SO4 (aq)  MgSO4 (aq) + H2 (g)
No. of moles of hydrogen = 1.00/24.79 = 0.0403 mol
Since 1 mol magnesium produces 1 mol hydrogen
Mol Mg required = 0.0403
Mass Mg required = 0.04 03 x 24.31 g = 0.981 g (to 3 s.f.)
3.
When sodium reacts with water, hydrog en gas and sodium hydroxide are formed. Calculate the
volume of hydrogen gas, at 0°C and 1 atm pressure, which could be produced from 3.46 g of
sodium.
2Na (s) + 2H2O (l)  2NaOH (aq) + H2 (g)
2(22.99) g sodium produces 1.00 0 mol hydrogen
Hence 45.98 g sodium produces 22.71 L hydrogen
3.46 g sodium produces 3.46/45.98 x 22.71 L = 1.709 L
30
Calculation of Density
Density is defined by:
Density(D) = mass(m) / Volume (V)
The units used for density are g/mL , g/cm3, kg/L and kg/m 3.
For water, at 25°C, 1 mL = 1 cm 3 and 1000 L = 1 m 3.
Water has an approximate density of 1.00 g/mL at 25°C. (Actually 0.9970 g/mL or 997.04 kg/m 3.)
1.
Calculate the density of an object that has a mass of 420 g and a volume of 52 cm 3.
D = m/V = 420/52 = 8.077 g cm -3
2.
What volume will 300 g of mercury occupy? The density of mercury is 13.6 g mL -1.
V = m/D = 300/13.6 mL = 22.1 mL
3.
A bottle has a mass of 220 g when empty, 380 g when fi lled with water and 351 g when filled
with kerosene. Determine the capac ity of the bottle and the density of kerosene.
Mass of water in bottle = 380 – 220 g = 160 g
Assume that the water has a density of 1 g/mL, hence the capacity of the bottle is 160 mL.
Mass of kerosene in bottle = 351 – 220 g = 131 g
Density of kerosene = m/V = 131/160 = 0.819 g/mL
4.
What is the density of a steel ball which has a diameter of 0.750 cm and a mass of 1.765 g?
(Volume of a sphere is 4/3π r3)
Volume of steel ball = 4/3 x 22/7 x (0.750/2) 3 = 0.221 cm 3
Density of ball = m/V = 1.765 g/ 0.221 cm 3 = 7.99 g/cm 3
5.
Exactly 40.0 mL of water was placed in a long graduated glass tube. The increase in mass of
the tube was 39.92 g. The tube was sealed to prevent evaporation and placed in a freezer
overnight at -2°C. The water froze to form ice. The ice had a volume of 43.5 mL. Calculate
the density of water at room temperature and of ice at -2°C.
Density water at room temperature = m/V = 39.92/40.0 = 0.998 g/mL
Density ice at -2°C = 39.92/43.5 = 0.918 g/ mL
31
Calculation of Concentrati ons of Solutions
1.
The concentration of a solution can be measured in several ways :
 grams of solute per litre of solution
 grams per 100 g solution (% by weight) (w/w)
 volume per 100 mL solution (% by volume) (w/v) or (v/v)
 moles per litre of solution
 ppm (parts per million)
For water, or aqueous solutions, s how that 1 mg/L is the same as 1 ppm
1 mg/L = 10 -3 g/L = 10 -3 g/1000g solution = 1 g/10 6 g solution = 1 g per 1 million grams
= 1 ppm
2.
3.2 g of copper sulfate was dissolved in w ater and the volume made up to 250 mL. Calculate
the concentration of this solution in g/ L and in g/100 mL.
3.2 g/250 mL = (3.2 x 4) g/L = 13 g/L = 1.3 g/100 mL (to 2 s.f.)
3.
What mass of sodium chloride must be dissolved in 250 mL water to make a 0.60% (w/v)
solution? Assume the solution volume is 250 mL.
0.60% means 0.60 g per 100 mL of solution
If 250 mL of solution is prepared, then 2.50 x 0.60 g of sodium chloride is dissolved = 1.5 g (to
2 s.f.)
4.
What volume of ethanol is present in 750 mL of a 13% (v/v) solution of alcohol (a bottle of
red wine)?
13% (v/v) means 13 mL per 100 mL of solution.
In a 750 mL bottle, the volume of ethanol is 750/100 x 13 = 98 mL (to 2 s.f.)
5.
Water in an industrial site was found to be polluted with mercury. The concentration was
reported as 5 ppm. Express this as % (w/w) and as mg/L.
5 ppm means 5 g per 10 6 g solution.
In 100 g solution, mass of mercury ion = 5 x 100/10 6 = 5 x 10 -4 g per 100 g solution
= 0.0005 g
Therefore the solution is 0.0005% (w/w)
0.0005 g/100 g solution = 0.005 g/1000 g solution = 0.005 g/L = 5 x 10-3 g/L = 5 mg/L
32
6.
Most chemists in a laboratory use molarity (moles per litre) as the measure of concentration of
solutions.
Describe how you would prepar e a solution of sodium hydroxide of concentration 0.1 00 mol/L.
If [NaOH] = 0.100 mol/L, then 0.100 moles must be weighed out, using an electronic balance,
and transferred into a 1.00 L volumetric flask. Distilled water is added until the bottom of th e
meniscus is level with the graduation mark on the volumetric flask.
Mass of 1 mole of NaOH is (22.99 + 16.00 + 1.0 08) g = 40.00 g (4 s.f.)
To make a 0.100 mol/L solution, 0.100 x 40.00 g = 4.00 g must be weighed out accurately and
transferred with washings into a beaker and then into the volumetric flask .
7.
Describe how you would dilute the solution in question 6 by a factor of 5. List the apparatus
you would use.
The solution prepared in question 6 was 0.100 mol/L.
To dilute it by a factor of 5:
50.0 mL of the 0.100 mol/L solution would be removed from the original volumetric flask and
transferred into a clean 250 mL volumetric flask. Distilled water would then be added until the
bottom of the meniscus was level with the 250 mL calibration mark on the neck of the flask.
8.
80.0 g of sodium hydroxide was dissolved in water and the solution produced was made up to
1000 mL. What is the theoretical concentration of the solution in mol L -1? (Theoretical
because sodium hydroxide absorbs water and carbon dioxide from the atmosphere as you weigh
it out, so it is impossible to make up a pure solution of known concentration)
No. of moles of sodium hydroxide = 80.0/40.0 = 2.00
c = n/V = 2.00/1.000 = 2.00 mol/L
9.
7.60 g of pure potassium hydroxide was dissolved in water and made up to 500 mL in a
volumetric flask. Calculate the molarity (concentration in mol/L) of the solution.
Molar mass of KOH = 56.11 g
Number of moles of KOH = 7.60/56.11 = 0.135 mol
[KOH] = 0.135 mol/0.5 L = 0.271 mol/L
33
10.
How many moles of NaCl are there in 23.6 mL of 0.145 mol/L sodium chloride solution? How
many grams of sodium chloride would there be in 23.6 ml of this solution?
n = cV = 0.145 x 23.6 /1000 = 0.00342 mol
mass = moles x molar mass = 0.00342 x 58.44 g = 0.200 g (3 s.f.)
11.
Calculate the number of moles of solute present in the following solutions.
 40 mL of 0.60 mol/L ZnCl 2
n = cV = 0.60 x 40/1000 = 0.024 mol (2 s.f.)
 300 mL of 2.0 mol/L KOH
n = cV = 2.0 x 300/1000 = 0.60 mol (2 s.f.)
 5.00 L of 0.12 mol L -1 HCl
n = cV = 0.12 x 5.00 = 0.60 mol (2 s.f.)
12.
Calculate the mass of solute present in the following solutions.
 100 mL of 0.50 mol/L NaCl
n = cV = 0.50 x 100/1000 = 0.050 mol
m = n x M = 0.050 x 58.44 g = 2.9 g (2 s.f.)
 250 mL of 0.25 mol/L H 2SO4
n = cV = 0.25 x 250/1000 = 0.06 25 mol
m = n x M = 0.0625 x 98.082 g = 6.1 g (2 s.f.)
 1.50 L of 4.0 mol L -1 HCl
n = cV = 4.0 x 1.50 = 6.0 mol
m = n x M = 6.0 x 36.46 g = 219 g = 2.2 x 10 2 g (2 s.f.)
13.
Calculate the moles of cations present in each of the following solutions.
 250 mL of 0.060 mol L-1 Li2SO4
moles lithium sulfate = cV = 0.060 x 250/1000 = 0.015 mol
moles lithium ions = 2 x moles of lithium sulfate = 0.030 mol (2 s.f.)
 5.0 mL of 2.0 mol L-1 Ba3(PO4)2
moles Ba3(PO4)2 = cV = 2.0 x 5.0/1000 = 0.010 mol
moles barium ions = 3 x moles Ba 3(PO4)2 = 3 x 0.010 = 0.030 mol (2 s.f.)
 34.7 mL of 0.15 mol/L (NH 4)2SO4
moles (NH 4)2SO4 = cV = 0.15 x 34.7/1000 = 0.005 2 mol
moles ammonium ions = 2 x moles (NH 4)2SO4 = 2 x 0.0052 = 0.010 mol (2 s.f.)
34
14.
Calculate the concentration (in mol/L) of solutions made by dissolving the following amounts
of solute and making up to the stated volumes.
 0.20 mol NaCl in 200 mL
c = n/V = 0.20/(200 x 10 -3) = 1.0 mol/L (2 s.f.)
 0.50 mol KNO 3 in 2.0 L
c = n/V = 0.50/2.0 = 0.25 mol/L (2 s.f.)
 0.67 mol H 2SO4 in 250 mL
c= n/V = 0.67/(250 x 10 -3) = 2.7 mol/L (2 s.f.)
 34.5 g sodium carbonate in 500 mL
n= 34.5/M = 34.5/(22.99 + 22.99 + 12.01 + 4 8.00) = 34.5/105.99 = 0.3255 mol
c = n/V = 0.3255 /0.500 = 0.651 mol/L (3 s.f.)
 5.05 g calcium hydroxide in 5000 mL
n=5.05/M = 5.05/(40.08 + 2(16.00 + 1.008) = 5.05/ 74.10 = 0.682 mol
c = n/V = 0.682/5.000 = 0.0136 mol/L (3 s.f.)
 11.9 g silver nitrate in 200 mL
n=11.9/(107.9 + 14.01 + 48.00) = 11.9/169.9 = 0.0700 mol
c = n/V = 0.0700/0.2 = 0.350 mol/L
15.
How many moles of

