1. C2.3 Atomic structure, analysis
and quantitative chemistry

2. C2.3.3 Quantitative chemistry

3. We will be calculating the percentage by mass of an
element in a compound. E.g. hydrogen makes up 11%
of the mass of water (water has a Mr of 18. The two
hydrogens have an Mr of 2).
We will be work out the ratio of mass of different
elements in a compound.

4. Percentage mass

5. Percentage mass
HT only – calculate the empirical formulae
either from a graph or from mass data.
calculate the percentage mass of a named
element in a formula.
BTEOTSSSBAT:

6. Fertiliser
• Farmer Giles wants to work out which
fertiliser has more nitrogen in it.
?
KNO3 NaNO3

7. How does he work it out?
Farmer Giles needs to work out how much
nitrogen there is per mass of fertiliser rather than
just how many nitrogen atoms are.
KNO3 has an Mr of
NaNO3 has an Mr of
101
85
So substance is better and why?
NaNO3 is better because it has a lower mass and
so more of the mass is nitrogen.

8. How is percentage by mass calculated?
The percentage by mass of an element in a compound is
sometimes known as the percentage composition.
Percentage by mass is calculated using Ar and Mr.
x 100% element =
Ar of element x number of atoms
Mr of compound

9. Calculating percentage by mass – example 1
What percentage by mass of nitrogen is in ammonia (NH3)?
(r.a.m.: H = 1, N = 14)
Step 1: Work out the relative formula mass (r.f.m.) of NH3.
Step 2: Work out the percentage by mass of nitrogen.
= 1 nitrogen atom + 3 hydrogen atoms
= (1 x 14) + (3 x 1)
= 17
r.f.m. of NH3
= 82%
r.a.m. x number of atoms
r.f.m. of compound
x 100=% of nitrogen in NH3
17
x 100=
(14 x 1)

10. How much oxygen?

11. Empirical formula (HT only)
The empirical formula is the ratio of moles of
elements in a compound but it is written as a
formula instead of a ratio.
E.g. The empirical formula of C2H6 is CH3

12. Empirical formula
Molecular formula Empirical formula
C2H2
MgO
C4H8
H2O2
C2H6O2
NaCl
C6H12O6
C10H22
H2SO4

13. Molecular formula Empirical formula
C2H2 CH
MgO MgO
C4H8 CH2
H2O2 HO
C2H6O2 CH3O
NaCl NaCl
C6H12O6 CH2O
C10H22 C5H11
H2SO4 H2SO4
Empirical formula

14. Burning magnesium planSources of risk:
• The crucible stays hot after being heated.
• Burning magnesium makes a bright white light.
Explain why they are sources of risk:
• Hot crucibles provide a burn risk.
• The bright light may damage eyes.
List safety measures to reduce the risk:
• Use tongs to handle the crucible.
• Do not look directly at the light.

15. Answers to the percentage mass
sheet
1) a) i) There are 9 atoms.
ii) Mass of nitrogen = 28.
Mass of ammonium nitrate = 80.
Percentage mass of nitrogen = (28/80) x 100 =
35%

16. Answers to the percentage
mass sheet
2) Mass of chlorine = 71.
Mass of calcium hypochlorite = 143.
% mass = (71/143) x 100 = 49.7%

17. Answers to the percentage
mass sheet
Mass of iron = 56.
Mass of iron sulphate = 56 (for iron) + 32
(for sulphur) + (4 x 16) = 64 (for oxygen) =
152
56/152 = 36.8%
Percentage of nitrogen in iron sulphate =
36.8%

18. Empirical formula answersElement name
Mass of element (g)
Mass ÷ Relative Atomic Mass
Divide by the smaller number
Ratio
Carbon
72
72 ÷ 12 = 6
6 ÷ 6 = 1
1
Hydrogen
12
12 ÷ 1 = 12
12 ÷ 6 = 2
2
The empirical formula of this compound is CH2.
Element name
Mass of element (g)
Mass ÷ Relative Atomic
Mass
Divide by the smaller
number
Ratio
Sodium
52.9
52.9 ÷ 23 = 2.3
2.3 ÷ 2.3 = 1
1
Chlorine
81.7
81.7 ÷ 35.5 = 2.3
2.3 ÷ 2.3 = 1
1
The empirical formula of this compound is NaCl.
1
1)
2)

