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NATIONAL 5 CHEMISTRY
A GUIDE TO CALCULATIONS
CALCULATIONS WILL MAKE UP ABOUT 25% OF YOUR OVERALL EXAM
GRADE IN NATIONAL 5 CHEMISTRY SO YOU MUST BE GOOD AT THEM!
THIS BOOKLET WILL GIVE YOU WORKED EXAMPLES OF ALL OF THE
DIFFERENT TYPES OF CALCULATIONS AND SOME TO TRY YOURSELF.
DO NOT WRITE ON THIS BOOKLET – IT WILL BE COLLECTED IN AT THE
END OF THE YEAR.
CONTENTS
1. Writing Chemical Formulae
Page 1
2. Writing Ionic Formulae
Page 6
3. Calculating the GFM
Page 7
4. Calculating Moles from Mass
Page 8
5. The Mole in Solutions
Page 10
6. Stoichiometry
Page 13
7. Balancing Equations
Page 13
8. Calculations From Equations
Page 15
9. Percentage Mass Calculations
Page 19
10. Average Rate Calculations
Page 21
11. Heat Energy Calculations
Page 24
12. Half Life Calculations
Page 27
13. Calculation Questions from Exam Papers
Page 31
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1. WRITING CHEMICAL FORMULAE
Back in Topic 1b you began to use VALENCIES to help you write CHEMICAL
FORMULAE for some simple 2 element compounds.
The valency of an element was the number of unpaired electrons in the outer
shell of one of its atoms.
There is a clear pattern to be seen as you go across a period in the Periodic
Table:
USING THE VALENCY OR COMBINING POWER
If we know the valencies of the atoms combining we can work out the formula
of the compound they form more easily than by drawing diagrams of the
molecules.
Follow these rules to find the formula of a compound:
1.SYMBOL
Write down the symbols of the elements involved e.g. water
2.VALENCY
Write the valency of each below the symbol.
3. SWAP
Swap the valency numbers over.
4. DIVIDE
Divide by a common factor if there is one present. e.g. divide by 2 with
Mg2O2or C2O4 to give MgO and CO2.
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5. FORMULA
Write the numbers at the bottom right of the symbols, miss out the number 1.
e.g. H2O
Here are some more examples written in a more compact form:
SWAP
Using the rules of writing formulae, work out the formulae of the following
compounds:
Magnesium oxide
Hydrogen sulphide
Aluminium chloride
Calcium chloride
Hydrogen chloride
Carbon hydride
Sodium phosphide
Nitrogen hydride
Phosphorus bromide
Silicon carbide
Aluminium oxide
2
METALS WITH MORE THAN ONE VALENCY
Metals with more than one valency are-usually transition metals. You are told
the valency in the name of the compound. eg. Iron(III) oxide is a compound
formed between iron (with a valency of 3) and oxygen. Iron(II) oxide is a
compound formed between iron (with a valency of 2) and oxygen.
When writing formulae for compounds containing transition metals the name
of the compound will always contain the valency of the metal listed as a
Roman numeral.
Roman Numerals
Valency
1
2
3
4
5
6
7
Roman Numeral
I
II
III
IV
V
VI
VII
(N.B. THE VALENCY OF ZINC IS ALWAYS 2)
Work out the formula of the compounds:
COMPOUND NAME
WORKING
FORMULA
Copper(I) chloride
Iron(III) oxide
Zinc bromide
Chromium(VI) oxide
Lead(IV) oxide
Tin(IV) iodide
3
IONS CONTAINING MORE THAN ONE TYPE OF ATOM
We found out in S1/S2 that compounds which have names ending in –ATE or
–ITE contain another element besides the ones obvious from the name.
Which additional element do these compounds contain? ______________
Compounds like this contain ions which have more than one type of atom in
them.
These ions can be found on Page 8 of your DATA BOOK in a table like this:
Valency = 1
Valency = 1
Valency = 2
Valency = 3
The size of the charge on the ion is the same as the valency.
Write in the valencies at the top of the correct column on page 4 of the Data
Book.
Example: Consider the carbonate ion which has carbon atom joined to 3
oxygen atoms.
It has the structure:
The whole “package” of atoms has valency of 2 (because it has a charge of
2-).