calcium chloride

calcium ions

chloride ions
are there in 18.5 mL of 0.0475 mol/L calcium chloride solution.
Number of moles of calcium chloride = cV = 0.0475 x 18.5/1000 = 8.79 x 10 -4 mol
Number of moles of calcium ions = 8.79 x 10 -4 mol (as formula for calcium chloride is CaCl 2)
Number of moles of chloride ions = 2 x 8.79 x 10 -4 mol = 1.758 x 10 -3 mol
35
16.
The graph shows the maximum dissolved ox ygen concentration in water as a function of
temperature at normal atmospheric pressure.
What is the volume of O 2 that can dissolve in 10.0 L of water at 25°C and normal atmospheric
pressure?
From the graph, 8 mg/L is the maximum concentration of dissolved oxygen at 25°C.
Hence the maximum mass of oxygen dissolved in 10.0 L water at 25°C is
10.0 x 8 mg = 80 mg
Moles of oxygen dissolved per 10.0 L = 80 x 10 -3 g/32.00 g = 2.5 x 10-3 mol
At 25°C, 1 mole of any gas occupies 24.79 L
Hence maximum volume occupied by oxygen gas is 2.5 x 10-3 x 24.79 L = 0.062 L = 62 mL
Since the data is only accurate to 1 significant figure, this should be expressed as 6 x 10 1 mL
36
17.
Calculate the concentration of sulfate ions (SO 42-) present in the solution resulting when the
following solutions are mixed.
 20 mL of 0.25 mol/L Na 2SO4
 100 mL of 0.15 mol/L MgSO 4
 200 mL of 0.08 mol/L Al2(SO4)3
 180 mL of 1.50 mol/L H 2SO4
Mol Na2SO4 = cV = 0.25 x 20/1000 = 0.0050 mol
Mol sulfate ions = 0.0050 mol
Mol MgSO 4 = cV = 0.15 x 100/1000 = 0.015 mol
Mol sulfate ions = 0.015 mol
Mol Al2(SO4)3 = cV = 0.08 x 200/1000 = 0.016 mol
Mol sulfate ions = 3 x 0.016 = 0.048 mol
Mol H2SO4 = cV = 1.50 x 180/1000 = 0.270 mol
Mol sulfate ions = 0.270 mol
Total moles of sulfate ions = 0.0050 + 0.015 + 0.048 + 0.270 = 0.306 mol
Total volume = 500 mL
Final concentration of sulfate ions = n/V = 0.3 06/0.5 = 0.612 mol/L = 0.61 mol/L
18.
400 mL of water is added to 600 mL of 0.18 mol L -1 sucrose solution. What is the
concentration of the diluted solution?
Moles of sucrose initially = cV = 0.18 x 600/1000 = 0.108 mol
This number of moles of sucrose is the same in the diluted solution.
Concentration of diluted solution = 0.108 mol/final volume = 0.108 mol /1.000 L = 0.108 mo l/L
19.
What volume of water must be added to 30 mL of 10.0 mol/L HCl solution to produce a 0.1 00
mol/L solution?
Moles HCl initially = 10.0 mol x 0.030 L = 0.300 mol
Volume of dilute solution = 0.300 mol/concentration of dilute solution = 0.300/0.100 = 3.000 L
Volume of water needed to be added = 3.000 L – 0.030 L = 2.970 L = 2,970 mL
20.
40 mL of water is added to 10 mL of 0.25 mol/L Al 2(SO4)3 solution. What is the concentration
of Al3+ in the diluted solution?
Mol Al2(SO4)3 = cV = 0.25 x 10/1000 = 0.0025 mol
In diluted solution (of total volume 50 mL = 0.050 L)
c = n/V = 0.0025/0.050 = 0.050 mol/L
37
Calculations Involving Precipitation Reactions
1.
What mass of lead iodide is formed when 25 mL of a 0.421 mol/ L solution of potassium iodide
is added to a solution containing excess Pb2+ ions?
Pb2+ (aq) + 2I- (aq)  PbI2 (s)
2.00 mol iodide ion produces 1.00 mol PbI2
Moles potassium iodide = cV = 0.421 x 0.025 = 0.01053 mol
Hence mol PbI 2
= 0.0105/2 = 0.00526 mol
Mass PbI2
= 0.00526 x (molar mass PbI 2)
= 0.00526 x 461.0
= 2.42 g
2.
What mass of barium sulfate forms when 3.4 g of barium nitrate reacts completely with excess
of a solution containing sodium sulfate?
Ba(NO3)2 (aq) + Na2SO4 (aq)  BaSO4 (s) + 2NaNO 3 (aq)
1.0 mol
1 mol
M=261.32 g
M=233.37 g
Since 261.32 g Ba(NO 3)2 forms 233.37 g BaSO 4
3.4 g Ba(NO 3)2 forms 3.4/261.32 x 233.37 g = 3.0 g BaSO 4
3.
How many moles of sodium carbonate are needed to produce 6.5 g of calcium carbonate by
precipitation?
Na2CO3 (aq) + Ca2+ (aq)  2Na+ (aq) + CaCO3 (s)
1.0 mol
1.00 mol
Molar mass of CaCO 3 = 100.09 g
Moles CaCO 3 = 6.5/100.09 = 0.065 mol
Hence moles of sodium carbonate needed is 0.065 mol
4.
Calculate the mass of precipitate formed when 50 mL of 0.15 mol/L HCl reacts with 100 mL of
0.10 mol/L AgNO 3 solution.
HCl (aq) + AgNO3 (aq)  AgCl (s) + HNO3 (aq)
1.0 mol
1.00 mol
1.00 mol
Moles HCl
= cV = 0.15 x 0.050 = 0.0075 mol
Moles AgNO 3 = cV = 0.10 x 0.10 = 0.010 mol
Hence AgNO 3 is in excess and HCl is the limiting reagent.
Moles of AgCl formed = moles HCl r eacting = 0.0075 mol
Mass of AgCl precipitate = 0.0075 x (107.9 + 35.45) = 0.0075 x 142.95 = 1.07 g = 1.1 g (2 s.f.)
38
5.
To measure the concentration of magnesium ions in a solution , a chemist took 50.0 mL of the
solution and added sodium hydroxide unti l no further precipitate of magnesium hydroxide
formed.
The precipitate was filtered, dried and weighed and found to have a mass of 1.85 g.

What was the molarity of the magnesium ions in the original solution?
Mg2+ (aq) + 2OH- (aq)  Mg(OH)2 (s)
Molar mass of magnesium hydroxide = (24.31 + 2(16.00 + 1.008)) = 58.33 g
Moles Mg(OH) 2 present = 1.85/58.33 = 0.0317 mol
Moles of magnesium hydroxide = moles of magnesium ions in original solution
Concentration of magnesium ions = n/V = 0.0317/(50.0 x 10 -3) = 0.634 mol/L
 How many grams of magnesium ions w ere there per litre of the original solution?
c = 0.634 mol/L = 0.634 x 24.31 g Mg ions /L = 15.4 g/L
 What was the concentration of magnesium ions in the original solution in g/100 mL?
If c = 15.4 g/L, then c = 1.54 g/100 mL
6.
What is the concentration (in mol L -1) of lead ions in a solution, 25 mL of which produced
3.19 g lead iodide when excess potassium iodide was added to it?
Pb2+ (aq) + 2I- (aq)  PbI2 (s)
1.00 mol lead ion produces 1.00 mol PbI 2
Molar mass PbI2
Moles PbI2
= 461.0 g
= m/M = 3.19/461.0
= 0.00692 mol
Hence moles of lead ion reacting
= 0.00692 mol
Volume of lead ion solution
= 25 mL
Concentration of lead ion
= n/V = 0.00692/0.025 = 0.28 mol/L
39
Calculations Involving Heat
1.
Explain the relationship H = -mCT. How did you use this relationship to determine the
heat energy changes in chemical reactions?
H = enthalpy change = change in hea t content of a system (measured in Joules, or kilojoules)
m = mass of material heated or cooled (measured in grams or kilograms)
C = specific heat capacity of the material , i.e. the amount of heat needed to change the
temperature of 1g or kg of the substance by 1 degree (measured in J g-1 K-1 or J kg–1 K-1).
∆T = change in temperature (positive if the temperature has increased or negative if the
temperature has decreased). It is normally the change in temperature of water that is measured.
If a reaction occurs in aqueous solution, the heat given out by an exothermic reaction can be
assumed to be absorbed totally by water.
Conversely, heat absorbed by an endothermic reaction can be assumed to be given out by the
water.
For an exothermic reaction, heat given out by the reaction is absorbed by water and the enthalpy
change for the exothermic reaction is calculated using the expression:
H = - mCT
where m is the mass of water (grams), C is the specific heat capacity of water (4.18 J g -1K-1)
and T is the change in temperature of the water.
Alternatively, using (from the Board of Studies Data Page) C = 4.18 × 10 3 J kg-1K-1. In this
case mass must be expressed in kg.
2.
The specific heat capacity of water is 4.18 J g -1 K-1, whereas the specific heat capacities of
octane and mercury are much smaller (2.22 and 0.14 J g-1 K-1g respectively).
 How does the heat capacity of a liquid influence its ability to absorb heat or release heat?
Water has a high specific heat capacity, which means that wate r can absorb large quantities of
heat without a large rise in temperature. It also means that water can lose heat without there
being a large change in temperature. This means that oceans and large bodies of water reduce
temperature fluctuations and this benefits living things. A habitat in or near water has less
fluctuation in temperature than a terrestrial habitat.
The liquids like octane and mercury, having smaller heat capacities, do not need to absorb as
much energy for their temperature to rise by a degree or conversely, lose heat more quickly.

