c2.3.3 quantitative chemistry

C2.3 Atomic structure, analysis

and quantitative chemistry

C2.3.3 Quantitative chemistry

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.

Percentage mass

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:

Fertiliser

• Farmer Giles wants to work out which

fertiliser has more nitrogen in it.

?

KNO3 NaNO3

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.

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

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 compoundx 100=% of nitrogen in NH3

17x 100=

(14 x 1)

How much oxygen?

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

Empirical formulaMolecular formula Empirical formula

C2H2

MgO

C4H8

H2O2

C2H6O2

NaCl

C6H12O6

C10H22

H2SO4

Molecular formula Empirical formula

C2H2 CH

MgO MgO

C4H8 CH2

H2O2 HO

C2H6O2 CH3O

NaCl NaCl

C6H12O6 CH2O

C10H22 C5H11

H2SO4 H2SO4

Empirical formula

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.

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%

Answers to the percentage

mass sheet

2) Mass of chlorine = 71.

Mass of calcium hypochlorite = 143.

% mass = (71/143) x 100 = 49.7%

Answers to the percentage

mass sheetMass 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%

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)


Mass of element (g)


Mass


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)


Mass of element (%)


Mass


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


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.

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

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.

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

Previous learning

Higher tier - Be able to do percentage

calculations.

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?

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.


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.


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.

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.

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

© Boardworks Ltd 200631 of 41

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.


Making copper

sulphateWhat 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


Answers to All about yields

questions1) 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.


Answers to All about yields

questions2) 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.


Answers to percentage yield

calculations1) 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%


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.


Homewor

kAnswer 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.


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


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.




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.



homework.

Q2) E.g. A low yield means wasted

chemicals which isn’t sustainable. Increasing

the yield would save resources for the future.

Balancing equations

Learning objectivesIdentify 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.


Previous learning

Higher tier – be able to balance symbol

equations and do mole calculations.

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)

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.

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.

Balanced equationsWe 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

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

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.

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.


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.


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.


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)


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.

Answers to the foundation balancing equations sheet.

1)80g

2)The reaction produces carbon

dioxide which is a gas.

3)Hydrogen peroxide water +

Oxygen


• 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.


• 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.

Answers to the June 2011 Foundation question

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.


Balance the following equations.

a N2 + 3H2 → 2NH3

b 2Ca + O2 → 2CaO

c Br2 + 2KI → 2KBr + I2d 3Fe + 4H2O → Fe3O4 + 4H2

e C3H8 + 5O2 → 3CO2 + 4H2O

f 4NH3 + 5O2 → 4NO + 6H2O


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.


What mass of carbon monoxide is needed to react with

480g of iron oxide?

Fe2O3 + 3CO 2Fe + 3CO2


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.

Complete the neutralization

reaction

What mass of carbon dioxide is formed

when 2kg of calcium carbonate reacts with

hydrochloric acid?

CaCO3 + 2HCl CaCl2 + H2O + CO2

Balancing equations higher

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.

HomeworkHigher 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.

Next lesson…

We will be writing equations for reversible

reactions such as reactions of indicators.

Lesson 8 3.3c – Reversible reactions– slide 164.

Homework answersQ1) 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


Homework answersQ3) 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

Homework answersQ5) 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

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

Reversible reactions

Explain what is meant by a reversible

reaction.

Name some reversible reactions.

Draw a symbol equation for a

reversible reaction.


Previous learning…

Indicators change colour when

added to acids and alkalis.

Changing colour

What reactions have you done where something changes

colour and then changes back?

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

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?

Litmus indicatorLitmus turns red when you add an acid to it.

It turns blue when you add alkali to it.

Blue litmus Red litmus

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

Reversible reactionsWrite these two reactions as a reversible

reaction:

NH4Cl NH3 + HCl

NH3 + HCl NH4Cl

NH4Cl NH3 + HCl

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)

Copper sulphateThe 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.

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%

Next lesson…There is a test on chapter 3.

c2.3.3 quantitative chemistry

Documents

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