When using such an ion in writing formulae it should be enclosed in brackets
i.e. (CO32-).
4
N.B. You NEVER change the numbers inside the brackets when you are
working out formulae.
Once enclosed in brackets the “package” of atoms can now be treated just
like a single atom and the previous rules for writing formulae now hold.
Here is an example:
Aluminium Carbonate
Step 1 - write out the formulae of each of the ions (you do not need to include
the
charges):
Al
(CO3)
Step 2 - write the valency of each ion above the symbol:
3
Al
2
(CO3)
Step 3 - “swap and drop” the valencies:
3
Al
2
(CO3)
Step 4 – rewrite formula without any “1”s and after cancelling any common
factors:
Al2(CO3)3
The formula tells us that in aluminium carbonate has 2 lithium ions and 3
carbonate ions in its formula.
Now try the following examples yourself:
Calcium Sulphate
Sodium Nitrate
Copper(II) Phosphate
Lithium Sulphite
Dichromate
Ammonium Hydroxide
Manganese(V)
Zinc Ethanoate
Tin(IV) Chromate
Potassium Permanganate
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2. WRITING IONIC FORMULAE
It is SOMETIMES useful to be able to recognise ionic formulae (Chemical
formulae which include ionic charges) or even to be able to write them.
Look at these examples showing how to use the valency method to write ionic
formulae.
Sodium Sulphate
Step 1 - write out the formulae of each of the ions with brackets around both
(you do not at this stage need to include the charges):
(Na)
(SO4)
Step 2 - write the valency of each ion above the symbol:
1
(Na)
2
(SO4)
Step 3 - “swap and drop” the valencies:
1
(Na)
2
(SO4)
Step 4 – rewrite formula without any “1”s and after cancelling any common
factors:
(Na)2(SO4)
Step 5 - Put the charges of each of the ions inside the brackets:
(Na+)2(SO42-)
NOTE: The charges are NOT swapped around.
Try writing out the following ionic formulae yourself:
sodium chloride
sodium phosphide
calcium chloride
iron(III) nitrate
ammonium oxide
aluminium phosphate
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3. CALCULATING GFM (THE MASS OF 1 MOLE)
The mass of 1 mole of a substance is known as its GRAM FORMULA MASS or
GFM.
It is calculated using the chemical formula of the substance and the table of
RELATIVE ATOMIC MASSES which can be found on page 7 of the National 5
Databook.
The relative atomic mass of every atom in the chemical formula is added together to
give an overall mass in grams – the GRAM FORMULA MASS or GFM.
Eg. Sodium Nitrate
NaNO3
3 oxygen atoms = 3 x 16 = 48
1 nitrogen atom = 1 x 14 = 14
1 sodium atom = 1 x 23 = 23
85g
Try these examples:
phosphorus chloride
nitrogen iodide
sulphur bromide
hydrogen fluoride
silicon carbide
germanium hydride
calcium nitrate
silicon fluoride
ammonium sulfate
phosphorus bromide
lead(IV) phosphate
potassium permanganate
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4. CALCULATING MOLES FROM MASS
Now that we are able to work out the mass of 1 mole (the GFM), we can use this to
work out the mass of any number of moles:
Eg.
2 moles will be 2 x GFM, 5 moles will be 5 x GFM and 0.0000023 moles will
be 0.0000023 x GFM.
In short, we can say:
or
Mass =
number of moles x Mass of 1 mole
m
n x GFM
=
This equation can be rearranged to allow you to work out the number of moles of a
substance if you are given a mass:
Number of moles
=
mass  GFM
Or
=
m  GFM
n
Both of these equations can be derived from
the following useful triangle:
m
= mass
manic
n
= number of moles
nuns
GFM = mass of 1 mole
m
n
GFM
Go For Men
Here are some examples to try – the first 2 are done for you:
1) How many moles are in 15 grams of lithium?
Ans:
n
=
m  GFM
=
15  7
=
2.14 moles
2) How many grams are in 2.4 moles of sulfur?
Ans:
n
=
n x GFM
=
2.4 x 32
=
76.8 g
3) How many moles are in 22 grams of argon?
4) How many grams are in 88.1 moles of magnesium?
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5) How many moles are in 2.3 grams of phosphorus?
6) How many grams are in 11.9 moles of chromium?