19. Empirical formula answersElement name
Mass of element (g)
Mass ÷ Relative Atomic
Mass
Divide by the smaller
number
Ratio
Aluminium
156
156 ÷ 27 = 5.8
5.8 ÷ 5.8 = 1
1
Oxygen
278
278 ÷ 16 = 17.4
17.4 ÷ 5.8 = 3
3
The empirical formula of this compound is AlO3.
Element name
Mass of element
(g)
Mass ÷ Relative
Atomic Mass
Divide by the
smaller number
Ratio
Carbon
25.2
25.2 ÷ 12 = 2.1
2.1 ÷ 2.1 = 1
1
Hydrogen
8.5
8.5 ÷ 1 = 8.5
8.5 ÷ 2.1 = 4.0
4
Oxygen
33.7
33.7 ÷ 16 = 2.1
2.1 ÷ 2.1 = 1
1
The empirical formula of this compound is CH4O.
3)
4)

20. Empirical formula answers
Element name
Mass of element (%)
Mass ÷ Relative Atomic
Mass
Divide by the smaller
number
Ratio
Magnesium
60
60 ÷ 24 = 2.5
2.5 ÷ 2.5 = 1
1
Oxygen
40
40 ÷ 16 = 2.5
2.5 ÷ 2.5 = 1
1
The empirical formula of this compound is MgO.
5)
6) Element name
Mass of element (%)
Mass ÷ Relative Atomic
Mass
Divide by the smaller
number
Ratio
Carbon
80
80 ÷ 12 = 6.7
6.7 ÷ 6.7 = 1
1
Hydrogen
20
20 ÷ 1 = 20
20 ÷ 6.7 = 3
3
The empirical formula of this compound is CH3.

21. Answers to the keyword
homework
Yield: The amount of product we get.
Predicted yield: The amount of product that you should get.
Actual yield: The yield that you actually get from the practical.
Percentage yield: (actual yield ÷ percentage yield) x 100
Contents page – slide 1

22. Answers to the keyword
homework
Reactants: The chemicals that you start with.
Products: The chemicals that you end up with.
Reversible reaction: A reaction where some of the
products can turn back into reactants.

23. Percentage yield
Explain why some reactions do not go to completion.
HT only – calculate the percentage yield.
Define the words yield and percentage yield.
BTEOTSSSBAT

24. Previous learning
Higher tier - Be able to do percentage
calculations.

25. Burning magnesium
If we burn 1g of magnesium, we make 1.67g of magnesium
oxide.
Multiply the mass of magnesium you used in last lesson by 1.67
to calculate how much magnesium oxide that you should have
made.
Did you make as much as you should have done?

26. Magnesium oxide yield
Why did we make less magnesium oxide than we should
have done?
Some of the magnesium vaporises
before it reacts with oxygen.
Some of the magnesium did not
react with oxygen because it wasn’t
in contact with the oxygen.

27. Magnesium oxide yield
How did putting the crucible on the balance make the yield
higher than taking the magnesium out of the crucible and
putting that on the balance?
If we would have taken the magnesium out of the crucible,
then some would have been left behind.

28. Magnesium oxide yield
Why did we clean the crucible out before we did the reaction?
If there were other substances in the crucible, then the
magnesium may have reacted in a different way and reduced
the yield.

29. Yield and percentage yield
• The yield is the amount of product that we get from a reaction.
• The percentage yield is the amount of product that we get
from a reaction as a percentage of how much we expected to
get from the reaction.

30. Percentage yield (HT)
Percentage yield = how much we get out of the reaction
how much we expect to get out of the reaction
X 100

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Things that affect percentage
yield
• A reaction may not go to completion
because it is reversible (the products turn
into reactants again)
• Some of the product may be lost when it is
separated from the reaction mixture.
• Some of the reactant may react in ways
different from the expected reaction.

32. © Boardworks Ltd 200632 of 41
Making copper
sulphate
What yield did you obtain? (H
only)
Why was the yield lower than
expected? (F and H)
Extension: Why do we expect to
make 1.6g of CuSO4 from 0.8g of
CuO? (H only

33. © Boardworks Ltd 200633 of 41
Answers to All about yields
questions
1) Reactants may not be completely pure.
• There may be other products made.
• It could be hard to separate the product from impurities.
• Some of the product may be left in the reaction vessel.