How does the low specific heat capacity of metals such as mercury influence the properties
and use of mercury?
Metals like mercury have low specific heats so only a small amount of heat energy is needed for
their temperature to rise by a degree. This means their temperature rises more rapidly than
water when both are exposed to the same amount of heat.
Mercury is used in thermometers since, when the metal absorbs heat , its temperature rises
significantly. The increase in temperature means the metal expands and the mercury moves up
the scale on the thermometer.
40
3.
Calculate the amount of heat needed to raise the temperature of 100 g of water from 16°C to
30°C.
Heat absorbed by water
4.
Given that the specific heat capacity of mercury is 0.14 J g-1K-1, calculate the amount of heat
needed to increase the temperature of 25 g of mercury from 18°C to 55°C.
Heat needed
5.
= mwater x 4.18 × 103 J kg-1 K-1 x T
= 0.100 kg x 4.18 × 10 3 x (30 – 16) = 0.100 x 4.18 x 10 3 x 14
= 5,852 J
= 5.9 kJ (2 s.f.)
= mmercury x 0.14 J g-1 K-1 x (55-18)
= 25 x 0.14 x 37 J
= 129 J
Calculate the change in temperature when 2 50 g of octane absorbs 15 kJ of heat energy. The
specific heat capacity of ethanol is 2.22 J g-1K-1.
Heat absorbed to change temperature = moctane x 2.22 Jg-1 K-1 x T
Hence T = 15000 / (2.22 x 250) = 15000/555 = 27°C
6.
A small heater was used to heat 125 g ethanol from 18.0°C to 23.6°C. The heater provided 1.73
kJ of energy. Calculate the specific heat capacity of ethanol. Assume no heat losses to the
surroundings.
Heat absorbed by ethanol
1.73 x103 J
Cethanol
7.
= methanol x Cethanol x Tethanol
= 125 g x Cethanol x (23.6 – 18.0)
= 1.73 x 10 3/(125 x 5.6)
= 2.47 J g-1K-1
When 10.9 g of sodium hydroxide was dissolved in 200 mL of water at 20.5°C, the temperature
rose to 32.5°C. Calculate the molar heat of solution of sodiu m hydroxide. Use the specific heat
capacity of water as 4.18 J g -1 K-1.
Heat released from 10.9 g NaOH
= mwater x 4.18 x (32.5-20.5)
= 200 x 4.18 x 12.0
= 10032 J
= 10.032 kJ
1 mole of NaOH has a mass of 40.00 g
Hence heat released from 1 mole of NaOH = 40.00/10.9 x 10.032 kJ
= 36.8 kJ
Hence molar heat of solution of NaOH is 36.8 kJ/mol (exothermic)
Hence H = - 36.8 kJ/mol
41
8.
When 5.3 g of calcium chloride was dissolved in 25 0 mL of water at 20.5°C, the temperature
rose to 24.5°C. Calculate the molar heat of solution of calcium chloride. Use the specific heat
capacity of the final solution as 4.18 J g -1 K-1.
Heat released from 5.3 g CaCl 2
= mwater x 4.18 x (24.5-20.5)
= 250 x 4.18 x 4.0
= 4180 J
= 4.2 kJ
1 mole of CaCl 2 has a mass of 111.0 g
Hence heat released from 1 mole of CaCl2 = 111.0/5.3 x 4.2 kJ
= 88 kJ
Hence molar heat of solution of CaCl2 is -88 kJ/mol (exothermic)
Hence H = - 88 kJ/mol
9.
Calculate the rise in temperature when 18.4 g (10 mL) of concentrated sulfuric acid is added to
100 mL of water, both at room temperature. The molar heat of solution of concentrated sulfuric
acid is -90 kJ/mol.
When 1.00 mol concentrated sulfuric acid dissolves in water , 90 kJ energy is released.
Molar mass H 2SO4 = 98.09 g
When 18.4 g concentrated H 2SO4 dissolves, the heat given out is
18.4/98.09 x 90 kJ = 16.9 kJ = 16,900 J
Heat is absorbed by (100 + 10) = 110 mL of solution, assumed to be equivalent to 100 g of
water
H
= msolution x 4.18 x ∆T
∆T
= 16900/(110 x 4.18)
= 37°C
10.
What will be the final temperature when 12.8 g ammonium nitrate is dissolved in 100 mL water
at 18.4°C? The molar heat of solution of ammonium nitrate is +26 kJ/mol. Assume the specific
heat capacity of the final solution as 4.18 J g -1 K-1.
When 1.00 mol ammonium nitrate dissolves in water , 26 kJ is absorbed (endothermic)
Molar mass ammonium nitrate (NH 4NO3) is 80.1 g
When 12.8 g ammonium nitrate dissolves in water, 12.8/80.1 x 26 kJ = 4.15 kJ are absorbed
4.15 x 103 J = 100 x 4.18 x ∆T
∆T
= 4150/(100 x 4.18)
= 9.9°C
If the initial temperature is 18.4°C, the final temperature will be lower, as the reaction is
endothermic, taking heat from the water.
Hence the final temperature is 18.4 – 9.9 = 8.5°C
42
Calculations Involving Hea ts of Combustion
1.
All combustion reactions are exothermic.
What does this mean?
The reaction gives out energy.
How do we show this information in an equation?
Show ∆H as a negative number.
Octane releases 5460 kJ per mole of octane. Rewrite the equation for the combustion of octane
to include the heat term. Ensure that 1 mole of octane is undergoing combustion.
C8H18 (l) +
25
2
O2 (g)  8CO2 (g) + 9H2O (l)
H = - 5460 kJ/mol
2.
When bonds are broken, energy is req uired. When bonds are formed, energy is released.
All combustion reactions are exothermic. Suggest a reason for this in terms of the bonds
broken and bonds formed.
Since combustion reactions are exothermic, the energy released when the bonds in carbon
dioxide and water are formed must be greater than the energy required to break the bonds in
oxygen and the fuel.
e.g. in the burning of octane (petrol) the bonds to be broken are C -C bonds and C-H bonds in
octane and the O=O bonds in oxygen mo lecules. Energy is required to break these bonds. This
energy is called the activation energy for the combustion reaction. Individual atoms of C (g),
H(g) and O(g) are formed. These atoms then recombine but in different arrangements in steps
which form 2 C=O bonds in CO 2(g) and 2 H-O bonds in water (l). The bond-forming steps
release more energy than the bond -breaking steps, so the overall H is –ve.
3.
The heat of combustion always measures the heat given out when one mole of a substance
burns completely with oxygen.
Explain why the heat of combustion of octane is greater than the heat of combustion of
methane.
The heat of combustion of octane is greater than that of methane because there are more
carbon and hydrogen atoms making up octane tha n methane. As a result, in the bond -forming
step, more moles of CO 2(g) and H2O(l) are formed, hence more bonds are made and more
energy is given out.
4.
Why does a candle become smaller as it burns? Identify the changes in state and the chemical
reactions which result in this decrease in mass (and hence size).
Candle wax is a mixture of long -chain hydrocarbons (compounds of carbon and hydrogen
only). When the candle wick is lit, heat is generated. This heat melts and vaporises some of the
hydrocarbons in the vicinity of the wick. The liquid hydrocarbons move up the wick as the
gaseous hydrocarbons are burnt with air and the carbon dioxide and water vapour diffuse into
the air. The burning fuel creates more heat to melt and vaporise more wax, and the process
continues. As a result of the wax being converted into gas and the gas undergoing combustion
with oxygen, the mass of wax and the size of the candle both decrease.
43
5.
An experiment was carried out to measure the heat of combustion of a fuel propanol, C3H8O.
The apparatus used is shown below:
The student found that when 1 .0 g of the fuel had been used up, the temperature of the water
had risen by 15°C. He knew the specific heat capacity of water and was able to calculate the
amount of heat absorbed by the 200 g of water.
 Calculate the amount of heat absorbed by the water .
H = m water x 4.18 x T = 200 x 4.18 x 15 = 12540 Joules = 13 kJ (only 2 s.f. justified)
He made an assumption that all the heat from the propanol had been t ransferred to the water.
 What conclusion would the student have reached about the amount of heat given out by
1.0 g of the propanol?
He would have concluded that 13 kJ had been given out by 1 g of the fuel.

