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Maths For AS Level Chemistry
Maths for AS Level
Chemistry
Instructions: please
complete question
sheets 1, 2 and 3.
-1Hanham Woods Academy
Maths For AS Level Chemistry
Maths for AS Level Chemistry
Aim
This is a short course to improve the maths skills that you will need for AS
level chemistry.
Objective
By the end of this short course you will be able to work out mathematical
answers using the methods given and apply them in the context of chemistry.
Resources
You will need to print off the appropriate 2 or 3 pages from the intranet each
week, or print them all at the start and keep them safe.
For each topic you are provided with a help sheet. On this help sheet you are
given a detailed method to work out answers. These are followed by
examples of how the maths is used in chemistry when appropriate.
You will then be asked to complete several questions, on the corresponding
question sheet. Each sheet contains 20 marks.
Resources are found on the intranet under,
Shared- Science – A level Chemistry – AS Level Maths Course
Contents
1. Standard Form
2. Significant Figures and Decimal Places
3. Ratios and Percentages
4. Rearranging Equations
5. Graphs and Gradients
6. Units
Submission Deadlines
It is crucial that you hand in your work to show that you are making progress
and committed to succeeding at AS level chemistry. Please hand in completed
work during the first lesson back at Hanham High School.
Feedback
If you have any queries please refer them to Dr Holland.
-2Hanham Woods Academy
Maths For AS Level Chemistry
HELPSHEET 1
Standard Form
What is standard form and why is it used?
In science we often need to use very small and very large numbers. It is
therefore useful to be able to write these down in an abbreviated notation.
This is known as standard form.
Example 1
30 000 = 3.0 x 104
The number is written as a single integer multiplied by 10 to a power. The
power is found by counting the number of places after the first number.
30 000 has four zeros therefore it is 3.0 x10 to the power 4 written as a small
number next to the ten.
Example 2
467 000 = 4.67 x 105
If after the first number there are not just zeros these are written after a
decimal point, as shown above, 4.67 in this case.
Therefore you can think about changing the number back as moving the
decimal place by the number shown as the power.
4.67
46.7
467
4670
46700 467000
move1
move2
move3
move4
move5
Example 3
0.000500 = 5.0 x 10-4
If the number is very small we still write the number as a single integer
multiplied by 10 but to a negative power. The power is found by counting the
number of zeros, or the number of places the number is after the decimal
point.
0.000500 has four zeros, or the 5 is the fourth decimal place, therefore it is 5
multiplied by 10 to the power of -4.
Example 4
0.00000451 = 4.51 x 10-6
As with large number if there are not only zeros present after the first number
they are put after the decimal point in standard form.
Now you have a go!!!!!
-3Hanham Woods Academy
Maths For AS Level Chemistry
QUESTION SHEET 1
Standard Form
Exercise 1 (1 mark for each)
Put each of these numbers into standard form.
A
B
C
D
300 000
5400
121 000
798 000 000
Exercise 2 (1 mark for each)
Put each of these numbers into standard form
A
B
C
D
0.007
0.00000098
0.00004567
0.06843
Exercise 3 (1 mark for each )
Write each of these numbers in full, not in standard form
A
6 x 106
B
3.45 x 109
C
7.0 x 10-3
D
9.792 x 10-5
Exercise 4 (1 mark for each)
Put each of these numbers into standard form
A
357 000 000 000
B
0.00 000 000 000 000 432
C
456 739 812
D
0.0063421
Exercise 5 (2 marks are awarded for each)
Write each of these constants used in chemistry in standard form
A
Avagadros constant - 602 200 000 000 000 000 000 000 mol-1
B
The gas constant - 8.3145 J mol-1K-1
-4Hanham Woods Academy
Maths For AS Level Chemistry
HELPSHEET 2
Significant Figures and Decimal Places
Why do we need significant figures and decimal places?