34. © Boardworks Ltd 200634 of 41
Answers to All about yields
questions
2) Reactants are not pure – purify the reactants before using them in
the reaction.
• Other products made – try to reduce the number of other products
made by altering the reaction, or separating these out.
• Some product left in the vessel – empty and rinse this out
thoroughly.
• Hard to separate the impurities – reduce the number of impurities
or use a better separation technique.
3) Recycle this reactant back into the reaction and use it again.

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Answers to percentage yield
calculations
1) Percentage yield = (36 ÷ 120)  100 = 30%
2) Percentage yield = (88 ÷ 110)  100 = 80%
3) Percentage yield = (2.3 ÷ 12.7)  100 = 18.1%
4) m/15 = 85/100
m = 0.85  15 = 12.8 g
5) 57.2 ÷ 13 = 4.4 g.
So 1% = 4.4 g.
143 ÷ 4.4 = 32.5%

36. © Boardworks Ltd 200636 of 41
Improvement
Write down:
• One way you could improve your practical
skills.
• One way the practical can improve to give
a better percentage yield.
• One way to improve the answers to your
questions.

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Homewor
k
Answer the questions in the white workbook page
entitled Percentage Yield and Reversible Reactions.
This is page 67 for additional science workbook and
page 71 for the chemistry workbook.
Foundation students answer Q1 a,c,d, Q2. Higher
students answer all questions.

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Next lesson…
Hier tier – we will be balancing equations.
Higher tier - we will be using equations to
calculate the amount of product you can get
if you are given the amount of reactants.
Lesson 7 3.3c –
Balancing equations–
slide 132

39. © Boardworks Ltd 200639 of 41
Answers to the Percentage Yield and Reversible Reactions
homework.
Q1 a) Yield, higher, percentage yield,
predicted.
b) (6 ÷ 15) x 100 = 40%
c) When the solution was filtered, a bit of
barium sulphate may have been lost. Less
product means lower percentage yield.
Contents page – slide 1

40. © Boardworks Ltd 200640 of 41
Answers to the Percentage Yield and Reversible Reactions
homework.
Q1) d) i) Not all the reactants turn into
products because the reaction goes both
ways. So the percentage yield is reduced.
ii) The unexpected reaction will use up
reactants, so there’s not as much left to make
the product you want. So the percentage
yield is reduced.

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Answers to the Percentage Yield and Reversible Reactions
homework.
Q2) E.g. A low yield means wasted
chemicals which isn’t sustainable. Increasing
the yield would save resources for the future.

42. Balancing equations
Learning objectives
Identify equations as balanced or not balanced
and write them out. Name substances from
their symbols.
HT only Balance symbol equations.
HT only Calculate the masses of individual
products from a given mass of a reactant
and the balanced symbol equation.
Contents page – slide 1

43. Previous learning
Higher tier – be able to balance symbol
equations and do mole calculations.

44. Neutralisation
If I have 25cm3 of sodium hydroxide
in the conical flask, predict how much
hydrochloric acid will be needed to
neutralise it.
The equation is:
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
Extension (H only):
How much sulphuric acid will be needed to neutralise the
alkali?
H2SO4(aq) + 2NaOH(aq)  H2O(l) + Na2SO4(aq)

45. Neutralisation
How did the equation help you to work out the amounts of
acid needed to neutralise the alkali?
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
This equation shows us that 1 mole of HCl reacts with 1
mole of NaOH so you need the same amount of HCl to
neutralise 25cm3 NaOH.

46. Neutralisation
H2SO4(aq) + 2NaOH(aq)  H2O(l) + Na2SO4(aq)
This equation shows us that 1 mole of H2SO4 reacts
with 2 moles of NaOH so you need half the amount of
H2SO4 to neutralise 25cm3 of NaOH.

47. Balanced equations
We need balanced equations to predict how much of each
substance will react.
An equation is balanced when there are the same numbers
of atoms of each side.
CH4 + 2 O2  CO2 + 2 H2O

48. Balancing equations
Carbon
Hydrogen
Oxygen
+ +
Are the carbons balanced?YesAre the hydrogens balanced?NoLet’s add more waterAre the oxygens balanced?NoLet’s add more oxygenIs the equation balanced?Yes.
CH4 CO2O2 H2O++ 22

49. Balanced equations
CH4 + 2 O2  CO2 + 2 H2O
This number is the
number of atoms in
the molecule. It
never changes.
This is the number
of molecules of
oxygen.
You can change
these numbers.
The combined mass of the
reactants is equal to the
combined mass of the
products.

