This page contains the AQA GCSE  Chemistry C4 Chemical Calculations Questions and kerboodle answers for revision and understanding Chemical Calculations.This page also contains the link to the notes and video for the revision of this topic. 

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C4.1 Relative Masses and Moles AQA GCSE Chemistry C4 Analytical Chemistry Kerboodle Answers  Page No 63

1. Answer.

Relative atomic mass is the ratio of the average mass of atoms of an element compared to 1/12 of the mass of 12C6   atom.

2. a. Answer.

Relative formula mass of Magnesium fluoride MgF2=  24 + (19 x 2) = 62 g

1. Answer.

Relative formula mass of glucose C6H12O6=  (12 x 6) + (1 x 12) + (16 x 6) = 180 g

3. a Answer.

In 0.02 g of helium

Moles = Mass/Mr

Mr of Helium = 4g

Moles = 0.02/4

= 0.005 moles are present in 0.02g of helium

3. b (i) In 9.6 g of Sulphur

Moles = Mass/Mr

Mr of Sulphur = 32 g

Moles = 9.6/32

= 0.3 moles of sulphur are present in 9.6 g of sulphur

b(ii) Moles = Mass/Mr

Mr of Sulphur = 32 g

Mass = 16 tonnes = 16 X 106 g

Moles =16 X 106 / 32

=  5 X 105 moles are present in 16 tonnes of sulphur

4. a. Answer.

Mass = Mr X Moles

= Mr of CaCO3 = ( 40+ 12 + 16X3)

= 100 g

Mass = 50 X 100

5000 g is the mass of 50 moles of calcium carbonate

1. Answer.

0.05*2 = 0.10 g is the mass of 0.05 moles of hydrogen.

1. Answer.

123.9*0.6 = 74.34 g is the mass of 0.6 moles of phosphorus P4.

5. Answer.

Atomic masses can be non whole numbers if the element has different isotopes with different masses and different relative abundances.

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C4.2 Equations and Calculations AQA GCSE Chemistry C4 Chemical Calculations Kerboodle Answers  Page No 6

       1.Answer

2HCl means there are 2 molecules of hydrochloric acid or 2 moles of hydrochloric acid is formed.

2. Answer.

According to the balanced chemical equation, with 2 moles of magnesium 1 mole of oxygen reacts. So, 48 g of magnesium will react with 32 g of oxygen. So 6 of magnesium will react with (32/48)*6 = 4 g of oxygen.

3. a. Answer.

Balanced equation for the decomposition of hydrogen peroxide to form hydrogen and oxygen:

2H₂O₂ (l) ⇌ 2H₂O (l) + O₂ (g)

1. Answer.

From the balanced chemical equation we can say that 1 mole of oxygen require 2 mole of hydrogen peroxide. Converting mass to moles give us the relationship that 32 g of oxygen require 68 g of hydrogen peroxide. So 1.6 g of oxygen will require (68/32)*1.6 = 3.4 g of hydrogen peroxide.

4. a. Answer.

Reaction of calcium with water to form calcium hydroxide and hydrogen

Ca (s) + 2H2O (l) = Ca(OH)2 (aq) + H2 (g)

1. Answer.

According to the balanced chemical equation, to produce one mole of calcium hydroxide one mole of calcium is needed. Hence, 74 g of calcium hydroxide requires 40 g of calcium. So, 3.7 g calcium hydroxide will require (40/74)*3.7 = 2 g of calcium.

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C4.3 From masses to balanced equations AQA GCSE Chemistry C4 Chemical Calculations Kerboodle Answers Page No 67

       1.Answer.

The limiting reagent in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete.

2. a. Answer.

Number of moles of each reactant and product when 6.35 g of copper, 1.60 g of oxygen and 7.95 g of copper oxide react.

Moles of Copper = 6.35/63.5 = 0.1 moles

Moles of Oxygen = 1.6/32 = 0.05 moles

Moles of Copper Oxide = 7.95/79.5 = 0.1 moles

1. Answer.

Relationship between the moles reacted and the balanced chemical equation.

Cu : O : CuO   =  2:1 :2

0.1/0.05 : 0.05/0.05 : 0.1/0.05

Balanced Chemical Equation,

2Cu + O2 = 2CuO

2:1:2

3. a. Answer.

