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GCSE CCEA Chemistry Quantitative Chemistry Complete Revision Summary

Quantitative Chemistry

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Quantitative Chemistry

1. a) Calculating Formulas Mass
2. b) Calculating Moles from Masses
3. c) Calculating Moles from Volume
4. d) Calculating Moles from Concentration
5. e) Calculations from Balanced Chemical equations
6. f) Percentage Yield
7. g) Atom Economy
8. h) Titrations

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RELATIVE FORMUALE MASS

It is the sum of relative atomic mass of all the atoms present in a compound.

The ratio of the average mass of one atom compared to 1/12th of the mass of C-12

Eg – CO₂ = Mass of C + 2(mass of O)

= 12+ (2× 16)

= 44g

CaCO₃ = Mass of Ca + Mass of C + 3(mass of O)

= 40+12+ (3× 16)

= 100g

Ca(PO₄)₃ = Mass of C + 3 × mass of P+ 12(mass of O)

= 310g

MOLES  – Amount of substance.

Examples

Q1 Calculate the moles of following

1. 22g of CO₂ Moles = Mass/Mr = 22/44 = 0.5 moles

1. 17g of NH₃ Mass/Mr = 24/17 =2 moles

1. 48dm³ of O₂ Volume/24 = 48/24 = 2 moles

1. 24000cm³ of CO₂ 24000/1000dm³ = 24dm³ = 24/24 = 1 mole

1. 20g of NaOH dissolved in 50cm³ of Solution Moles = 20/40 = 0.5 moles

Q2 Calculate the mass of :-

1. a) 2 moles of calcium carbonate Mass = Moles × Mr =Mr CaCO₃ =100 = 2 × 100g= 200g
2. b) 0.1 moles of hydrochloric acid Mr of Hcl = 36.5 = 36.5×0.1 = 0.365g

Q3 Calculate the concentration of the following

1. a) 2 moles of NaOH dissolved in 10 dm³ of solution = Concentration = Moles/V(dm)³ = 2/10 = 0.2 mol dm^-3
2. b) 20 g of NaOH dissolved in 50 cm³ of solution = Moles of NaOH = 20/40 = 0.5 moles

C = 0.5/0.05 moldm^-3 = 10moldm^-3

V= 50/1000 = 0.05dm³

Avogadros Constant

1 mole = 6.02 x 10²³ atoms

Q1 Calculate the number of molecules in the following :

1. 49 g of sulphuric acid

Mr of H₂SO₄ = 98g

=98/49 =0.5 moles

= 0.5 x 6.02 x 10²³ atoms

= 3.01 x 10²³ atoms

1. 8 g of oxygen gas

Mr of O₂ = 32g

=8/32 =0.25 moles

= 0.25 x 6.02 x 10²³ atoms

= 1.50 x 10²³ atoms

1. 48 dm³ of Nitrogen

Mr of N₂ =48/24 = 2 moles

= 2 x 6.02 x 10²³ atoms

= 1.20 x 10²⁴ atoms

1. d) 50 cm³ of 0.1 mol dm³ of sodium hydroxide

Moles = C x V = 0.1 x 50/1000 = 5 x 10^-3 moles

=5 x 10^-3 x 6.02 x 10²³ atoms

= 3.01 x 10²¹ atoms

Quantitative Chemistry

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Balanced Chemical Equations

C + O₂                        CO₂

One mole of carbon is reacting with one mole of oxygen to form one mole of carbon dioxide.

2Al + 3O₂                       2Al₂O₃

2 moles aluminium reacts with 3 moles of oxygen to form 2 moles of aluminium oxide.

Mg + 2HCl                       MgCl₂ + H₂

One mole of magnesium is reacting with 2 moles of hydrochloric acid to form one moles of magnesium chloride and one mole of hydrogen gas.

 
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CALCULATIONS FROM EQUATIONS

Calculate the mass of Magnesium oxide produced from 6 g of Magnesium when burned complete y in air ?

STEP 1: Write the balanced chemical equation.

2Mg + O₂                        2MgO

STEP2: Write known to the left and unknown to the right

Mg = MgO

STEP3: Write the moles relationship from the balanced chemical equation

2moles = 2 moles

STEP4: Convert moles into mass

48g = 80g

STEP5: Do the Maths.

48/48 = 1g                                                 80/48 = 1.66g

1g× 6 = 6g                                                 1.66g × 6 = 10g

Calculate the mass of alumium needs to produce 306 g of Aluminium Oxide ?

STEP 1: Write the balanced chemical equation.

4Al + 3O₂                        2Al₂O₃

STEP2: Write known to the left and unknown to the right

2Al₂O₃                         Al

STEP3: Write the moles relationship from the balanced chemical equation

2moles = 4 moles

STEP4: Convert moles into mass

204g = 108

STEP5: Do the Maths.

306g                       108/204 x 306 = 162g
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LIMITING REAGENTS

To work out the limiting reagent work out moles of all the reagents and then work from the balanced chemical equation which reagent is in excess and which is in limiting

The reactant that get all used up completely is the limiting reagent. The other reagent which is present in greater quantity than required is in excess.