The molar mass of propanol is 60 g. What c onclusion would the student have reached
about the heat of combustion of propanol?
He would have concluded that 60 x 13 kJ = 780 kJ would be given out – i.e. that the heat of
combustion of propanol is 780 kJ/mol.
After completing the experiment the stu dent noticed black soot on the bottom of his beaker.
 What conclusion could he have drawn about the combustion reaction?
The combustion reaction was incomplete; i.e. carbon, rather than carbon dioxide , had been
formed.
The student decided his answer cou ld not be correct because his results were inaccurate. The
teacher told him his experiment was invalid but not inaccurate.
 Clarify the difference between the terms accuracy and validity.
Accuracy is a measure of how the measurements are made; whether t he mass of water and of
the fuel had been measured correctly, whether the temperature change had been measured
correctly.
Validity measures how well the experimental design/method could achieve the desired result.
In this case, the method c ould not guarantee complete combustion, yet the experiment had been
designed to measure the heat given out when complete combustion occurred . The experimental
method could not guarantee that all the heat released from the fuel was transferred to the water,
yet the calculation of results depended on this (invalid) assumption.
The design of the experiment was invalid, as it did not allow the desired result to be achieved.
44
6.
The molar enthalpy change for the reaction of butane with excess oxygen is -2880 kJ/mol.
How much heat is absorbed or released when 1.8 g butane is burnt in excess oxygen?
When 1 mole of butane, C 4H10, of molar mass 58.12 g, burns completely with oxygen , 2880 kJ
of heat energy is released.
When 1.8 g butane undergoes complete combustion , 1.8/58.12 x 2880 = 89 kJ of energy is
released.
7.
Calculate the heat released per gram for the combustion of each of the following fuels. The
heats of combustion for each fuel (in kJ/mol) is shown.
Fuel
hydrogen
ethanol
ethyne
propane
sucrose
octane
Molecular
formula
H2
C 2 H6 O
C 2 H2
C 3 H8
C12H22O11
C8H18
Molar heat of
combustion
(kJ/mol)
285
1360
1300
2220
5650
5460
Heat released per gram
(kJ/g)
285/2.016 = 141
1360/46.07 = 29.5
1300/26.04 = 49.9
2220/44.09 = 50.4
5650/342 = 16.5
5460/114 = 47.9
 Arrange these fuels in decreasing order of heat released per gram.
hydrogen
propane
ethyne
octane
ethanol
sucrose
 Comment on the values for the heats of combustion for the hydrocarbons.
The heats of combustion (on a per mole basis) for the alkanes inc rease as the number of carbons
increase. On a per gram basis, the values for the 3 hydrocarbons, propane, ethyne and octane
are similar, ranging from 47.9 kJ/mol to 50.4 kJ/mol.
 Comment on the values for the heats of combustion of fuels containing oxy gen.
Fuels which contain oxygen have a lower heat of combustion per gram than fuels which do not
contain oxygen. The oxygen component of a fuel does not combine with oxygen from the air,
so does not contribute to the energy released.
45
8.
The heat of combustion of ethanol is 1350 kJ/mole, while the enthalpy change for the
combustion of ethanol is quoted as -1350 kJ/mole.
Explain this difference.
The heat of combustion is defined as the absolute value of the amount of heat released and
hence is quoted as a positive value.
The enthalpy change = H = chemical energy of products – chemical energy of the reactants.
As this measure is a negative quantity (the products store less chemical energy than the
reactants), the overall change is negative. It is an exothermic reaction.
9.
Explain why the heat of combustion of alkanols increa ses as the length of the carbon chain
increases.
As the length of the carbon chain of alkanols increases, there are more carbons and hydrogens
present. As a result, when these carbon and hydrogen atoms recombine with oxygen atoms to
form carbon dioxide and water, more bonds are formed. Note: It is the bond -forming steps (not
the bond-breaking steps) which must be stressed in this answer.
10.
A student wished to determ ine the heat of combustion of ethanol.
He used a spirit burner containing ethanol to heat 250 g of water in a beaker. The temperature
rose from 15°C to 31°C. During combustion, the burner lost 0.90 g in mass, due to the ethanol
burning. Calculate the heat of combustion of ethanol, in kJ mol -1 and in kJ g -1.
Heat absorbed by water = mass
water
x T x 4.18 Joules = 250 x 16 x 4.18 = 16,720 J
= 17 kJ
Assume this heat is given out by 0.90 g ethanol
Heat of combustion ethanol
Molar heat combustion
= 17/0.9 kJ/g
= 19 kJ g-1
= 19 x 46.068 kJ/mol = 874 kJ mol -1
Only 2 significant figures are justified, so
Heat of combustion = 8.7 x 10 2 kJ mol-1
46
11.
A student was investigating the heat of combustion of ethanol. She used an ethanol
burner that had an initial mass of 28.0 g. She lit the burner and placed it under a beaker
containing 500 mL (500 g) of water. After a few minutes she noticed that the water
temperature had risen from 24 °C to 38°C and the burner now weighed 26.5 g.
Assuming that only the water was heated in this experiment, what would be the heat of
combustion for ethanol?
Heat absorbed by water = mass
water
x T x 4.18 Joules = 500 x 14 x 4.18 = 29,260 J
= 29 kJ
Assume this heat is given out by (28.0 – 26.5) = 1.5 g ethanol
Heat of combustion ethanol
Molar heat combustion
= 29/1.5 kJ/g
= 19 x 46 kJ/mol
= 19 kJ g-1
= 874 kJ mol -1
Only 2 significant figures are justified, so
Heat of combustion = 8.7 x 10 2 kJ mol-1
12.
The heat of combustion of ethanol is 1367 kJ mol-1. A student carried out an investigation to
determine the heat of combustion of ethanol. The experimental value he determined differed
from the theoretical value.
Identify ONE reason for this difference in values.
The combustion was incomplete, rather than complete (or heat was lost to the surroundings).
Calculate the theoretical mass of ethanol required to heat 50 mL of water from 21.0°C to 30.0°C.
Energy required to heat water from 21.0°C to 30.0°C
= mwater x 4.18 x ∆T
= 50 x 4.18 x 9
= 1881 Joules
= 1.9 kJ
If 1367 kJ is released from 1 mole (46 g) ethanol
Then 1.9 kJ would be (theoretically) released from 1.9/ 1367 x 46 g = 0.064 g ethanol
47
13.
Consider the table below:
Which of the fuels is most efficient as a source of energy per gram?
Methane
Calculate the heats of combustion of the 4 fuels.
Methane:
Heat of combustion = 55.6 x 16.04 = 892 kJ/mol
Propane:
Heat of combustion = 50.3 x 44.1 = 2218 kJ/mol
Octane:
Heat of combustion = 47.9 x 114 = 5461 kJ/mol
Ethanol:
Heat of combustion = 29.7 x 46.1 = 1369 kJ/mol
14.
Disposable cigarette lighters generally contain butane, C 4H10, as the fuel. A lighter contains 3.0 g
butane. The heat of combustion of butane is 2.88 x 10 3 kJ/mol.
Write the equation for the complete combustion of butane.
C4H10 (g) + 13/2 O2 (g)  4CO2 (g) + 5H2O (l)
If this lighter were used to heat 1.0 kg of water at 20°C until all the butane had been used up, what
would be the final temperature of the water?
If 1.00 mol butane (molar mass = 58.12 g) releases 2.88 x 10 3 kJ
Then 3.0 g butane releases 3.0/58.12 x 2.88 x 10 3 kJ = 149 kJ
∆H = m water x 4.18 x10 3 x ∆T
∆T = 149 x 103 J / (1.00 x 4.18 x 10 3) = 149000/4180 = 35.6°C
Final temperature of water = (20 + 35.6) °C = 56°C
48
15.
Heat absorbed by water = mass
water
x T x 4.18 Joules = 300 x 20.5 x 4.18 = 25,707 J
= 25.7 kJ
Assume this heat is given out by 1.15 g ethanol
Heat of combustion ethanol = 25.7/1.15 kJ/g
= 22.3 kJ g-1
Molar heat combustion
= 22.3 x 46.1 kJ/mol = 1027 kJ mol-1
Only 3 significant figures are justified, so
Heat of combustion = 1.03 x 103 kJ mol-1
16.
Calculate the mass of ethanol that must be burnt to increase the temperature of 210 g of water by
65°C, if exactly half of the heat released by this combustion is lost to the surroundings.
The heat of combustion of ethanol is 1360 kJ mol -1.
Heat absorbed by water (assuming no heat losses)
= mass water x T x 4.18 Joules
= 210 x 65 x 4.18
= 57057 J
= 57 kJ
1360 kJ is released from 46.01 (1 mole) ethanol
57 kJ is released from 57/1360 x 46.01 g ethanol = 1.9 g (if no heat losses had occurred).
If 50 % of the heat had been lost to surroundings, then 2 x 1.9 g = 3.8 g of ethanol must be
burnt.
49
Calculations Involving Fermentation of Glucose
1.
In a fermentation experiment 6.50 g of glucose was completely converted to ethanol and carbon
dioxide.
 What mass of carbon dio xide is produced?
Yeast
C6H12O6 (aq)
2C2H6O (aq) + 2CO2 (g)
When 1 mol glucose is fermented , 2 mol carbon dioxide form s
When 180.2 g glucose is fermented, 88.02 g carbon dioxide forms
When 6.50 g glucose is fermented, 6.50/180.2 x 88.02 g = 3.17 g carbon dioxide forms
 What volume of carbon dioxide is formed at 25°C and 1 atm?
No. of moles of carbon dioxide = 6.50/180. 2 x 2 = 0.0721 mol
Volume of carbon dioxide at 25 °C and 1 atm = 0.0721 x 24.79 L = 1.79 L
 What mass of ethanol is formed?
Mol ethanol = moles carbon dioxide = 0.072 1 mol
Mass ethanol = 0.0721 x 46.01 g = 3.32 g
2.
A student studying the mass change that occurs during fermentation added glucose, water
and yeast to a flask and stoppered the flask with some cotton wool.
The student measured the mass of the flask daily for seven days. The table shows the data
collected.
Calculate the mass of glucose fermented. Explain your reasoning.
Assume fermentation complete after 7 days. (Some evaporation may have occurred. Air
enters the flask as carbon dioxide leaves. These factors are neglected in calculation.)
Assume the loss in mass of flask over 7 days = Mass of carbon dioxide formed = 10.66 g
Moles carbon dioxide formed = 10.66/44.0 1 = 0.2422 mol
Moles glucose fermented = 0.2422/2 = 0.1211 mol
Mass of glucose fermented = 0.1211 x 180.2 g = 21.82 g (4 s.f - justified by the numerical
data but not by the validity of the experiment).
50
Calculations Involving Oxidation-Reduction Reactions
1.
Draw a diagram for the galvanic cell formed using Ag and Cu electrodes.
 Label the anode, cathode, direction of electron flow, direction of ion movement in the salt
bridge.
 Write equations for the half cell reactions and for the overall reaction.
 Predict the voltage of this cell under standard conditions.
electrons
NO3
-
K+
copper
anode
silver
cathode
KNO3 (aq)
Cu(NO3)2 (aq)
AgNO3(aq)
electrolytes
The anode half-reaction is:
Cu (s)
Cu2+ (aq) + 2eThe cathode half-reaction is:
Ag + (aq) + eAg (s)
-0.34 V
+0.80 V
The overall equation is: Cu (s) + 2Ag+ (aq)
Cu2+ (aq) + 2Ag (s)
E = +0.46 V
The voltage of the cell under standard conditions = Eoverall = +0.46 V
2.
A galvanic cell is constructed as follows: Br 2(l)/Br-//Cl2(g)/Cl-.