7) How many moles are in 9.8 grams of calcium?
8) How many grams are in 238 moles of arsenic?
9) How many grams are in 4.5 moles of sodium fluoride, NaF?
10) How many moles are in 98.3 grams of aluminum hydroxide, Al(OH)3?
11) How many grams are in 0.02 moles of beryllium iodide, BeI2?
12) How many moles are in 68 grams of copper (II) hydroxide, Cu(OH)2?
13) How many grams are in 3.3 moles of potassium sulfide, K2S?
14) How many moles are in 1.2 grams of ammonia, NH3?
15) How many grams are in 2.3 moles of calcium phosphate, Ca3(PO4)2?
16) How many moles are in 3.4 grams of silicon dioxide, SiO2?
17) How many grams are in 1.11 moles of manganese sulfate, Mn2(SO4)7?
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5. THE MOLE IN SOLUTIONS
INTRODUCTION
A solution is made up of two parts, the SOLVENT and the SOLUTE. The solution
could be CONCENTRATED or DILUTE.
SOLVENT
The liquid which dissolves the solute. It is usually water.
SOLUTE
The substance which dissolves - often a solid, but not always.
CONCENTRATED
There is a lot of solute and a little water.
DILUTE
There is a lot of water and only a little solute.
When we measure out 50cm3 of a solution, e.g., hydrochloric acid, we measure out
some solvent (water) and some acid together. In any reactions of the acid, it is the
HCl particles which are involved. The water is just a carrier for the acid, so when we
measure out a volume of the solution, we want to know how much acid it contains.
To do this, we need to define the CONCENTRATION of the solution. This tells us
how much solute there is in a fixed volume of the solution. In chemistry, “how
much” means MOLES and the “fixed volume” is taken as 1 litre (1l).
[NB: 1l is 1000cm3]
DEFINITION
Concentration of solution
=
number of moles of solute
volume of solution (in
litres)
 You must learn this relationship and be able to use it and rearrange it to calculate
moles of solute or volume of solution, given the other two quantities.
 Remember that the volume must be in litres, so convert from cm3 if necessary.
 The units for concentration are moles per litre which is written mol l-1
 Lab bottles use the letter M (meaning molar) and this is the same thing as mol l-1
RELATIONSHIPS
Use the triangle opposite to help you answer the
questions which follow. Some worked examples
are given to help you.
n = number of moles
nuns
C = concentration
chase
V(l) = Volume (in litres)
vicars (in lycra)
n
C
V(l)
10
Examples:-
1) What is the concentration of a solution which contains 2 moles of
copper(II) sulfate in 4 litres of solution?
Ans:
concentration = moles  volume = 2  4 = 0.5 mol l-1
2) How many moles of sodium hydroxide are there in 200cm 3 of a solution of
concentration 0.2 mol l-1 ? (Remember that 200cm3 = 200  1000 = 0.2
litres)
Ans:
moles = concentration  volume = 0.2  0.2 = 0.04 moles
QUESTIONS
1) Calculate the concentration in mol l-1 of each of the following
solutions:a)
b)
c)
d)
e)
10 moles of potassium hydroxide in 5 litres of solution.
1.5 moles of hydrochloric acid in 3 litres of solution.
0.5 moles of calcium chloride in 500cm3 of solution.
0.2 moles of sulphuric acid in 100cm3 of solution.
0.075 moles of magnesium sulphate in 25cm3 of solution.
2) Find the number of moles of solute in each of the following
solutions:a)
b)
c)
d)
e)
2 litres of 2 mol l-1 nitric acid.
0.25 litres of 0.4 mol l-1 ammonium chloride solution.
200cm3 of 0.8 mol l-1 sodium carbonate solution.
40cm3 of 0.2 mol l-1 hydrochloric acid.
300cm3 of 4 mol l-1 sodium hydroxide solution.
3) Find the concentration in mol l-1 of these solutions:In question 3 you have to find the concentration of a solution starting from a
mass of solute. The first step is to use the GFM to convert the mass to moles.
(Remember that number of moles = mass in grams  gram formula mass)
Then continue as in question 1 above.