If you were asked to measure the length of your page with a ruler you would
probably suggest a length of 29.4cm or something similar, this would appear
reasonable. You would be unlikely to say that the length was 29.38467cm,
this would be unrealistic as you could not possibly measure to that accuracy
with a ruler. The same needs to be applied when calculating answers. If you
start from numbers of 2 significant figures you should only quote your answer
to 2 significant figures as you cannot be more accurate than the information
you are working with.
What is a significant figure or a decimal place?
If a number is quoted to 3 significant figures, abbreviated to 3.s.f., it will
show only three integers other than zero.
Example 1
134 267 000 = 134 000 000 to 3.s.f.
We round the number to the nearest 3rd significant figure. The number above
is nearer to 134 000 000 than 135 000 000 as the fourth number was a 2.
Example 2
986 732 134 = 987 000 000 to 3.s.f.
Here the number has been rounded up as the fourth number was a 7.
If a number is quoted to 2 decimal places, abbreviated to d.pl., it will show
only 2 numbers after the decimal place, if there are no numbers after the
decimal place then zeros are added.
Example 3
234.56111 = 234.56 to 2 d.pl.
The same application as above works with rounding up or down. Note
however that the above number is shown to 2.d. pl. but is in this case 5.s.f.
Example 4
2100 = 2100.00 to 2.d.pl.
Your answers should never be quoted to higher accuracy than the
information you started with in a calculation.
Example 5
Density = mass/volume If Mass = 10.0159g and volume = 2.3cm3
Therefore density = 10.0159/2.3 = 4.35473913 g cm-3
However the answer should only be quoted to 2.s.f as the volume quoted was
to 2.s.f. So, density = 4.3 g cm-3
Now you have a go!
-5Hanham Woods Academy
Maths For AS Level Chemistry
QUESTION SHEET 2
Significant Figures and Decimal Places
Exercise 1 (1 mark each)
Write each of the following to the number of significant figures quoted in
brackets.
A
234 667 (3.s.f)
B
134 000 829 (5.s.f)
C
126 456 090 (2.s.f)
D
128.765 (4.s.f.)
Exercise 2 (1 mark each)
Write each of the following to the correct number of decimal places, quoted in
brackets.
A
213.456 (1 d.pl.)
B
4.567 898 (4 d.pl.)
C
1200 (2d.pl.)
D
63.214 567 32(3 d.pl.)
Exercise 3 (1 mark each)
Write true or false for each one, are they to the stated number of significant
figures or decimal places?
A
23.442 12 = 23.44 (2.d.pl.)
B
965 000 = 970 000 (2.s.f)
C
526.398 769 = 526.399 (3 d.pl.)
D
126 456 = 127 000 (3.s.f)
E
357.432 117 = 357.5 (1 d.pl.)
F
76.98 = 77.0 (1.d.pl)
Exercise 4 (3 marks)
Carry out the appropriate calculation and quote the answer to the correct
number of significant figures.
Density = mass/volume
Mass- 12.3602g
Volume – 4.9cm3
Exercise 5 (3 marks)
Carry out the appropriate calculation and quote the answer to the correct
number of decimal places.
Number of moles = mass/molar mass
Mass of copper = 248.538g
Molar mass of chlorine = 35.5gmol-1
-6Hanham Woods Academy
Maths For AS Level Chemistry
HELPSHEET 3
Ratios and Percentages
Why do we need ratios and percentages?
Ratios and percentages are two methods of expressing mathematical
relationships. In chemistry ratios and percentages are integral to balancing
chemical equations and understanding their relationships.
Example 1
In the compound methane, CH4, the ratio of carbon to hydrogen, C to H, is 1 to
4. This is usually written, C:H, 1:4. As for every carbon there are four hydrogens
present.
Example 2
If the expected amount is 50g but only 40g is obtained, this can be expressed as
a percentage. Percentage obtained = (obtained mass/theoretical mass)
x 100 = 40/50 x100 = 80%
In chemistry,
 The ATOMIC MASS of a compound is the mass of one mole of
atoms of a compound compared to the mass of one twelfth of a
carbon 12 atom. (Abbreviated to Ar, found on the periodic table)
 The EMPIRICAL FORMULA of a compound is the simplest ratio of
elements in a compound.