Use the table of standard potentials to determine the voltage of this cell under standard
conditions.
The anode half-reaction is:
2Br- (aq)
Br2 (l) + 2e-1.08 V
The cathode half-reaction is:
Cl2 (g) + 2e2Cl- (aq)
+1.36 V
The overall voltage of the cell is
Eoverall = +0.28 V
51
3.
Identify the anode in this diagram.
Write the equation for the overall cell reaction and calculate the overall cell voltage (assume
standard conditions).
Lead is the anode. (Lead is more active than silver and so is oxidised (loses electrons) more
readily than silver.)
The anode half-reaction is:
Pb (s)
Pb2+ (aq) + 2e- +0.13 V
The cathode half-reaction is:
Ag + (aq) + eAg (s)
+0.80 V
+
The overall equation is: Pb (s) + 2Ag (aq)
Pb2+ (aq) + 2Ag (s)
E = +0.93 V
4.
A beaker contained 250 mL of a solution of lead (II) nitrate of concentration 0.105 mol/L.
A strip of magnesium of initial mass 2.40 g was placed in the beaker and left to react for 5
minutes. A grey solid formed on the surface of the magnesium. The magnesium was removed
from the beaker and the solution tested to determin e the concentration of lead ions remaining,
by adding excess sulfuric acid to precipitate the lead ions as lead sulfate. The precipi tate was
dried and weighed. The mass of preci pitate was found to be 5.750 g. Determine the final
concentration of the lead (II) nitrate solution and the mass of magnesium which had reacted.
Mg (s) + Pb2+ (aq)
Mg2+ (aq) + Pb (s)
Initial no. of moles of Pb(NO 3)2 = cV = 0.105 x 0.250 = 0.02625 mol
Pb2+ (aq) + SO42- (aq) → PbSO 4 (s)
Molar mass PbSO4 = 303.3 g
No. of moles of PbSO 4 precipitate = 5.750/303.3 mol = 0.01896 mol
No. of moles of Pb 2+ remaining in beaker after reaction with magnesium = 0.01896 mol
Hence no. of moles of Pb 2+ which had reacted with Mg = 0.02625 – 0.01896 = 0.007292 mol
Final concentration of Pb(NO 3)2 solution = 0.01896/0.250 = 0.0758 mol/L (3 s.f.)
Moles of magnesium reacted = moles of Pb 2+ reacted = 0.00729 mol
Mass of magnesium which had reacted = 0.00729 x 24.31 g = 0.177 g (3 s.f.)
52
5.
Students performed the following first-hand investigation.
The beaker initially contained 500.0 mL of 0.025 mol L -1 copper sulfate solution. After several
hours the colour of the solution had become lighter and a reddish deposit had formed on the
surface of the zinc. The reddish deposit was removed from the zinc and dried. The deposit was
found to weigh 0.125 g.
Explain the observations and write a balanced overall equation for the reaction which
occurred.
An oxidation-reduction reaction has occurre d, with transfer of electrons from the zinc to the
copper ions. A reddish deposit of solid copper metal has formed. The solution became lighter
as the concentration of the blue copper ions decreased as the ions were converted into copper.
Zn (s) + Cu2+ (aq)
Zn2+ (aq) + Cu (s)
E = + 1.10 V
Calculate the concentration of the copper sulfate solution remaining after the reaction.
Initial moles of copper sulfate in solution = c x V = 0.025 x 0.5000 mol = 0.0125 mol
Moles of copper formed = 0.125 g/63.55 g = 0.00 1967 moles
Since 1 mole of copper is formed from 1 mole of copper ions,
Moles of copper ions remaining in solution = 0.0125 – 0.001967 = 0.01053 mol
Volume of solution = 500.0 mL
Hence concentration of final solution = 0.01053/0.5000 = 0.021 mol/L (2 s. f.)
53
Molar Volume of Gases, Part 2 (Part 1 in Preliminary Course)
1.
The Data Sheet gives the following information
Volume of 1 mole ideal gas: at 100 kPa and
at 0°C (273.15 K) ...................... 22.71 L
at 25°C (298.15 K) .................... 24.79 L
Calculate the volume occupied by :
3.00 mol of hydrogen gas at 100 kPa and at 0°C
3.00 x 22.71 L = 68.1 L (3 s.f.)
0.75 mol of carbon dioxide at 100 kPa and at 0°C
0.75 x 22.71 L = 17 L (2 s.f.)
3.3 g of carbon dioxide at 100 kPa and at 25°C
No. of moles CO 2 = 3.3/44.01 = 0.75 mol
Volume CO 2 = 0.75 x 24.79 L = 1.9 L (2 s.f.)
2.
S (s) + O2 (g)  SO2 (g)
What volume of SO 2 is produced when 8.4 g of O 2 is combusted with excess su lfur at 100 kPa
and 25oC?
1 mole O 2 (g) forms 1 mole SO 2 (g)
 8.4/32.0 mol O 2 (g) forms 8.4/32.0 mol SO 2 (g)
Volume of 8.4/32.0 mol any gas at 100 kPa and 25 oC = 8.4/32.0 x 24.79 L = 6.5 L (2 s.f.)
3.
Calculate the mass of carbon dioxide formed when 15.6 L of oxygen gas at 100 kPa and 25 oC
is used to completely combust a sample of ethanol (CH 3CH2OH).
(Hint: write a balanced chemical equation first .)
CH3CH2OH (l) + 3O2 (g)  2CO2 (g) + 3H2O (l)
15.6 L oxygen gas is 15.6/24.79 moles at 100 kPa and 25 oC
Since 3 moles O 2 forms 2 moles CO 2, then moles CO 2 = 2/3 x 15.6/24.79 = 0.420 mol
Mass of carbon dioxide = moles x molar mass = 0.420 x 44.1 = 18.5 g
What volume of carbon dioxide will be formed at 100 kPa and 25 oC?
Volume carbon dioxide = moles x 24.79 L = 0.420 x 24.79 = 10.4 L (3 s.f.)
54
4.
A 3.065 L flask contains chlorine gas at 0°C and 100.0 kPa. What mass of gas is in the flask?
1 mol of chlorine occupies 22. 71 L at 0°C and 100.0 kPa
No. mol of chlorine in flask = 3.065/22.71 = 0.1350 mol
m = n x M = 0.1350 x 70.9 g = 9.5 69 g
5.
What volume would 1.66 g carbon dioxide occupy at 25 °C and 100.0 kPa?
1.66 g carbon dioxide contains 1.66/44.01 mol = 0.0377 2 mol
Volume of 0.3772 mol at 25°C and 100.0 kPa is 0.0377 2 x 24.79 L = 0.935 L
6.
When 1.0 g aluminium dissol ves in excess hydrochloric acid, what volume of hydrogen
(measured at 0°C and 100 kPa pressure) is formed?
2Al (s) + 6HCl (aq)  2AlCl3 (aq) + 3H2 (g)
2 mol
3 mol
Mol of Al reacting = 1.0/26.98 mol = 0.03706 mol
Mol of hydrogen formed = 3/2 x 1.0/26.98 mol = 0.05560 mol
Volume of hydrogen formed at 0°C and 100 kPa pressure = 0.05560 x 22.71 L = 1.263 L
= 1.3 L (2 s.f.)
7.
What volume of carbon dioxide, measured at 25°C and 100 kPa, is formed when 0.66 g calcium
carbonate is decomposed to form calcium oxide and carbon dioxide?
CaCO3 (s)  CaO (s) + CO2 (g)
Molar mass CaCO 3 = 100.1 g
100.1 g CaCO 3 forms 1 mol CO 2
0.66 g CaCO 3 forms 0.66/100.1 = 0.006593 mol
Volume of CO 2 = 0.006593 x 24.79 L = 0.1635 L = 0.16 L (to 2 s.f.)
8.
Potassium hydroxide reacts with carbon dioxide to form p otassium carbonate and water. What
volume of carbon dioxide (measured at 0°C and 100 kPa pressure) will 250 mL of 0.850 mol/L
solution of potassium hydroxide absorb?
2KOH (aq) + CO2 (g)  K2CO3 (aq) + H2O (l)
Mol KOH = cV = 0.850 x 0.250 = 0.212 5 mol
Moles CO 2 reacting = 0.2125/2 = 0.1062 5 L
Volume CO 2 reacting = 0.10625 x 22.71 L = 2.413 L = 2.41 L (3 s.f.)
9.
1.22 g of an unknown gas has a volume of 15.0 L at 100 kPa and 25°C.
 Calculate the molar mass of the gas.
No. of mol of gas = 15.0/24.79 = 0.605 mol
0.605 mol has a mass of 1.22 g
1 mol has a mass of 1/0.605 x 1.22 = 2.02 g (3 s.f.)
 Identify the gas.
Hydrogen (H 2)
55
pH calculations
1.
By definition, pH = –log10[H+]. Show mathematically that a change of 1 in pH means a tenfold
change in [H +].
pH is defined as –log10[H+].
If [H+] = 0.1 mol/L
pH = -log10 [H+] = 1.0
+
If [H ] = 0.01 mol/L
pH = 2.0
+
If [H ] = 0.001 mol/L
pH = 3.0
Hence a change of 1 pH unit means a ten fold change in [H +].
2.
Explain the self-ionisation of water and write the equilibrium equation for this ionisation
reaction.
Water is capable both of gaining and donating a proton. As a result , it is amphiprotic and 2
water molecules can react together to form hydronium and hydroxide ions.
H2O (l) + H2O (l)
H3O+ (aq) + OH- (aq)
The molecules are in equilibrium with their ions. The equilibrium lies to the left, with only a
very small fraction of the water molecules ionised at any time.
3.
Use Le Chatelier’s Principle to explain why the concentration of hydroxide ion decreases as
acid is added to water.
Using the equilibrium equation in 2 above, if the hydronium ion concentration in water is
increased by addition of acid, the equilibrium shifts to the left to compensate . As a result, the
concentration of hydroxide ions decreases.
4.
Predict what happens to the hydrogen ion concentration if a base (like sodium hydroxide) is
added to water.
Using the equilibrium equation in 2 above, if the hydroxide ion concentration in water is
increased by addition of base the equilibrium shifts to the left to compensate. As a result, the
concentration of hydronium ions decreases.
5.
In aqueous solutions, at 25 oC, the product of the concentrations of the hydrogen and hydroxide
ions always equals 10 -14.
Write this statement as a mathematic expression.
[H+] x [OH -] = 10-14
6.
By definition pOH = –log10[OH-]. Show mathematically that a change of 1 in pOH means a
tenfold change in [OH-].
pOH is defined as –log10[OH-]
If pOH = 4
[OH-] = 10-4 = 0.0001 mol/L
If pOH = 3
[OH-] = 10-3 = 0.001 mol/L
If pOH = 2
[OH-] = 10-2 = 0.01 mol/L
Hence, a change of 1 unit in pOH means a tenfold change in [OH -].
56
7.
Complete the table showing the relationship between [H+], [OH-], pH, pOH, acidic solution,
alkaline solution, neutral solution at 25 oC.
[H+] (mol/L)
1.0
1.0 x 10 -1
1.0 x 10 -2
1.0 x 10 -3
1.0 x 10 -4
1.0 x 10 -5
1.0 x 10 -6
1.0 x 10 -7
1.0 x 10 -8
1.0 x 10 -9
1.0 x 10 -10
1.0 x 10 -11
1.0 x 10 -12
1.0 x 10 -13
1.0 x 10 -14
[OH-] (mol/L)
1.0 x 10 -14
1.0 x 10 -13
1.0 x 10 -12
1.0 x 10 -11
1.0 x 10 -10
1.0 x 10 -9
1.0 x 10 -8
1.0 x 10 -7
1.0 x 10 -6
1.0 x 10 -5
1.0 x 10 -4
1.0 x 10 -3
1.0 x 10 -2
1.0 x 10 -1
1.0
pH
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
pOH
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Acidic/alkaline/neutral
Acidic
Acidic
Acidic
Acidic
Acidic
Acidic
Acidic
Neutral
Alkaline
Alkaline
Alkaline
Alkaline
Alkaline
Alkaline
Alkaline
When calculating pH values, ensure that you are careful to use the correct number of significant
figures. The rule is:
Determine the number of significant figures of the H3O+ concentration. Express your pH value to
this same number of decimal points; e.g. if a concentration is given to 3 significant figures, the pH
would be reported to 3 decimal place s.
1.
Calculate the pH of a 0.01 mol/L HCl solution.
pH = -log10[H+] = -log10 0.01 = -log10 (1 x 10-2) = 2.0 (1 d.p.)
2.
Calculate the pH of a 0.0 005 mol/L H2SO4 solution.
Assume complete ionisation
[H+] = 0.001 mol/L
pH = -log10 (1 x 10-3) = 3.0 (1 d.p.)
3.
Calculate the H 3O+ concentration of a solution if the pH is 4.0.
[H3O+] = 10-pH = 10-4.0 = 1 x 10-4 mol/L (1 s.f.)
4.
Calculate the pH of a 0.0135 mol/L HCl solution.
Assume complete ionisation
pH = -log10[H+] = -log10 0.0135 = 1.870 (3 d.p.)
57
5.
Calculate the pH of a 0.0 15 mol/L H2SO4 solution.
Assume complete ionisation
[H+] = 0.030 mol/L
pH = -log10[H+] = -log10 0.030 = 1.52 (2 d.p.)
6.
Calculate the H 3O+ concentration of a solution if the pH is 2.50.
[H3O+] = 10-pH = 10-2.50 = 3.162 x 10-3 mol/L = 3.2 x 10 -3 (2 s.f.)
7.