Part a) is worked out for you as an example.
a) 20g of sodium hydroxide (NaOH) in 2 litres of solution. (GFMs: Na = 23, O
= 16, H = 1)
Ans:
1 mole of NaOH = 23 + 16 + 1 = 40g : n = m/GFM = 20/40 = 0.5 mol
 there is 0.5 mole of NaOH in 2 litres
 concentration = n / V(l) = 0.5  2 = 0.25 mol l-1
11
b) 10.6g of sodium carbonate (Na2CO3) in 1 litre of solution.
c) 0.98g of sulphuric acid (H2SO4) in 200 cm3 of solution.
d) 117g of sodium chloride (NaCl) in 5 litres of solution.
e) 0.83g of potassium iodide (KI) in 25 cm3 of solution.
4) Calculate the mass of solute in the following solutions:In question 4 you have to work out the mass of solute contained in the solution.
First find the number of moles of solute, as in question 2 above, then convert the
moles to a mass. Part a) has been done as a worked example for you.
a) 0.25 litres of 2 mol l-1 calcium chloride solution. (GFM of Ca = 40)
Ans:
n = C x V(l) = 2  0.25 = 0.5 mole
1 mole of CaCl2 = 40 + (35.5  2) = 111g
 0.5 mole = 0.5  111g = 55.5g of solute
b) 2 litres of 0.2 mol l-1 potassium hydroxide (KOH) solution.
c) 200 cm3 of 0.1 mol l-1 sodium carbonate (Na2CO3) solution.
d) 25 cm3 of 0.05 mol l-1 copper (II) sulphate (CuSO4) solution. (GFM of Cu =
64)
e) 1.5 litres of 0.4 mol l-1 nitric acid (HNO3). (GFM of N = 14)
12
6. STOICHIOMETRY – FUN TO SAY, FUN TO DO!
You’re possibly reading this booklet because you’re being forced to do so by an
authority figure of some kind. Or perhaps you’re reading it for the pure love and
wonder of science. Most likely, you’re reading it because some teacher keeps saying:
“Calculations are going to be worth about 25% of your
final grade in National 5 Chemistry – so you would be a
complete idiot not to try to learn how to do them
properly.”
Well, don’t panic! We’re going to talk about the magical world
of stoichiometry (“stoy-key-o-met-rey”) and how it can be
used to enrich your life, etc.
What on Earth is Stoichiometry?
The word stoichiometry is just a fancy way of saying “the method you use to figure
out how much of a chemical you can make, or how much you need, during a
reaction.” For example, if you are carrying out a reaction and want to make 88.5
grams of the product, you could do a bunch of calculations to figure out how much of
each reactant you would need. Those calculations are stoichiometry!
7. BALANCING EQUATIONS
Here is a link to a beginner’s guide to balancing equations on youtube:
http://www.youtube.com/watch?v=_B735turDoM
Atoms are not lost or gained during a chemical reaction.
Scientists know that there must be the same number of atoms on each side of
the equation. To balance the chemical equation, you must add numbers in front
of the chemical formulae in the equation. You cannot add or remove subscripts!
1.
2.
3.
4.
Determine number of atoms for each
element.
Pick an element that is not equal on
both sides of the equation.
Add a number in front of the
formula with that element and adjust
your counts.
Continue adding numbers to get the
same number of atoms of each element
on each side.
1.
___Mg + ___O2  ___MgO
Mg = 1
Mg = 1
O= 2
O= 1
2Mg
+
O

___MgO
2
3.
Mg = 2
Mg = 1
O= 2
O= 1
2Mg
+
O

2MgO
2
4.
Mg = 2
O= 2
Mg = 2
O= 2
13
Balance each equation. Be sure to show your lists! Remember you cannot add
subscripts or place numbers in the middle of a chemical formula.
1. Na + MgF2  NaF + Mg
2. Mg + HCl  MgCl2 + H2
3. Cl2 + KI  KCl + I2
4. NaCl  Na + Cl2
5. Na + O2  Na2O
6. Na + HCl  H2 + NaCl
7. K + Cl2  KCl
Challenge: This one is tough!
C2H6 + O2  CO2 + H2O
14
8. CALCULATIONS FROM EQUATIONS
As you have probably already discovered in your chemistry class, there are about a
bazillion types of stoichiometric calculation out there. The good news, however, is
that none of them are all that difficult. Seriously.