50. Working out masses from balanced equations.
Work out the mass of carbon dioxide made from 24g of
carbon.
C + O2  CO2
1) Work out the number of moles of carbon using moles = mass/Ar.
Moles = 24/12 = 2 mol.

51. Working out masses from balanced equations.
2) Using the equation, look at how many moles of CO2 are
made from 1 mole of C.
Both substances have no numbers in front of
them, so 1 mole of carbon makes 1 mole of CO2.

52. Working out masses from balanced equations.
3) Work out how many moles of CO2 you have.
.
4) Using mass = moles x Mr work out the mass of CO2.
We have 2 moles of carbon, so we make 2 moles of CO2
The Mr of CO2 is 44. 2 x 44 = 88g.

53. Working out masses from balanced equations.
What mass of sulphur trioxide is formed from 96g of sulphur
dioxide?
2SO2 + O2  2SO3
SO2 has an Mr of 64 (32 for sulphur + 16 x 2 for the oxygens).
There are 1.5 moles of sulphur dioxide (96g/64 = 1.5)

54. Working out masses from balanced equations.
2 moles of sulphur dioxide form 2 moles of sulphur trioxide so
Sulphur trioxide has an Mr of 112.
1.5 x 112 = 168g of sulphur trioxide produced.
1.5 moles of sulphur dioxide form 1.5 moles of sulphur
trioxide.

55. Answers to the foundation balancing equations sheet.
1)80g
2)The reaction produces carbon
dioxide which is a gas.
3)Hydrogen peroxide  water +
Oxygen

56. Answers to the foundation balancing equations sheet.
• There are six carbons on the left hand side of the
equation and six carbons on the right hand side.
• The equation has twelve hydrogens on the left hand
side of the equation and twelve hydrogens on the right
hand side.
• There are six oxygens on the left hand side of the
equation and six oxygens on the right hand side.

57. Answers to the foundation balancing equations sheet.
• There are four carbons on the left hand side of the equation and four
carbons on the right hand side.
• The equation has twelve hydrogens on the left hand side of the
equation and twelve hydrogens on the right hand side.
• There are fourteen oxygens on the left hand side of the equation and
fourteen oxygens on the right hand side.

58. Answers to the June 2011 Foundation question

59. Answers to the June 2011 Foundation question

60. Answers to the higher balancing equations sheet.
1) This is incorrect because the candidate
has changed the small numbers. The
correct answer is
Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g)
2) This answer is correct.

61. Answers to the higher balancing equations sheet.
Balance the following equations.
a N2 + 3H2 → 2NH3
b 2Ca + O2 → 2CaO
c Br2 + 2KI → 2KBr + I2
d 3Fe + 4H2O → Fe3O4 + 4H2
e C3H8 + 5O2 → 3CO2 + 4H2O
f 4NH3 + 5O2 → 4NO + 6H2O

62. Answers to the higher balancing equations sheet.
4) The Ar of magnesium is 24.
192/24 = 8 moles of magnesium.
2 moles of magnesium react with 1 mole of
oxygen so we have 4 moles of oxygen.
The Mr of oxygen is 32. 4 x 32 = 128.
128g of oxygen reacts with 192g of
magnesium.

63. Answers to the higher balancing equations sheet.
What mass of carbon monoxide is needed to react with
480g of iron oxide?
Fe2O3 + 3CO  2Fe + 3CO2

64. Answers to the higher balancing equations sheet.
The Ar of iron oxide is 160.
480/160 = 3 moles of iron oxide.
1 moles of iron oxide reacts with 3 moles of carbon
monoxide so we have 9 moles of carbon monoxide.
The Mr of carbon monoxide is 28. 9 x 28 = 252.
The mass of carbon monoxide is 252g.

65. Complete the neutralization
reaction

66. What mass of carbon dioxide is formed
when 2kg of calcium carbonate reacts with
hydrochloric acid?
CaCO3 + 2HCl  CaCl2 + H2O + CO2
Balancing equations higher

67. Answer to the balancing equations problem.
The Ar of calcium carbonate is 100.
There are 2000g in 2kg
2000/100 = 20 moles of calcium carbonate .
1 moles of calcium carbonate forms 1 mole of carbon
dioxide, so we have 20 moles of carbon dioxide.
The Mr of carbon dioxide is 44. 20 x 44 = 880.
The mass of carbon dioxide is 880g.