What is the pH of a solution of pure water at 25°C? What is the [H3O+] of pure water at 25 oC?
pH = 7.0
[H3O+] = 1 x 10 -7 mol/L
8.
Calculate the OH - concentration and pOH of a 0.1 mol/L solution of NaOH. Hence determine
the pH of this solution.
If [NaOH] = 0.1 mol/L, the [OH -] = 0.1 mol/L and [H +] = 1 x 10 -14/0.1 = 1 x 10-13 mol/L
pOH = 1.0
pH = 14.0 - 1.0 = 13.0
9.
Calculate the OH - concentration and pH of a 0. 005 mol/L solution of Ba(OH) 2.
Since 1 mole Ba(OH) 2 forms 2 moles OH [OH-] = 0.01 mol/L = 1 x 10 -2 mol/L
pOH = -log10[OH-] = -log10 0.01 = 2.0 (1 d.p.)
pH = 14.0 – 2.0 = 12.0
10.
Calculate the H 3O+, Cl- and OH- concentrations in a 1.0 x 10-2 mol/L solution of HCl.
Since HCl is completely ionised
[H3O+] = 1.0 x 10 -2 mol/L
[Cl-] = 1.0 x 10 -2 mol/L
[OH-] = 1.0 x 10 -12 mol/L
11.
Calculate the H 3O+ and OH- concentrations in a sample of orange juice with a pH of 3.17.
If pH = 3.17, then [H 3O+] = 10-3.17 = 6.8 x 10-4 mol/L (2 s.f.)
pOH = 14.00 – 3.17 = 10.83
[OH-] = 10-10.83 mol/L = 1.5 x 10-11 mol/L (2 s.f.)
58
12.
Phenolphthalein is colourless when the pH of solution is 8.5 or lower and changes to purplepink over the pH range 8.5 – 10.0. By what factor does the hydroge n ion concentration of the
solution decrease over this range?
[H+] at a pH of 8.5 = 10 -8.5 mol/L
[H+] at a pH of 10.0 = 10 -10.0 mol/L
10-8.5/10-10.0 = 101.5 = 31.6
Answer should be expressed to 1 significant figure (as only 1 decimal point in pH).
The hydrogen ion concentration decreases by a factor of 30.
13.
Calculate the pH of a solution containing a mixture of 25.0 mL of 1.50 mol/L HNO 3 and
25.0 mL of 0.550 mol/L KOH.
Moles HNO 3 = 1.50 x 25.0/1000 = 3.75 x 10-2 mol
Moles KOH = 0.550 x 25.0/1000 = 1.38 x 10-2 mol
HNO3 is in excess. All KOH will be neutralised by 1.38 x 10-2 moles of HNO 3 and
2.37 x 10-2 moles of HNO 3 will remain, in 50.0 mL of solution.
Final [H+] = 2.37 x 10-2 /50.0 x 10 -3 = 0.474 mol/L (3 s.f.)
Final pH = -log10 0.474 = 0.324 (3 d.p.)
14.
Hydrochloric acid ionises completely. Calculate the pH of a 0.1 25 mol/L HCl solution.
pH = -log10 0.125 = 0.903 (3 d.p.)
15.
A 0.10 mol/L acetic acid solution is 1.5% ionised (at 25C).
 Are there more molecules o f acetic acid or acetate ions in the resulting equilibrium mixture?
Acetic acid
 Calculate the hydrogen (hydronium) and acetate ion concentration s in this solution.
[H3O+] = [CH3 COO-] = 1.5/100 x 0.10 = 1.5 x 10-3 mol/L (2 s.f.)
 Calculate the pH of a 0.10 mol/L solution of acetic acid.
pH = -log10 (1.5 x 10 -3) = 2.82 (2 d.p.)
59
16.
If a student mixed 20 mL of 1 mol L -1 HCl with 20 mL of 1 mol L -1 NaOH, what would be the
approximate pH of the final solution?
The solution should be neutral as t he moles of HCl = moles NaOH. At 25°C, pH = 7.0
17.
A solution of HCl is changed from a pH of 1 .0 to a pH of 2.0. Determine the change in the
concentration of hydronium ions.
If pH = 1.0, [H3O+] = 1 x 10 -1 mol/L
If pH = 2.0, [H 3O+] = 1 x 10 -2 mol/L
Hence the new concentration of hydronium ions is
18.
1
10
of the original concentration.
HCl is a strong acid. A 0.1 mol L-1 solution of HCl produces a pH of 1. What proportion of
the HCl molecules has been left unconverted to io ns?
0%
(All molecules have been ionised.)
19.
Acetic acid is a weak acid. A 0.1 mol L-1 solution produces a pH of 3.0. What proportion of
the acetic acid molecules remains unconverted to ions?
If pH = 3.0, then [H3O+] = 10-3.0 mol/L = 0.001 mol/L (1 s.f.)
0.1 mol of acid has produced 0.001 mol H 3O+.
1% of the acid molecules have been converted into ions.
99% of acetic acid molecules remain as molecules (unconverted to ions).
20.
A 196 g sample of H 2SO4 is dissolved in 500 mL of water. Write an ionic equation for the
reaction which occurs and determine the H 3O + concentration.
H2SO4 (aq) + 2H2O (l)  2H3O+ (aq) + SO42- (aq)
Moles H 2SO4 = 196/98.1 = 2.00
[H2SO4] = 2.00/0.500 = 4.00 mol/L
[H3O+] = 8.00 mol/L (given H 2SO4 is diprotic and strong and assume completely ionised).
60
21.
Calculate the pH of a 0.02 mol L-1 solution of hydrochloric acid.
pH = -log10[H+] = -log10(0.02) = 1.7 (1 d.p.)
Calculate the pH after 20.0 mL of 0.015 mol L-1 sodium hydroxide is added to 40.0 mL of 0.02
mol L-1 hydrochloric acid. Include a balanced chemical equation in your answer.
NaOH (aq) + HCl (aq)  NaCl (aq) + H2O (l)
Mol NaOH = cV = 0.015 x 20.0/1000 = 0.00030 mol = 3.0 x 10 -4 mol
Mol HCl = cV = 0.020 x 40.0/1000 = 0.00080 mol = 8.0 x 10 -4 mol
All NaOH will be neutralised and ((8.0 x 10 -4) – (3.0 x 10 -4)) = 5.0 x 10 -4 mol HCl remain
Volume of final solution = 60.0 mL
[ H+] = 5.0 x 10 -4/0.0600 = 0.0083 mol L-1 = 8.3 x 10 -3 mol L-1 (2 s.f.)
pH = -log10(8.3 x 10 -3) = 2.08 (2 d.p.)
22.
20 mL of 0.080 mol L–1 HCl is mixed with 30 mL of 0.05 0 mol L–1 NaOH.
What is the pH of the resultant solution?
NaOH (aq) + HCl (aq)  NaCl (aq) + H2O (l)
Mol HCl = cV = 0.080 x 20/1000 = 0.0016 mol = 1.6 x 10 -3 mol
Mol NaOH = cV = 0.050 x 30/1000 = 0.0015 mol = 1.5 x 10-4 mol
All NaOH will be neutralised and 0.1 x 10-3 mol HCl = 1 x 10 -4 mol HCl remains
Volume of final solution = 50.0 mL
[ H+] = 1 x 10-4/0.0500 = 0.002 mol L-1 = 2 x 10-3 mol L-1 (1 s.f.)
pH = -log10(2 x 10-3) = 2.7 (1 d.p.)
61
Calculations Involving Titrations
1.
15.43 g of pure barium hydroxide was dissolved in water and made up to exactly 500 mL in a
volumetric flask. Calculate the concentration of the solution.
Molar mass barium hydroxide ( Ba(OH)2 ) = 171.32 g
Moles Ba(OH) 2 = 15.43/171.32 = 0.09007 mol
c = n/V = 0.09007/0.500 mol/L = 0.180 mol/L (3 s.f.)
2.
What mass of pure sulfuric acid must be dissolved in 100 mL to make a 0.56 mol/L solution?
Molar mass (M) H2SO4 = 98.08 g
n = m/M = cV
Hence m = cVM = 0.56 x 0.100 x 98.08 g = 5.5 g (2 s.f.)
3.
How many moles of HCl are there in 24.3 mL of 0.168 mol/L hydrochloric acid sol ution?
n = cV = 0.168 x 24.3/1000 mol/L = 0.00408 mol = 4.08 x 10 -3 mol (3 s.f.)
4.
25.0 mL of a solution of sodium hydroxide was pipetted into a flask, a few drops of a suitable
indicator were added, and the solution titrated with 0. 205 mol/L sulfuric acid solution from a
burette. 22.4 mL of sulfuric acid was required to reach the equivalence point. Calculate the
concentration of the sodium hydroxide solution in mol/L and g/L.
2NaOH (aq) + H2SO4 (aq)  Na2SO4 (aq) + 2H2O (l)
2 mol
1 mol
n sulfuric acid
n NaOH
[ NaOH]
= cV = 0.205 x 22.4/1000 mol
= 2 x 0.205 x 22.4/1000 = 0.00918 mol
= n/V = 0.00918/0.025 = 0.367 mol/L
Concentration in g/L = 0. 367 mol/L x molar mass NaOH
= 0.367 x (22.99 + 16.00 + 1.01)
= 14.7 g/L
62
5.
6.74 g of anhydrous (without water) sodium carbonate was dissolved in water in a volumetric
flask and the volume made up to 250 mL. 10.0 mL of this solution was pipetted into a conical
flask and titrated with HCl. 20.6 mL was required to reach the endpoint. Calculate the
concentration of the HCl solution.
Na2CO3 + 2HCl  2NaCl + H 2O + CO2
1 mole
2 moles
Moles anhydrous sodium carbonate = 6.74/(45.98 + 12.01 + 48.00)
= 6.74/105.99
= 0.0636 mol
Concentration sodium carbonate = 0.0 636/0.250 = 0.254 mol/L
Moles Na 2CO3 in 10.0 mL of solution
Hence moles of HCl = 2 x 0.00254 mol
= 10.0/1000 x 0.254 mol = 0.00254 moles
= 0.00508 mol
Concentration HCl = n/V = 0.00 508/0.0206 = 0.247 mol/L
This solution was then used to determine the concent ration of an unknown barium hydroxide
solution. 25.0 mL of the barium hydroxide required 27.4 mL HCl for exact neutralisation.
Calculate the concentration of the barium hydroxide solution.
2HCl
2 mol
+
Ba(OH)2  BaCl2 + 2H2O
1 mole
Moles HCl used in titration
Hence moles Ba(OH) 2
Concentration Ba(OH) 2
6.
= cV = 0.247 x 0.0274 = 0.00677 mol
= 0.5 x 0.00677 = 0.00338 mol
= 0.00338/0.025 = 0.135 mol/L
In order to determine the concentration of an HCl solution, a student carefully diluted 10.0 mL
to 250.0 mL, then titrated 10.0 mL of the diluted solution with 0. 185 mol/L sodium hydroxide
solution. 16.5 mL was needed to reach the endpoint. Calculate the concentration of H Cl.
Original concentrated HCl solution was diluted by a factor of 25.
NaOH + HCl  NaCl + H 2O
1 mol
1 mol
Moles NaOH used in titration = cV = 0.1 85 x 0.0165 = 0.00305 mol
Hence moles HCl reacted = 0.003 05 mol
Concentration of diluted HCl = 0.00305/0.0100 mol/L = 0.305 mol/L (to 3 s.f.)
Hence concentration of the original HCl solution = 25 x 0.3 05 mol/L = 7.63 mol/L
63
7.
Vinegar is a solution of acetic acid in water. The concentration of the acetic acid in the vinegar
was determined by titration. The vinegar was systematically diluted by a factor of 5.
(a)
State the equipment needed for this systematic dilution.
Pipette (say 20.0 mL), volumetric flask (100.0 mL)
(b)
25.0 mL of the diluted vinegar was placed in a conical flask. What pi ece of glassware
would be used?
Pipette (25.0 mL)
The dilute acetic acid was titrated with 0.15 0 mol/L NaOH. An endpoint was reached
after a titre of 22.1 mL.
(i)
What indicator would be used for this titration?
Phenolphthalein
(ii)
Calculate the concentration of acetic acid in both the diluted and original
vinegar solutions.
NaOH + CH 3COOH  NaCH3COO + H 2O
1 mole
1 mole
(c)
Moles NaOH used in titration = cV = 0.1 50 x 0.0221 = 0.003315 mol
Hence moles CH 3COOH = 0.003315 mol
Concentration of diluted CH3COOH = 0.003315/0.0250 = 0.1326 mol/L
= 0.133 mol/L (3 s.f.)
Since the original vinegar was diluted by a factor of 5, concentration of original
acetic acid solution = 5 x 0.1 326 = 0.663 mol/L (3 s.f.)
(iii)
(d)
Calculate the concentration of acetic acid in the original vinegar in grams per
100 mL.
0.663 mol/L = 0.663 x (molar mass CH 3COOH) g/L = 0.663 x 60.24 = 39.9 g/L
= 3.99 g/100 mL
Suggest a sequence of steps that would have been done prior to this tit ration to
standardise the sodium hydroxide solution. Include in your explanation reasons why the
sodium hydroxide could not be a primary standard.
Sodium hydroxide cannot be weighed out accurately, as it absorbs water and carbon
dioxide from the atmosphere and hence its purity and mass change when exposed to
air. Hence it cannot be used as a primary standard.
Oxalic acid is a primary standard and can be weighed out accurately.
The concentration of an oxalic acid solution is determined by calculation
(moles/volume).
The oxalic acid can then be titrated against the sodium hydroxide.
25.0 mL of the sodium hydroxide solution is pipetted into a conical flask and the oxalic
acid titrated from the burette until the phenolph thalein indicator changes from pink to
colourless.
Oxalic acid is diprotic, so 2 moles of NaOH are needed for 1 mole of oxalic acid.
The concentration of the NaOH is determined by calculation.
64
8.
To determine the mass of calcium carbonate (CaCO3) in an antacid tablet, the following
procedure was used:

The tablet was crushed and then placed in a beaker.

A pipette was used to add 25.0 mL of 0.1 20 mol L-1 hydrochloric acid to the crushed tablet
in the beaker.

Once the reaction between the calcium carbonate a nd hydrochloric acid had stopped,
phenolphthalein indicator was added to the reaction mixture.

A burette was then used to add 0.05 60 mol L-1 sodium hydroxide to the beaker to neutralise
the excess hydrochloric acid.

The phenolphthalein changed from colourless to pink after 27.4 mL of the sodium
hydroxide solution had been added.
(a)
Write a balanced chemical equation for the reaction that occurred between the calcium
carbonate in the tablet and the hydrochloric acid.
CaCO3 (s) + 2HCl (aq)  CaCl2 (aq) + CO2 (g) + H2O (l)
(b)
How many moles of hydrochloric acid were added to the tablet?
Moles HCl added to the tablet = c V = 0.120 x 0.0250 mol = 0.00300 mol
(c)
Explain why the solution was colourless after the phenolphthalein w as first added.
The solution had an excess of hydrochloric acid and phenolphthalein is colourless in
acidic solutions (in any solution below pH of 8) .
(d)
Explain why the colour of the solution changed to pink after 27.4 mL of the sodium
hydroxide solution had been added.
The excess hydrochloric acid had been just neutralised by the sodium hydroxide. With
one extra drop, the solution was now alkaline. Hence the phenolphthalein turned pink .
(e)
Calculate the mass of calcium carbonate in the original antacid tablet.
Moles of NaOH required to neutralise the excess acid = 0.05 60 x 0.0274 = 0.00153 mol
Hence moles of excess hydrochloric acid = 0.00153 mol
Hence moles of HCl used to react with tablet = 0.00300 – 0.00153 = 0.00147 mol
Since 2 moles of HCl react with 1 mole of CaCO 3 (molar mass 100.1 g)
Moles CaCO 3 in original tablet = 0.5 x 0.001 47 = 0.000735 mol = 0.000735 x 100.1 g
=0.0736 g (3 s.f.)
65
9.
The nitrogen content of bread was determined using the following procedure:

A sample of bread weighing 2. 6 g was analysed.

The nitrogen in the sample was converted into ammonia.

The ammonia was collected in 50.0 mL of 0.12 4 mol L-1 hydrochloric acid. All of the
ammonia was neutralised, leaving an excess of hydrochloric acid.