Before I show you some examples of stoichiometry, let me show you a handy picture
that you will be using a lot:
info about
substance
A
moles of
substance
A
moles of
substance
B
info about
substance
B
The best way to teach you how to use this picture to do stoichiometry is to simply
give you a stoichiometry problem and solve it:
Problem:
Given the equation:
2H2 + O2  2H2O
What mass of water can be made from 50.0 grams of oxygen?
Answer:
Let’s go through the following steps to solve this problem:
1.
Determine what you’re trying to figure out, and what you’ve been
given:
In this problem, you’ve been given the information “50.0 grams of
oxygen”, so we’ll call O2 “substance A.” The problem tells you that
you’re trying to find the “mass of water”, so we’ll call H2O “substance
B.”
2.
Figure out where you are on this table, and where you’re trying to
get:
Since we’re given “50.0 grams of oxygen”, we start in the “info about
substance A” box. Because we’re trying to find out how many grams
of water we’ll be
making,
we endofinOthe
“info about substance B” box.
Number
of moles
2 = m / GFM
3.
= 50 / 32
Convert to moles :
Looking at the diagram, to get the answer, we have to get from the
= 1.56
box on the left hand side to the box on the
rightmoles
hand side. To do this
we must convert any information we are given into moles.
15
So, if we add this information to the diagram:
Mass of
O2
Moles of
O2
50g
1.56 moles
Moles of
H2O
Mass of
H2O
Find number of moles of substance B from “mole ratio” :
4.
The number of moles of substance A and substance B are linked by
what we call a “mole ratio”. The reason that we balance equations is
so that we can find this ratio between the quantities of reactants and
products in a chemical reaction.
2H2 + O2  2H2O
The balanced equation for this reaction tells us that 2 moles of H2
reacts with 1 mole of O2 to produce 2 moles of H2O.
So, if :
1 mole of O2

2 moles of H2O
then,
1.56 moles of O2 
3.12 moles of H2O
This information can be added to the diagram:
Mass of
O2
Moles of
O2
Moles of
H2O
Mass of
H2O
50g
1.56 moles
3.12 moles
?
5.
The final step involves calculating the mass of substance B from
the number of moles of substance B you have just worked out :
Mass of H2O = number of moles x GFM
= 3.12 x 18
= 56.16g
Mass of
O2
Moles of
O2
Moles of
H2O
Mass of
H2O
50g
1.56 moles
3.12 moles
56.16g
16
EQUATIONS INVOLVING VOLUMES AND CONCENTRATIONS
The same rules apply to any calculation from a balanced equation. If you
convert the quantities given into moles, then you should be able to work out
the quantity of any reactant or product in the reaction.
Eg.
What mass of iron reacts with 50 cm3 of 2 mol l-1 hydrochloric acid in the
following equation:
Fe(s) + 2HCl(aq)

FeCl2(aq) + H2(g)
Ans : Substance A = hydrochloric acid
Substance B = iron
Info from
the
question
50cm3
2 mol l-1
Number of moles of hydrochloric acid
= C x V(l)
= 2 x 0.05
= 0.1 moles
50cm3
2 mol l-1
0.1 moles
Mole ratio from balanced equation:
Fe(s) + 2HCl(aq)
1 mole

2 moles
FeCl2(aq) + H2(g)
1 mole
1 mole
So,
2 moles HCl
0.1 moles HCl


1 mole of Fe
0.05 moles of Fe
17
50cm3
2 mol l-1
0.1 moles
0.05 moles
?
The question asks for mass of iron, so:
Mass of iron
= n x GFM
= 0.05 x 56
= 2.8g
Try these examples:
1.
Use the following equation to determine the mass of carbon dioxide
formed when 26.125 g of sodium carbonate reacts with excess
hydrochloric acid?
Na2CO3 + 2 HCl  2 NaCl + CO2 + H2O
2.
The following equation shows the reaction between carbon monoxide
and iron(III) oxide, what mass of iron can be obtained from 12 tonnes
of iron(III) oxide?
Fe2O3 + 3 CO  2 Fe + 3 CO2
3.
What mass of ammonia can be made from 70 kg of nitrogen using the
Haber Process?
N2 + 3 H2 
4.
What mass of magnesium chloride can be produced by the reaction between
3.6 g of magnesium and excess hydrochloric acid?