68. Homework
Higher students do questions on the white
workbook on the pages entitled calculating
masses in reactions and percentage yield
and reversible reactions.
This is pages 69-70 of the
chemistry workbook.
This is pages 66-67 of the
additional science workbook.

69. Next lesson…
We will be writing equations for reversible
reactions such as reactions of indicators.
Lesson 8 3.3c – Reversible reactions– slide 164.

70. Homework answers
Q1) a) 2Mg + O2  2MgO
b)
Q2)
2Mg 2MgO
2x24 = 48 2x(24+16) = 80
48÷48 = 1g 80÷48 = 1.67g
4Na 2Na2O
4x23=92 2x[(2x23)+16] = 124
92÷124 = 0.74g 124÷124 = 1g
0.74x2 = 1.5g 1x2 = 2g
Contents page – slide 1

71. Homework answers
Q3) a) 2Al + Fe2O3  Al2O3 + 2Fe
b)
Q4) CaCO3  CaO + CO2
Fe2O3 2Fe
[(2x56)+(3x16)]=160 2x56=112
160÷160=1 112÷160=0.7
1x20=20g 0.7x20 = 14g
CaCO3 CaO
40+12+(3x16)=100 40+16=56
100÷56 =1.786 56÷56=1
1.786x100=178.6kg 1x100=100kg

72. Homework answers
Q5) a)
b) It could be recycled and used in stage B
– all this is used in stage C.
C 2CO
12 2x(12+16)=56
12÷12 =1g 56÷12=4.67g
1x10=10g 4.67x10=46.7g

73. Homework answers
Q6) a) 2NaOH + H2SO4  Na2SO4 + 2H2O
b)
c)
2NaOH Na2SO4
2x(23+16+1)=80 (2x23)+32+4(4x16)=142
80÷142 = 0.56g 142÷142 = 1g
0.56x75=42g 1x75=75g
H2SO4 2H2O
(2x1)+32+(4x16)=98 2x[(2x1)+16]=36
98÷98=1g 36÷98=0.37g
1x50=50g 0.37x50=18.5g

74. Reversible reactions
Explain what is meant by a reversible
reaction.
Name some reversible reactions.
Draw a symbol equation for a
reversible reaction.
Contents page – slide 1

75. Previous learning…
Indicators change colour when
added to acids and alkalis.

76. Changing colour
What reactions have you done where something changes
colour and then changes back?

77. Reversible reactions
A reversible reaction is a reaction where the products
can turn back into the reactants.
We could write the reactions as:
Reactants  Products
Products  Reactants
But we could use a reversible reactions symbol to
write it as one equation.
reactants products

78. Litmus indicator practical• Add 2cm3 of HCl to a test tube.
• Add a few drops of blue litmus indicator.
• Add sodium hydroxide solution until the indicator
changes colour.
• When it has changed colour, add acid again.
• Does the colour change back? Is this an example of
a reversible reaction? What is causing the colour
change?

79. Litmus indicatorLitmus turns red when you add an acid to it.
It turns blue when you add alkali to it.
Blue litmus Red litmus

80. Equation for litmusAll acids release Hydrogen ions (H+)
The formula for blue litmus is shortened to Lit-.
When we add acid to blue litmus, the hydrogen bonds to the
litmus forming red litmus with the formula HLit.
Lit- + H+ HLit

81. Reversible reactions
Write these two reactions as a reversible
reaction:
NH4Cl  NH3 + HCl
NH3 + HCl  NH4Cl
NH4Cl NH3 + HCl

82. Copper sulphate
How can one of these reactions be used to test for the presence of
water?
Hydrated
copper
sulphate
Anhydrous
Copper
Sulphate
(copper sulphate
without water)

83. Copper sulphate
The reaction can be used to test for water if
we put a liquid on the anhydrous copper
sulphate. If it turns blue, there is water
present.

84. Answers to the extracting copper sheet.
1) 2CuO + C  2Cu + CO2
2) CuO 80 C 12 Cu 64 CO2 44
3) 171
4) I can make 2 moles of copper.
5) 160/80 = 2 moles
6) I can make 2 moles of copper.
7) I can make 128g of copper.
8) (102.4 ÷ 128) x 100 = 80%

85. Next lesson…
There is a test on chapter 3.