13.5 gm Aluminium reacts which 32 g of Oxygen

1. a) Work out the moles of Aluminium and Oxygen
2. b) Which reagent is limiting and which is in excess
3. c) Calculation the mass of aluminium oxide produced ?

4Al + 3O₂                        2Al₂O₃

13. Moles of Al = 13.5/27 = 0.5 moles

Moles of O₂ = 32/32 = 1 moles

1. 4 moles of Alumunium = 3 moles of oxygen

1 moles of alumnum = ¾ O₂

0.5 mol of aluminium = ¾ x 0.5 = 0.375

So 0.375 moles oxygen is required hence oxygen is in excess and alumunium is limiting

1. Al = A₂O₃

4 moles = 2 moles

• 2/4

0.5 – 2/4 x 0.5

=0.25 x 102

=25.5g
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PERCENTAGE YIELD

% yield = observed mass/expected mass x 100

= Actual yield/Theoretical yield x 100

When 28 g of nitrogen combined with hydrogen 30 g of ammonia is made ? Calculate the percentage yield ?

N₂ + 3H₂                        2NH₃

N₂ = NH₃

1mole = 2moles of ammonia

28g = 32g of ammonia

Expected = 32g

Observed = 30g

% yield = 30/32 x 100 = 93.75%

When 80 g of iron oxide reacts with carbon, 20 g of iron is produced. Calculate the % yield ?

2Fe₂O₃ + 3C               Fe + 3CO₂

Fe₂O₃ = Fe

2 moles = 4 moles

320g = 224 g

1g =224/320

80g = 224/320 x 80 = 56g

20/56 x 100 = 35.71%

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WHY PERCENTAGE YEILD IS NOT 100 %

The percentage yield can never by greater than 100%. It is very difficult to get 100% percentage yield but scientist always look for that route that gives maximum percentage yield.

• Reason for not getting 100% yield
• The reaction does not go to completion so complete products are not formed.
• Some of the reaction can start moving to the reverse direction if they are reversible
• Some of the reaction can go and form alterative or unwanted product.
• During the reaction some of the reactants can get lost or stick to the reaction vessel so do not react.
• If the reaction involves gaseous reactants they can escape.
• Some of the products can also get lost in the reaction vessel. If the reaction involves gaseous products they can also escape.
• The reagent might not be pure therefore did not react completely to give the desired yield.

Atom Economy

Atom economy = Mr of desired product/Total Mr of all the Products x 100

Calcium Oxide is produced using the following reaction

CaCO₃                   CaO + CO₂

Calculate the atom economy

Desired Product = CaO =56g

Mass of reactants = CaCO₃ = 100g

Atom Economy = 56/100 x 100 = 56%

Iron oxide is reduced by carbon to form iron and carbon monoxide

Fe₂0₃ + 3C                   2Fe +3CO

Calculate the atom economy

Desired Product = 2Fe =56 x 2 = 112g

Mass of reactants = Mass of Fe₂CO₃ + 3(Mass of C) = 196g

% Yield = 112/196 x 100 = 57%

REACTIONS WITH 100 % ATOM ECONOMY

• Addition reaction
• Reactions with only one product have 100 atom economy
• How to increase atom economy?
• Chemist should look for reaction that produce single product
• If the by products are produced they should look for recycling the byproducts or use them in some other reactions to increase the atom economy.

Titration

It is the technique used to determine the exact volume of acid and base to carry out a neutralization reaction.

TITRATION PROCEDURE

• Known Concentration solution either acid or base is measured by the pipette and is added into the conical flask. For this solution both concentration and volume is known.
• The indicator is also added in the conical flask. When the indicator changes colour the end point is reached i.e the solution gets completely neutralised.
• The unknown concentration solution is added into the burette. The starting volume is noted from the burette. The tap is then opened and the unknown solution is added dropwise into the conical flask with regular mixing.
• As soon as the indicator changes colour, the tap is closed and the final reading from the burette is noted.
• The entire process is repeated three times and the values are noted in the following format.
• The concordont reading are taken. The anamolous results are not taken into account.
• The means of the concordont readings are noted and used in the calculation

Titration Example

10 cm³ of 0.5 mol dm³ of sulphuric acid is titrated with sodium hydroxide. The results are given in the table.

2NaOH + H₂SO₄                  Na₂SO₄ + H₂0

Calculate the concentration of sodium hydroxide required to completely neutralize the acid ?

Steps

1. Write the balanced chemical equation

2NaOH + H₂SO₄                  2NaOH + H₂0

1. Underneath each equation write the numerical value given for each

Moles = 5 x 10^-3 x 2 = 10^-2 moles

1. The quantity that has two value use the concentration triangle to find the moles

V = 10cm³

= 0.01dm³

C = 0.5 mol dm^-3

Moles = C x V

0.01 x 0.5=5 x 10^-3 moles

1. Use the molar ratio to find the moles of unknown quantity
2. Use titration volume and find the concentration

Mean Titre = 25.24+25.34+ 25.29/3

= 25.29cm3

Q1 In a Titration 50 cm³ of 0.1 mol dm³ of potasisum hydroxide is neutralized by 20 cm3 of sulphuric Acid.