Mg + 2 HCl
5.
2 NH3

MgCl2 + H2
What mass of carbon dioxide can be produced by heating 15g of calcium
carbonate?
CaCO3  CaO + H2O
18
9. PERCENTAGE COMPOSITION
Percentage composition
Percentage composition is just a way to describe what proportions of the different
elements there are in a compound.
If you have the formula of a compound, you should be able to work out the
percentage by mass of an element in it.
Example
The formula for sodium hydroxide is NaOH. It contains three different elements:
Na, O and H. But the percentage by mass of each element is not simply 33.3 per
cent, because each element has a different relative atomic mass. You need to use
the Ar values to work out the percentages. Here is how to do it:
Question
What is the percentage by mass of oxygen (O) in sodium hydroxide
(NaOH)?
1. First, work out the Gram Formula Mass (GFM) of the compound,
using the Relative Atomic Mass values for each element from the
Data Book. In the case of sodium hydroxide, these are Na = 23, O =
16, H = 1.
2. Next, divide the total mass of oxygen in the formula by the GFM of
NaOH, and multiply by 100 to get a percentage.
ie. Use the formula : % mass = mass / GFM x 100
Answer
1. GFM of NaOH is 23 + 16 + 1 = 40g
2. (16 ÷ 40 ) × 100 = 0.4 × 100 = 40%
So the percentage by mass of oxygen in sodium hydroxide is 40%.
19
Find the percent compositions of all of the elements in the following compounds:
1)
CuBr2
Cu: ___________
Br: ___________
2)
NaOH
Na: ___________
O: ___________
H: ___________
3)
(NH4)2S
N: ___________
H: ___________
S: ___________
4)
N2S2
N: ___________
S: ___________
5)
KMnO4
K: ___________
Mn: ___________
O: ___________
6)
HCl
H: ___________
Cl: ___________
7)
Mg(NO3)2
Mg: ___________
N: ___________
O: ___________
20
10. AVERAGE RATE CALCULATIONS
Reaction Rates
The rate of a chemical reaction is a measure of how fast the reactants are being
used up and how fast the products are being made.
The rate can be determined by measuring:

Changes in the concentration of the reactants or products

Changes in the mass of the reactants or products

Changes in the volume of the reactants or products
Rate Graphs
In chemistry, graphs can be used to follow the course of a reaction. A graph can
tell us many things about a reaction.
Over the course of a reaction, the average rate can be calculated. Whichever
quantity is measured during the course of a reaction, the average rate of reaction
over is always given by:
NB change in measurable quantity = concentration or mass or volume of
reactants or products.
21
Example:
A student reacted some pure and some impure copper(II) carbonate with
hydrochloric acid and measured the mass of the flask and its contents over time:
(a)
For the sample of pure copper(II) carbonate, calculate the
average rate over the first 10 seconds in g s-1.
Ans:
= (165.00 – 164.83)
10
= 0.17 / 10
= 0.017 g s-1
(b) For the impure sample of copper(II) carbonate, calculate the
average rate, in g s-1 for the first 20 seconds.
22
23
11. HEAT ENERGY CALCULATIONS
Energy changes take place during chemical reactions. Exothermic reactions give out
thermal energy and endothermic reactions take in thermal energy. These changes
can be measured experimentally or calculated before being analysed. Knowing the
amount of energy involved in a reaction can be used to ensure that resources are
used efficiently.
MEASURING ENERGY TRANSFERS
Heat energy can be given out or taken in from the surroundings during chemical
reactions. The amount of energy transferred can be measured. This is called
calorimetry.
Energy changes from combustion
The diagram shows a simple calorimetry experiment to measure the heat energy
released from burning a fuel. You should be able to recognise and label apparatus
like this.
To do the experiment:
1.
2.
3.
4.
measure cold water into a calorimeter (a metal or glass container)
record the starting temperature of the water
heat the water using the flame from the burning fuel
record the final temperature of the water
The biggest source of error is usually heat loss to the surroundings. This can be
reduced by insulating the sides of the calorimeter and adding a lid.
24
CALCULATING ENERGY CHANGES
The amount of energy transferred during a chemical reaction (either from the burning
of a fuel or a chemical reaction in solution) can be calculated using the equation:
Eh = Cm ΔT
Where:
Eh = the heat energy transferred (kJ)
m = the mass of the water being heated (kg).