 The MOLECULAR FORMULA of a compound is the actual number
of atoms of each element in a compound. It can be the same as
the empirical formulae but it may not be.
Example 3
From the percentage composition of the compound the empirical formula can be
found.
In a substance containing hydrogen, oxygen and sulphur only, hydrogen is 2%,
oxygen 65.3%. Therefore the percentage of sulphur is 100 – 65.3 -2 = 32.7%.
If the percentage is turned into mass;
Hydrogen is 2g, oxygen 65.3g and sulphur32.7g.
Divide each mass by the elements Ar, (found on the periodic table)
H 2/1 = 2
O 65.3/16 = 4.08
S 32.7/32 = 1.02
Divide each of the above by the smallest number
H 2/1.02 = 1.96
O 4.08/1.02 = 4.0 S 1.02/1.02 = 1
Therefore the ratio is H:O:S = 2:4:1 The empirical formula is H2SO4
Example 4
If the empirical formula of a compound is known and the Mr or weight of the
compound then the molecular formula can be found.
If the empirical formula is CH3, the mass of the empirical formula can be found
by adding the Ar of the elements present. 12(C ) + (3 x 1)(3 x H) = 15.
The Mr of the compound is given as 30. Therefore 30/15 = 2.
The molecular formula is therefore, 2 x CH3, therefore it is C2H6.
Now you have a go!!!
-7Hanham Woods Academy
Maths For AS Level Chemistry
QUESTION SHEET 3
Ratios and Percentages
Exercise 1 (1 mark each)
Write down the ratio (remember the simplest relationship) for each of the
following.
A
C4H10 ratio of C:H.
B
If there are 2 blue crayons , 6 green crayons and 1 red crayon. What is
The ratio of blue: green :red.
C
If there are 9 fairy cakes 6 chocolate rolls and 3 smarties, what is the
Ratio of fairy cakes to smarties to chocolate rolls.
D
H2O2 ratio of H:O
Exercise 2 (1 mark each)
Work out the percentage obtained for each of the following.
A
30g from a possible 60g
B
45g from a possible 54g
C
90g from a possible 100g
D
25g from a possible 75g
Exercise 3 (1 mark each)
Work out the amount obtained in each of the following.
A
50% of a possible 40grams
B
27% of a possible 90g
C
91% of a possible 63g
D
29% of a possible 27tonnes
Exercise 4 (4 marks)
Work out the empirical formula (simplest possible ratio) for a compound
containing sodium, sulphur (S) and oxygen only, sodium (Na) 32.4%, and
oxygen (O) 45.0%.
Exercise 5 (4 marks)
If the empirical formula is CH2, and the total mass of the compound (Mr) is 70
work out the molecular formula, (actual numbers of each type of atom in the
compound).
-8Hanham Woods Academy
Maths For AS Level Chemistry
HELPSHEET 4
Rearranging Equations
Why do we need to be able to rearrange equations?
In science, including chemistry many equations are used to calculate different
quantities. Although equations are often learnt in one particular format, it is
essential to be able to rearrange them to find any one of the quantities within
the equation.
Example 1
Density = Mass\Volume
But what if you know the volume and density and want to find the mass.
We move volume to the other side of the equation but instead of dividing by
volume we multiply.
Density x Volume = Mass
Therefore if we wanted to calculate volume
Volume = Mass/Density
At GCSE you may have used triangles to make this easier, covering up the
quantity you would like to calculate, this is fine for equations with three
variables but will not work for equations with more than this.
Example 2
The ideal gas equation is;
PV = nRT
P = pressure of the gas in Pascals
R = the gas constant 8.3145Jmol-1K-1
V = volume of gas in m3
T = temperature in K
n = number of moles of gas
If we wanted to calculate the number of moles of gas, n.
n = PV\RT
R and T have been taken to the other side of the equation and now we divide
by them rather than multiply.
We would then plug in the numbers.