The excess hydrochloric acid was titrated with 2 8.5 mL of 0.112 mol L-1 sodium hydroxide
solution.
(a)
Write balanced equations for the TWO reactions involving hydrochloric acid.
HCl (aq) + NH3 (g)  NH4Cl (aq)
HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)
(b)
Calculate the moles of excess hydrochloric acid.
Moles of HCl added to the ammonia = cV = 0.124 x 0.0500 = 0.00620 mol
Moles of NaOH used to react with excess HCl = cV = 0.112 x 0.0285 = 0.00319 mol
Hence moles of excess HCl = 0.00319 mol (1:1 mol ratio in equation)
(c)
Calculate the moles of ammonia.
Moles of HCl reacted with NH 3
Moles of ammonia
(d)
= Initial moles HCl – moles excess HCl
= 0.00620 – 0.00319
= 0.00301 mol
= 0.00301 (1:1 mole ratio in equation)
Calculate the percentage by mass of nitrogen in the bread.
Moles of nitrogen = moles of ammonia produced
Mass of nitrogen atoms = n x 14.01 = 0.003 01 x 14.01 = 0.0422 g
Percentage by mass of nitrogen in bread sample
= 0.0422/2.6 x 100 = 1.62%
= 1.6% (2 s.f.)
66
10.
Phosphorus pentoxide reacts with water to form phosphoric acid according to the following
equation.
P2O5 (s) + 3H2O (l)  2H3PO4 (aq)
Phosphoric acid reacts with sodium hydroxide according to the following equation.
H3PO4 (aq) + 3NaOH (aq)  Na3PO4 (aq) + 3H2O (l)
A student reacted 2.57 g of phosphorus pentoxide with excess water.
What volume of 0.215 mol L-1 sodium hydroxide would be required to neutralise all the
phosphoric acid produced?
Mol P2O5 = 2.57/141.94 = 0.0181 mol
Mol phosphoric acid produced from reaction with water = 2 x 0. 0181 = 0.0362 mol
Mol of sodium hydroxide required for complete neutralisation = 3 x 0.0362 = 0.109 mol
Volume of NaOH = n/c = 0.109/ 0.215 = 0.505 L
11.
Citric acid, the predominant acid in lemon juice, is a triprotic acid. A student titrated
25.0 mL samples of lemon juice with 0. 660 mol L-1 NaOH. The mean titration volume was
39.5 mL. The molar mass of citric acid is 192.12 g mol -1.
What was the concentration of citric acid in the lemon juice?
H3X + 3NaOH  3H2O + Na3X
(where citric acid, since it is triprotic , is represented as H 3X)
Average mol of NaOH used = cV = 0.660 x 39.5/1000 = 0.0261 mol
Average mol of citric acid in the sample = 0.0261/3 = 0.00869 mol (1:3 ratio in equation)
Concentration citric acid in lemon juice = n/V = 0.00869/0.0250 = 0.348 mol L -1
67
12.
A standard solution was prepared by dissolving 1.314 g of sodium c arbonate in water. The
solution was made up to a final volume of 250.0 mL.
Calculate the concentration of the sodium carbonate solution.
Molar mass sodium carbonate
= 105.99 g
Mol Na2CO3 = 1.314/105.99
= 0.01240 mol
[Na2CO3)] = 0.01240/0.250
= 0.04959 mol/L
This solution was used to determine the concentration of a solution of hydrochloric acid.
Four 25.00 mL samples of the acid were titrated with the sodium carbonate solution. The
average titration volume required to reach the end point was 23.45 mL.
Write a balanced equation for the titration reaction.
Na2CO3 (aq) + 2HCl (aq)  2NaCl (aq) + H2O (l) + CO2 (g)
Calculate the concentration of the hydrochloric acid solution .
Mol Na2CO3 = cV = 0.04959 x 23.45/1000 = 0.001163 mol
Mol HCl = 2 x Mol Na 2CO3 (from balanced equation)
Mol HCl = 0.002326 mol
[HCl] = n/V = 0.002326/0.02500 = 0.9304 mol/L
13.
Aspirin is a weak monoprotic acid with a formula of HC 9H7O4.
To determine the amount of aspirin in a headache tablet, a chemist ground up the tablet and
dissolved it in 25.0 mL of 0.125 mol/L sodium hydroxide solution.
After complete reaction, the left -over hydroxide was titrated with 0.0975 mol/L hydrochloric
acid. 11.56 mL was required.
Calculate the mass of the aspirin in the tablet.
Moles NaOH added initially to react and dissolve aspirin = cV = 0.125 x 0.025 = 0.003125 mol
Moles HCl required for titration excess sodium hydroxide = 0.0975 x 0.01156 = 0.001127 mol
Hence moles of excess NaOH = 0.001127 (as NaOH and HCl react in 1:1 ratio)
Moles of NaOH which had reacted with aspirin = 0.003125 – 0.001127 = 0.001998 mol
Hence mol aspirin in tablet = 0.001998 mol (as aspirin is monoprotic and reacts with NaOH in
1:1 ratio)
Mass of aspirin in tablet = n x M = 0.001998 x180.2 = 0.3599 g = 0.360 g (3 s.f.)
68
Calculations Involving Gravimetric Analysis
1.
A student carried out an investigation to analyse the sulfate content of lawn fertiliser. The
student weighed out 1.0 g of fertiliser into a beaker, added dilute hydrochloric acid solution and
then stirred. 50.0 mL of 0.350 mol L-1 barium chloride solution was then added. A white
precipitate of barium sulfate formed. The student washed the precipitate, filtered it, left it to dry
and then determined the weight of the precipitate to be 1.14 g.
Determine the percentage of sulfate ion, by mass, in the f ertiliser sample.
Ba2+ (aq) + SO42- (aq)  BaSO4 (s)
Mass of barium sulfate precipitate
Molar mass BaSO 4
Moles BaSO 4 formed
Moles sulfate ion in sample
= 1.14 g
= 233.37 g
= 0.004885 mol
= 0.004885
Mass of 1 mol sulfate ion (SO 42-)
= 96.07 g
Mass of sulfate ion in sample
= 0.004885 x 96.07 g
= 0.4693 g
Hence percentage of sulfate ion in sam ple = 0.4693/1.0 x 100 = 46.9% = 47% (2 s.f.)
2.
A 2.45 g sample of lawn fertiliser was analysed for its sulfate content. After filtration and drying,
2.18 g of barium sulfate was recovered.
What is the % w/w of sulfate in the lawn fertiliser?
Moles barium sulfate precipitate = 2.18/233.37 = 0.009341 mol
Moles of sulfate ion in sample of fertiliser = 0.009 341 mol
Mass of sulfate ion = 0.0092341 x 96.07 g = 0.8974 g
Hence percentage by mass of sulfate in sample = 0.897 4/2.45 x 100 = 36.7% (3 s.f.)
69
3.
A student collected a 500 mL sample of water from a local creek for analysis. It was filtered
and the filtrate evaporated to dryness. The following data were collected.
(a)
(b)
Mass of filter paper
0.20 g
Mass of filter paper and solid
0.28 g
Mass of evaporating basin
46.33 g
Mass of basin and solid remaining
46.64 g
Calculate the mass of suspended solids in the 500 mL sample .
Suspended solids are filtered out of the sample.
Mass of suspended solids
= (Mass of filter paper + solid) – mass of filter paper
= 0.28 – 0.20 = 0.08 g
Calculate the mass and hence the percentage of dissolved solids in the sample.
Dissolved solids are collected in the filtrate and the water removed by evaporation.
Mass of dissolved solids
= (Mass basin + solid) – mass basin
= 46.64 – 46.33 g = 0.31 g
Percentage of dissolved solids = 0.31 g per 500 mL
= 0.31/5 g/100 mL
= 0.062 g/100 mL
= 0.062 g/100 g solution
= 0.062% (w/w)
4.
To determine the concentration of magnesium ions in sea water, a chemist measured out
250 mL exactly of seawater, put it in a beaker and gently heated it to evapo rate about half the
water. Excess sodium hydroxide was then added slowly to precipitate all the magnesium ions
as magnesium hydroxide. The mixture was cooled, filtered and the precipitate dried and
weighed. Its mass was found to be 0.850 g.
Calculate the percentage (w/w%) of magnesium ions in sea water.
Mass of magnesium hydroxide
= 0.850 g
Moles of magnesium hydroxide
= 0.850/58.33 = 0.0146 mol
Moles of magnesium ions in 250 mL = 0.0146 mol
Mass of magnesium ions in 250 mL = 0.0146 x 24.31 g = 0.354 g
Mass of magnesium ions in 100 mL = 0.354/2.50 = 0.142 g/100 mL
Mass of magnesium ions per 100 g sea water = 0.142 g/100 g seawater or 0.142% (w/w)
70
5.
The phosphate content of a laundry powder was determined as follows. 6.85 g o f the powder
was dissolved in approximately 500 mL of water. A slight excess of a solution of magnesium
chloride in an ammonia/ammonium chloride buffer was added, with stirring, until precipitation
was complete. The precipitate was filtered, dried and its mass determined. The mass of
precipitate was found to be 1.89 g
The precipitate is magnesium ammonium phosphate hexahydrate (Mg (NH4)PO4.6H2O).
Calculate the number of moles of phosphorus in the precipitate and hence the percentage
phosphorus in the original laundry powder.
Molar mass of magnesium ammonium phosphate hexahydrate (Mg(NH 4)PO4.6H2O) = 245.42
Moles precipitate = 1.89/245.42 = 0.0077 01 mol
Moles P in the precipitate = 0.007701 mol
Mass P = 0.007701 x 30.97 = 0.2385 g per 6.85 g powder
Hence % P in powder = 0.23 85/6.85 x100 = 3.48%
71
Analysis of Water Quality
1.
The Winkler method is used to determine the amount of dissolved oxygen in a sample.
In this procedure, oxygen reacts with Mn 2+ under alkaline conditions to produce a
precipitate of MnO(OH) 2.
2Mn2+ (aq) + O2 (aq) + 4OH- (aq)  2MnO(OH) 2 (s)
The precipitate is then dissolved in acid and reacted with iodide, forming iodine and Mn2+.
MnO(OH) 2 (s) + 2Ι- (aq) + 4H+ (aq)  I2 (aq) + Mn2+ (aq) + 3H2O (aq)
Finally, the amount of iodine produced is determined by reaction with thiosulfate.
I2 (aq) + 2S2O32- (aq)  2Ι- (aq) + S4O62- (aq)
When a sample of water was analysed using the Winkler method, a total of 0.60 mol of
thiosulfate was used in the reaction. How many moles of oxygen were present in the original
sample?
If 0.60 mol thiosulfate reacts, 0. 30 mol I2 must be present; 0.30 mol I2 is formed from 0.30 mol
MnO(OH) 2. 0.30 mol MnO(OH) 2 is formed from 0.15 mol O 2
Hence 0.15 mol oxygen was present in the origina l sample.
2.
The treated waste from a food processing factory had a BOD of 50 ppm. This waste was
diluted by a factor of 20 with pure water and added to a river which had a DO level of 9 ppm.
What would be the DO concentra tion at sufficient distance downstream of the discharge point
for all the waste to be decomposed? Would there be any effect of this discharge on the river
water quality?
After dilution BOD = 50/20 ppm = 2.5 ppm. This means the DO level would be reduced by 2.5
ppm. Hence DO level in river at sufficient distance downstream of the discharge point for all
the waste to be decomposed = 9 – 2.5 = 6.5 ppm. There is still sufficient DO for the water
quality to support life. The water would be likely to be turbid near the waste entry point but
downstream there would be little impact on the water quality except for slightly higher levels of
suspended or dissolved solids.
If untreated waste, with a BOD of 240 ppm, had been discharged at a dilution factor of 20,
what would be the DO level some distance downstream from the point of discharge?
After dilution BOD = 240/20 = 12 ppm. This means that the DO level would be reduced by 12
ppm. This means that there would not be any DO in the river, as the initial DO level was only 9
ppm. The river would not support life and would be polluted by dead organisms, would be
discoloured and turbid.
3.
Water from a particular source was quoted as having a hardness of 54 mg/L. Express this in
total moles of magnesium and calcium ions per litre .
A hardness of 54 mg/L means 54 mg CaCO 3 per litre.
[CaCO3] = 54 x 10 -3 g per litre or 54 x 10 -3/100.1 mol/L = 0.00054 mol/L
Hence the total concentration of Ca 2+ ions and Mg 2+ ions is 5.4 x 10 -4 mol/L
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4.
What is the hardness of water in which the calc ium ions and magnesium ions have
concentrations of 3.7 x 10 -4 and 0.9 x 10 -4 mol/L respectively?
The total concentration of Ca 2+ and Mg2+ = (3.7 + 0.9) x 10 -4 mol/L = 4.6 x 10 -4 mol/L
Equivalent concentration of CaCO 3
= 4.6 x 10 -4 x 100.1 g/L
= 46 x 10-3 g/L
= 46 mg/L
= 46 ppm
5.
The concentration of chloride ion in water is determined by formation of a silver chloride
precipitate. If the volume of the water sample being tested is 50.0 mL and the mass of the dried
precipitate obtained is 3.65 g, calculate the chloride ion concentration in the water sample in
ppm.
Moles AgCl = 3.65/143.35 = 0.02546 mol
Moles Cl = 0.02546 mol
Mass Cl
= 0.02546 x 35.45 g = 0.9026 g
Volume of solution = 50.0 mL
[Cl-]
= 0.9026 g /50.0 mL
= 0.9032 x 1000/50 g/L
= 18.05 g/L = 18,050 mg/L = 18.1 x 103 ppm
6.
One of the most common methods for determining the concentration of metal ions in water
samples involves titration with a reagent called EDTA. In alkaline solution EDTA is present as
an anion with a -4 charge. In this form it reacts with metal ions such as calcium and magnesium
in a 1 : 1 ratio.
Ca2+ + EDTA 4-  Ca(EDTA) 2When the reaction between the metal ions and EDTA 4- is complete, an indicator also present in
the solution changes colour.
A student used the following procedure to determine the concentration of calcium in a sample
of water: 50.0 mL of water sample was pipetted into a conical flask.
5.0 mL of ammonia/ammonium ion buffer and two drops of indic ator were added.
The sample was titrated with 0.01880 mol L -1 EDTA4- until indicator changed colour. The
above procedure was repeated a further three times.
The average volume of EDTA 4- used in the four titrations was 22.55 mL
(a)
What is the average number of moles of EDTA4- added to reach the end point?
n= cV = 0.01880 x 22.55/1000 = 0.0004240 mol = 4.240 x 10-4 mol EDTA 4-
(b)
Calculate the concentration of Ca 2+ in the water sample in mg L -1.
1 mol EDTA combines with 1 mol Ca 2+.
Mol Ca2+ = 4.240 x 10-4
Assume all calcium ion was sourced from the 50.0 mL water sample (i.e. none was in
the 5 mL of buffer solution)
[Ca2+] = 4.240 x 10-4/0.0500 mol/L = 8.48 x 10-3 mol/L = 8.48 x 10 -3 x 40.08 g/L
= 0.340 g/L = 340 mg/L
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HSC CHEMISTRY
STUDENT WORKBOOKS
About the author:
Bronwen Hegarty is an experienced teacher of HSC Chemistry and has been the convenor of Examination
Committees and Chemistry writing teams. She continues to teach Intensive Chemistry Revision courses and is
working to assist Chemistry teachers and their students in a number of schools. She has recognised the need for
these workbooks in schools, not only to provide revision for students throughout their Year 11 and HSC courses,
but also to reduce the workload on teachers in preparing course material s and sequencing lesson plans. The
workbooks are designed to be a low cost alternative to creating and photocopying teacher-designed student
worksheets, assignments and revision questions. The format will faci litate student-centred teaching and learning
and provide a framework for students to monitor their progress through the HSC syllabus. Teacher Editions of the
workbooks include a CD and are available only to teachers, to provide them with fully worked solu tions for each
workbook.
Student Workbooks available:
Year 12:
Core Module 1: Production of Materials
Core Module 2: The Acidic Environment
Core Module 3: Chemical Monitoring and Management
Option: Industrial Chemistry
Option: Shipwrecks, Corrosion and Conservation
Option: Forensic Chemistry
Year 11: Preliminary Chemistry
Years 11 and 12: Calculations in Chemistry for the Preliminary & HSC Courses
Note: Teacher Editions of the workbooks (including worked solutions to questions on
CD) are available only to teachers.
To obtain copies of these workbooks, contact Bronwen Hegarty:
Mobile 0402 890 724 or [email protected]
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