Balanced chemical equation when aluminium reacts with iron(III) oxide to form aluminium oxide and iron

2Al(s) + Fe2O3 (s) = Al2O3 (s) + 2Fe (l)

1. Answer.

When 32 g of iron (III) oxide was reacted with 16.2 g of aluminium then aluminium will be in excess and iron oxide will be the limiting reagent.

moles of aluminium = 16.2/27 = 0.6 moles

moles of iron oxide = 32/159.7 = 0.2 moles

For one mole of iron oxide 2 moles of aluminum will react. So for 0.2 moles of iron oxide, 0.4 moles of aluminium is required. Hence, aluminium is in excess and iron oxide is limiting.

1. Answer.

When 32 g of iron (III) oxide was reacted with 16.2 g of aluminium then aluminium will be in excess and iron oxide will be the limiting reagent and 22.4 g is the maximum mass of iron that can be obtained.

According to the balanced chemical equation, 2 moles of aluminium gives 2 moles of iron. Hence it is 1:1 ratio. So 0.4 moles of aluminium when reacts give 0.4 moles of iron = 0.4*56= 22.4 g.

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C4.4 The yield of a chemical reaction AQA GCSE Chemistry C4 Chemical Calculations Kerboodle Answers  Page No 69

1. Answer.

If the industry find ways to make products using high yield reactions  and minimize the waste than

It will reduce the waste of reactants and it will also result in lower cost of the manufacturing processes.

  2.Answer.

Factors that affect the percentage yield are :

1. a) The reactants or the products are lost during the reaction during separation or stick to the vessel which separating the product.
2. b) Reaction is reversible and resulted in backward reaction reducing the yield.
3. c) Some reactions proceed with some alternative route and make some other undesired products.
4. d) Some of the reactant might not be pure resulting in decrease in yield of the reaction.
5. Answer.

According to the balanced chemical equation, 1 moles of N2 gives 2 moles of ammonia. So, 28 g of nitrogen forms 34 g of ammonia. So 7 g of nitrogen will form  (34/28)*7 = 8.5 g of ammonia

So Percentage Yield

=(1.8/8.5)*100 = 21.17%

2 a) Balanced chemical equation for the thermal decomposition of sodium bicarbonate to give sodium carbonate water and carbon dioxide is :

2NaHCO3  =   Na2CO3 + H2O + CO2

16. b) 2 moles of sodium bicarbonate yields one mole of sodium carbonate. So 168 g of sodium bicarbonate gives 106 of sodium carbonate. So 16.8 g of hydrogen carbonate should yield

106/168*16.8 = 10.6 g of sodium carbonate. But the sodium carbonate produced is 9.2g.

So percentage yield = 9.2/10.6 = 86.8%.

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C4.5 Atom Economy AQA GCSE Chemistry C4 Chemical Calculations Kerboodle Answers  Page No 71

1. Answer.

The formula for  atom economy is = ^{mass of atoms in product}/_{mass of atoms in reactant} x 100

2. Answer.

The thermal decomposition of calcium carbonate, with the desired product being calcium oxide and waste product being carbon dioxide is shown like this:

CaCO3 (s) → CaO (s) + CO2 (g)

The total mass of all the reactants = mass of CaCO3 = 100 g

The mass of desired product = mass of CaO = 56 g

So Atom Economy = (56/100)*100 = 56%

3. a. Answer.

Atom Economy of reaction 1  = 100%  ((64.5/ 64.5)*100)

Atom Economy of Reaction 2 =  78.18%  (64.5/64.5+18)*100.

1. Answer.

Reaction 1 is favored as it has no by-product, therefore, has 100% atom economy. All the reactants are used up to make the direct product.

1. Answer.

The two major factors that the chemical company should consider are the energy cost and the atom economy. The energy cost should be minimum and the atom economy of a reaction should be highest.

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 C4.6 Expressing Concentrations AQA GCSE Chemistry C4 Chemical Calculations Kerboodle Answers Page 73

1. a. Answer.

50 g of sodium chloride when dissolved in 2.5 dm3 of water

50/2.5 = 20 g/dm3

1. Answer.

1.8 g of sodium carbonate in 862 cm3 of water = 862 cm3 = 0.862 dm3

1.8/0.862 = 2.08  g/dm3

2. Answer.

7 g of potassium hydroxide when added to 100 cm3 of water = 0.1 dm3 of water

7/0.1 = 70 g/dm3

3. Answer.

Concentration = Mass/ Volume. So increases in the mass of solute will increase the concentration of a solution. However, increase in volume will make the solution dilute and will decrease its volume. Hence concentration and mass of solutes are directly proportional and the concentration and volume of the solute is inversely proportional to the concentration of the solution.