Example 3
Using the ideal gas equation (above), to calculate the volume of gas when;
P = 101325 Pa
T = 293K
n = 1.0mol R = 8.3145Jmol-1K-1
V = nRT/P
= (1.0 x 8.3145 x 293)/ 101325
= 0.024m3
Now you have a go!!!!!
-9Hanham Woods Academy
Maths For AS Level Chemistry
QUESTION SHEET 4
Rearranging Equations
Exercise 1 (1 mark each)
Rearrange the following equations to show how you would calculate the
quantity in bold after each equation.
A
Speed = Distance/Time
TIME
B
Moles = Mass/ Molar Mass
MASS
C
Concentration = Moles/Volume
VOLUME
D
Density = Mass/Volume
VOLUME
Exercise 2 (1 mark each)
Rearrange the ideal gas equation to show the calculation needed to find each
of the following quantities.
Ideal gas equation PV = nRT
A
P – Pressure
B
V – Volume
C
n – Number of moles
D
T – Temperature
Exercise 3 (2 marks each)
Using the equation Number of moles = Mass/Molar mass calculate
A
Mass, when number of moles = 3 and molar mass =m 18gmol-1
B
Number of moles when, mass = 784g and molar mass = 98gmol-1
C
Molar mass when, number of moles = 4 and mass = 64g
D
Molar mass when, number of moles = 0.5 and mass = 17g
Exercise 4 (4 marks )
Calculate the number of molecules of methane, using the ideal gas equation,
if
P = 200 000Pascals
V = 0.5m3
R = 8.3145Jmol-1K-1
T = 300K.
- 10 Hanham Woods Academy
Maths For AS Level Chemistry
HELPSHEET 5
Graphs and Gradients
When do we use graphs and gradients?
Graphs have been used throughout scientific history to illustrate data and
draw conclusions on mathematical relationships.
Graph Basics
When drawing a graph you need to follow the simple rules listed below.
1. Always use at least half of the piece of graph paper.
2. Put the variable you have measured, the independent variable, on the
x axis (horizontal).
3. Put the variable being measured, the dependent variable, on the y axis
(vertical).
4. Label both axis including the units with sensible scales.
5. Plot the points carefully with a sharp pencil.
6. Draw a smooth pencil line of a line of best fit with a ruler.( Never just
join the point to give a jagged line.)
Gradients
The gradient of a graph can be used to find the relationship between 2
measurable quantities. A gradient can be calculated by finding the change in
the y axis and dividing by the change in the x axis.
The Equation of a Straight Line Graph Equations of straight lines are in
the form y = mx + c (m and c are numbers). m is the gradient of the line and
c is the y-intercept (where the graph crosses the y-axis).
Example 1
The above graph has equation y = (4/3)x – 2. The gradient = change in y /
change in x = 4 / 3 (m). It cuts the y-axis at -2, and this is the constant in
the equation, (c).
- 11 Hanham Woods Academy
Maths For AS Level Chemistry
Finding the gradient of a curve. To find the gradient of a curve, you must
draw an accurate sketch of the curve. At the point where you need to know
the gradient, draw a tangent to the curve. A tangent is a straight line which
touches the curve at one point only. You then find the gradient of this
tangent.
Example 2 Find the gradient of the curve y = x² at the point (3, 9).
Gradient of tangent = (change in y)/(change in x)
= (9 - 5)/ (3 - 2.3)
= 5.71
Note: this method only gives an approximate answer. The better your graph
is, the closer your answer will be to the real answer.
Now you have a go!!!!
- 12 Hanham Woods Academy
Maths For AS Level Chemistry
QUESTION SHEET 5
Graphs and Gradients
Exercise 1 (12 marks)
Plot a graph of the following data. In the experiment the strength of the acid
was changed, the temperature, pressure, time and amount were kept
constant throughout. The volume of gas produced was recorded each time,
this is a reflection of the reaction rate.
Strength 0.1
of acid
(M)
Volume
3.0
of gas
produced
(cm3)
0.2
0.4
0.8
1.0
2.0
5.8
12.3
23.5
30.3
59.4
Ensure you have joined the data in the most appropriate way.