NOTE: 1kg = 1000g = 1000cm3
C = the specific heat capacity of water (kJ kg-1°C-1)
ΔT = the change in temperature of the water (degree Celsius, °C)
The specific heat capacity of water is 4.18 kJ kg-1°C-1.
Worked example
In an experiment, ethanol was burnt from a spirit burner and the energy released was
used to heat 50 cm3 of water.
The starting temperature of the water was 19°C but by the end of the reaction, the
temperature had risen to 41°C.
Calculate the energy change in kJ.
Step 1: Calculate the temperature change, ΔT
41 – 19 = 22 °C
Step 2: Use Eh = Cm ∆T
Remember that C = 4.18 kJ kg-1°C-1 for water.
Eh = Cm ΔT
Eh = 0.05 × 4.18 × 22 = 4.598kJ
Eh = 4.6 kJ
25
A student investigated the amount of energy released when an alkane
burns using the apparatus shown.
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12. HALF LIFE CALCULATIONS
Half-life
The nuclei of radioactive atoms are unstable. They break down and change into a
completely different type of atom. This is called radioactive decay. For example,
carbon-14 decays to nitrogen-14 when it emits beta radiation.
It is not possible to predict when an individual atom might decay. But it is
possible to measure how long it takes for half the nuclei of a piece of radioactive
material to decay. This is called the half-life of the radioactive isotope.
Two definitions
There are two definitions of half-life, but they mean essentially the same thing:
1.
the time it takes for the number of nuclei of the isotope in a sample to
halve
2.
the time it takes for the count rate from a sample containing the
isotope to fall to half its starting level
Different radioactive isotopes have different half-lives. For example, the half-life
of carbon-14 is 5,715 years, but the half-life of francium-223 is just 20 minutes.
Graphs
It is possible to find out the half-life of a radioactive substance from a graph of
the count rate against time. The graph shows the decay curve for a radioactive
substance.
The decay curve for a radioactive substance
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The count rate drops from 80 to 40 counts a minute in two days, so the half-life is
two days. In the next two days, it drops from 40 to 20 - it halves. In the two days
after that, it drops from 20 to 10 - it halves again - and so on.
Half Life Calculations
Example:
A radioactive chemical has an activity of 100 counts min-1. What is the activity of this
chemical after 2 half-lives have passed?
Solution:
1 half-life
2 half-lives
100
50
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activity ÷ 2
activity ÷ 2
After 2 half-lives, the activity is 2500 counts min-1.
Notice that our example didn’t even say how long the half life was. We just used the
fact that activity is halved each time a half-life passes!
A radioactive sample in a laboratory has a half-life of 10 days. If the sample has an
activity of 4800 counts min-1 when it is returned to the safe, what will be it’s activity
when it is next used 40 days later?
Solution:
Number of half-lives = half life
time
= 10 days
40 days
= 4.
So 4 half-lives have passed.
Now we calculate the activity 4 half-lives later.
Half-lives:
1
4,800
activity ÷ 2
2
2,400
activity ÷ 2
3
1,200
4
600
activity ÷ 2
300
activity ÷ 2
So after 40 days, the activity of the sample is 300 counts min-1.
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Now try these questions:
1.
A scientist prepares a radioactive solution with a half-life of 36 hours. Using a
GM tube, the scientist says the liquid has a count rate of 72 counts min-1.
What will be the activity of the solution (in counts min-1) if it is measured again
in 108 hours?
2.
A radioactive material with a half-life of 2 days is found to have an activity of
1500Bq. What was the activity of this material 4 days ago?
(use the same method but work backwards)
Calculating the half-life
If you have a lot of information about the radioactive decay of a material, such as
times and count rates, then it is possible to find the material’s half-life by drawing a
graph, similar to the graph shown earlier.
However, we normally don’t have enough information to draw a graph and so we
need another method to calculate the half-life.
To see how it works, let’s try an example.
The activity of a radioactive material is 1728Bq. After 24 hours, the activity
has fallen to 27 counts min-1. Calculate the half life of the radioactive material.
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Examples:
1.