Calculate the concentration of sulphuric acid.

1. a) Write the balanced chemical equation

2KOH + H₂SO₄                        K₂SO₄ + 2H₂O

1. b) Underneath each equation write the numerical value given for each
2. c) The quantity that has two value use the concentration triangle to find the moles
3. d) Use the molar ratio to find the moles of unknown quantity
4. e) Use titration volume and find the concentration

V = 0.05dm³ V = 0.02 dm³

C = 0.1 mol dm^-3

= 0.005 moles

m = 0.0025 mole

C = 0.125 mol dm^-3

Key Terms

• Relative atomic mass — It is the ratio of the average mass of an atom compared to one twelfth of the mass of carbon-12
• Relative formula mass — It is the sum of relative atomic masses of all the atoms present in a formulae
• Moles – It is the amount of substance that has the same number of particles found in 12 g of carbon-12.
• Avogadros Constant —Number of particles present in one mole of the substance.
• 1mole = 6.02 x 10²³ atoms
• Limiting Reagent – It is the reagent that is completely used up in the reaction
• Yield— The mass of desired product obtained in a chemical reaction
• Percentage yield = Actual yeild /Theoretical Yield x 100
• Atom Economy = Mass of the desired product/total mass of all the reactants x 100

Concentration = Mass of Solute(g)/Volume of Solution(dm³)

Titration — It is the technique used to determine the exact volume of acids and alkali required to carry out complete neutralization.

Neutralization Reaction Reaction in which acid and base react to form salt and water.

Acids—Substance that has pH less than T

Alkali — Soluble bases that has pH greater than 7

Pipette – It is a glass tube with a bulge in the middle that is used to take out the exact volume of known concentration solutiom

Burette— It is a long tube with the.tap at the bottom that is used to measure the titre

Concordant – Values which are in the range of difference 0.1 to 0.2 cm³

TEST YOURSELF

Q1 Calculate the number of moles in the following

1. 10 g of calcium carbonate

Mr = CaCO₃ = 100g = 10/100 = 0.1 moles

1. 98 g of sulphuric acid

Mr = H₂SO₄ =98/98 = 1 mole

1. 18 g glucose

C₆H₁₂0₆

Mr = 180

18/180 = 0.1 moles

1. 90 cm3 of oxygen gas

Firstly convert 90cm³ in to dm³

90/1000 = 0.09 dm³

0.09/24 = 0.00375 moles

1. 10 cm3 of 0.2 mol dm3 sodium hydroxide solution

0.2 x 10/1000 = 0.002 moles

Q2 What is the mass of magnesium required to form 160 g of magnesium oxide ?

MgO    =       Mg

Mg + O₂                   2MgO

MgO    =       Mg

Moles     1                  1

Mass 4Og               24g

160g              96g

= 96g

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Q3 Hydrogen react with chlorine to form hydrogen chloride gas. When 71 g of chlorine reacts 70 g of hydrogen chloride is obtained. Calculate the percentage yield.

H₂ + Cl₂                             2HCl

Cl₂                            HCl

Cl₂    =       HCl

Moles     1                  2

Mass   71g               73g

160g              96g

% yield = 70/73 x 100 = 95.8%

Q4 Calculate the atom economy of both of these reactions:-

N₂ + 3H₂                             2NH₃

34/28+6 x 100 = 100%

CaCO₃                             CaO + CO₂

40+16/100 x100 = 56%

Q5 50 cm3 of 0.2 mol dm3 of HCI reacts with 10 cm3 of NaOH. Determine the concentration of NaOH ?

NaOH + HCl                             NaCl + H₂O

V = 50/1000 = 0.05dm³

C = 0.2

Moles = 0.1

NaOH V = 10/1000 = 0.01dm³

Moles = 0.1 moles

C = 10 moldm^-3

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Disclaimer:

I have tried my level best to cover the maximum of your specification. But this is not the alternative to the textbook. You should cover the specification or the textbook thoroughly. This is the quick revision to help you cover the gist of everything. In case you spot any errors then do let us know and we will rectify it.

References:

BBC Bitesize

Wikipedia

Wikimedia Commons

Image Source:

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Commons

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Make sure you have watched the above videos and are familiar with the key definations before trying these questions. It is also good to time yourself while doing these questions so that you can work on the speed as well.

Quantitative Chemistry

• Amount of Substance 1 MS
• Amount of Substance 1 QP
• Amount of Substance 2 MS
• Amount of Substance 2 QP
• of Mass, Chemical Measurements & Eqns 1 MS
• of Mass, Chemical Measurements & Eqns 1 QP
• of Mass, Chemical Measurements & Eqns 2 MS
• of Mass, Chemical Measurements & Eqns 2 QP
• of Mass, Chemical Measurements & Eqns 3 MS
• of Mass, Chemical Measurements & Eqns 3 QP
• Using Conc. of Solutions in mol(dm)-3 MS
• Using Conc. of Solutions in mol(dm)-3 QP
• Yield & Atom Economy MS
• Yield & Atom Economy QP