Exercise 2 (8 marks)
For the above graph can you produce an equation in the form y = mx +c, to
show the mathematical relationship.
Using your equation can you calculate the volume of gas you would expect to
produce if the strength of the acid was;
A
0.5M
B
5.0M
- 13 Hanham Woods Academy
Maths For AS Level Chemistry
HELPSHEET 6
Units
Why do we need units?
You may have one. One what? A pencil? A sweet? Numbers are not
meaningful unless we know what they are in reference to.
Changing units
In chemistry we often work on very small or very large scales that would not
be appropriate for a biologist or physicist. Those scales can be converted and
the relationships between them are well documented. It is essential however
that you can apply that information. There are also many specialist units,
these are always related to more well known units so that scientists can
correspond, however again it is important that you understand and can work
within those units and their conversions.
The SI Units
SI units are the Standard International Units, those accepted by the
international societies, the standard. The bas SI units are;
Property
Base unit
Symbol for property
Mass
Kilogram
kg
Length
Metre
m
Time
Second
s
Temperature
Kelvin
K
Amount of substance
Mole
mol
SI Prefixes
SI prefixes are used to denote multiples of ten of the above units.
Value
Prefix
Symbol
9
10 billion
Giga
G
106 million
Mega
M
3
10 thousand
Kilo
k
-1
10 tenth
Deci
d
-2
10 hundreth
Centi
c
10-3thousanth
Milli
m
-6
10 millionanth
Micro
µ
-9
10 trillionanth
Nano
n
-12
10
Pico
p
Example 1
10 km = 10 000m
k stands for kilo meaning a thousand so there are 10 x 1000 metres.
Example 2
9.8µg = 0.0000098g
µ stands for micro so there are 9.8 x 10-6grams
- 14 Hanham Woods Academy
Maths For AS Level Chemistry
Deriving Units
Mathematical relationships can be used to find the units of a quantity.
Example 3
Density = Mass/Volume
Therefore the units are;
= kilograms/m3
Or we can write this as kg m-3. By changing the 3 to -3 we denote that it is
actually divide rather than times, therefore the division sign is no longer
needed.
Example 4
The ratio E/RT is an expression regularly seen in chemistry.
E= energy per mole of substance (units joules per mole, Jmol-1)
T = temperature (units Kelvin, K)
R = universal gas constant (units joules per mole per Kelvin, Jmol-1K-1)
So the units of the ratio are
Jmol-1 \ (K x Jmol-1K-1)
If you cancel each of the above units that are repeated e.g. J/J can be
cancelled and K x K-1 as this is the same as K/K, you will see that you are left
with no units. Therefore the ratio E/RT is unitless, but then it is a ratio so you
would expect it to be unitless.
Now you have a go!!!!!
- 15 Hanham Woods Academy
Maths For AS Level Chemistry
QUESTION SHEET 6
Units
Exercise 1(1 mark each)
Write each of the following amounts in base SI units, without the prefixes.
A
55 grams
B
237 km
C
3.4 µs
D
345cm
Exercise 2 (1 mark each)
Write each of these amounts with a suitable prefix.
A
0.000 007 897 m
B
234 456 000 kg
C
345 798 000 000 m
D
0.00 000 00098 s
Exercise 3 (4 marks)
Work out the units of molar mass.
Mass = number of moles x molar mass
Mass is measured in grams and number of moles in moles.
Exercise 4 (4 marks)
Work out the units of concentration.
Concentration = number of moles/volume
Number of moles is measured in moles and volume is measured in dm3.
Exercise 5 (4 marks)
Work out the units of specific heat capacity, C.
q = m C ΔT
q = energy (units in joules, J)
m = mass (units in grams, g)
ΔT = change in temperature (units of Kelvin used for temperature, K)
You will need to rearrange the equation first!
Congratulations you have completed your maths for AS Chemistry
course!!!!!!!!!!!!!!!!!!!!
- 16 Hanham Woods Academy