A radioactive chemical is prepared in a laboratory. The chemical has an initial
activity of 272 counts min-1. The activity of the chemical is rechecked 3 hours
later and found to be 17 counts min-1. What is the half-life of the chemical?
2.
The activity of a radioactive material is 344 counts min-1. Two days later, the
activity of the same material has fallen to 43 counts min-1.
Calculate the half-life of the material.
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13. CALCULATIONS QUESTIONS FROM EXAM PAPERS
1.
2.
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3.
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4.
A student investigated the reaction of carbonates with dilute hydrochloric
acid.
(a)
In one reaction lithium carbonate reacted with dilute hydrochloric acid.
The equation for the reaction is:
Li2CO3(s) + HCl(aq) → LiCl(aq) + CO2(g) + H2O(l)
(i) Balance this equation.
(ii) Identify the salt produced in this reaction.
(b)
In another reaction 1 g of calcium carbonate reacted with excess
dilute hydrochloric acid.
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)
Calculate the mass, in grams, of carbon dioxide produced.
5.
Rhubarb contains oxalic acid, C2H2O4. Oxalic acid decolourises acidified
potassium permanganate solution.
An experiment was carried out to time how long it takes to decolourise the
solution using different numbers of rhubarb cubes.
(a)
Calculate the time taken for 10 cubes of rhubarb to decolourise
the solution.
1
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(b)
The equation for the reaction is:
2MnO4–(aq) + 5C2H2O4(aq) + 6H+(aq) XMn2+(aq) + 10CO2(g) + YH2O(l)
2moles
(i)
5moles
Balance the above equation to find X and Y.
X = _________
Y = _________
1
(ii)
Calculate the number of moles of permanganate ions (MnO4–)
in 100cm3 of a 1.0 mol l–1 solution.
1
(iii)
The above equation shows that 2 moles of permanganate ions
react with 5 moles of oxalic acid.
Calculate how many moles of oxalic acid (C2H2O4) react with
100 cm3 of 1.0 mol l–1 permanganate (MnO4–) solution?
1
(c)
A strip of rhubarb was found to contain 1·8 g of oxalic acid.
How many moles of oxalic acid, C2H2O4, are contained in 1·8 g.
(Formula mass of oxalic acid = 90)
1
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6.
Potassium sulphate can be produced by titrating potassium hydroxide
solution with dilute sulphuric acid.
(a)
Name the chemical added to the conical flask to show the end-point of
the titration?
1
(b)
The student carried out three titrations. The results are shown in the
table.
Rough
First titre
Second titre
Initial burette
reading(cm3)
0.0
20.1
0.0
Final burette
reading (cm3)
20.1
40.6
20.7
Volume used
(cm3)
20.1
20.5
20.7
(i)
State why the student repeated the titration a number of times.
1
(ii)
Calculate the average volume of acid used to neutralise the
potassium hydroxide.
1
(iii)
Using your answer from part (b) (ii) calculate the number of
moles of sulphuric acid used.
2
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(iv)
The equation for the reaction is:
H2SO4 + 2KOH K2SO4 + 2H2O
Calculate the concentration of the potassium hydroxide
solution.
3
7.
The concentration of chloride ions in water affects the ability of some plants
to grow.
A student investigated the concentration of chloride ions in the water at
various points along the river Tay.
The concentration of chloride ions in water can be determined by reacting
the chloride ions with silver ions.
Ag+(aq) + Cl–(aq) → AgCl(s)
A 20 cm3 water sample gave a precipitate of silver chloride with a mass of
1·435 g.
(a)
Calculate the number of moles of silver chloride, AgCl, present in this
sample.
Show your working clearly.
(b)
Using your answer to part (a), calculate the concentration, in mol l–1,
of chloride ions in this sample.
Show your working clearly.
8.
Urea, H2NCONH2, can be used as a fertiliser.
Calculate the percentage of nitrogen in urea.
Show your working clearly.
3
9.
Ammonia can be used to produce ammonium nitrate NH4+NO3-.
Calculate the percentage of nitrogen in ammonium nitrate.
Show your working clearly.
3
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10.
11.
12.
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13.
75% of the potassium-40 atoms originally present in the rock sample were
found to have undergone radioactive decay.
The half-life of potassium-40 is 1,260,000,000 years.
Calculate the age of the rock, in years.
14.
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