4. Answer.

Concentration = Mass /volume

Mass= Concentration * Volume

= 93.6*0.025 = 2.34 g.

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 C4.7 Titrations AQA GCSE Chemistry C4 Chemical Calculations Kerboodle Answers Page No 75

1. a. Answer.

The curved surface of a liquid in a measuring cylinder is called a meniscus.

1. i. Answer.

The measuring instruments used to measure the volume of liquid in a titration are Burette and volumetric flask.

1. (ii) The correct way to read the volume of the liquid is to read the lower meniscus.

2 a. To carry out the titration between dilute nitric acid of known concentration and sodium hydroxide of unknown concentration we would require methyl orange or phenolphthalein as a suitable indicator. We will start with pipetting out 25 cm3 of known concentration of acid and will add it to the conical flask. Inside the conical flask we will also add a few drops of the indicator. We will then fill the burette with unknown concentration of sodium hydroxide. We will note the starting volume of the burette. We will then slowly release the tap and allow the sodium hydroxide to be added dropwise to the conical flask and we will add until the indicator changes colour. As soon as indicator changes colour we will stop the tap and note the final reading of the burette. We will repeat all the steps three times and take concordant readings.

1. HNO3 + NaOH = NaNO3 + H2O
2. a. Answer.

Two acid/base indicators are Universal indicator and Ordinary single indicator.

1. Answer.

Litmus and phenolphthalein are single indicators used to show whether a solution is acidic or alkaline, while the universal indicator is a mixed indicator that can give an estimate of the pH. A pH curve shows how the pH changes when an acid and alkali are mixed together.

1. Concordant titres are the results that agree within 0.1 mL of each other, with the aim of each titration being to get concordant results.

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 C4.8 Titration Calculations AQA GCSE Chemistry C4 Chemical Calculations Kerboodle Answers  Page No 77

1. Answer.

Moles = Concentration * Volume

Concentration = 0.2 mol/dm3

Volume = 10 cm3 = 10/100 = 0.01 dm3

= 0.2 * 0.01 = 0.002 moles.

2. Answer.

Reaction of potassium hydroxide with nitric acid

KOH(aq) + HNO3 (aq) —–> KNO3 (aq) + H2O (l)

3. Answer.

Moles = Concentration * Volume

= 0.4 * 0.01 = 0.004 moles.

4. Answer.

There is 1:1 ratio of moles of KOH and HNO3 so moles of nitric acid is also 0.004 moles.

5. Answer.

Moles/Volume =  0.004/0.0125 = 0.32 mol/dm3.

6. Answer.

0.32*63 = 20.16  g/dm3.

 C4.9 Volumes of gases AQA GCSE Chemistry C4 Chemical Calculations Kerboodle Answers  Page 79

1. Answer.

Molar volume of gas, V_m, is defined as the volume of 1 mole of the gas at a specified temperature and pressure.

2. a.  i. Answer.

1 moles of each gas at r.tp occupies 24 dm3.

Moles in 36 dm3 of carbon dioxide will be 36/24 = 1.5 moles.

1. Answer.

10,000 dm3 of hydrogen will occupy

10,000/24 = 416.67 moles

1. i. Answer.

1 moles of each gas at r.tp occupies 24 dm3.

36 g of helium will be  Moles of Helium = 36/4 = 9 moles

Volume of Helium = 9*24 = 216 dm^3.

1. Answer.

Moles of NO2=  13.8/46 = 0.3 moles

Volume of NO2 = 0.3 * 24 = 7.2 dm^3.

1. Answer.

24 000 cm3 of each gas = 1 moles

48 cm3 of each gas = 0.002 moles

Mass of O2 = Moles *Mr = 0.002*32 g = 0.064 g.

3. Answer.

Moles of Methane = 150/24 = 6.25 moles

1 moles of methane requires 2 moles of oxygen. So 6.25 moles of methane will require 12.5 moles of oxygen which is equal to 12.5*32 g of Oxygen = 400 g

4. Answer.

0.80 of Calcium = 0.8/40  = 0.02 moles

1 moles of calcium forms one mole of hydrogen. 0.02 moles of calcium will produce 0.02 moles of hydrogen  = 0.02*24 dm3 = 0.48 dm3.

C4 SUMMARY QUESTIONS AQA GCSE Chemistry C4 Chemical Calculations Kerboodle Answers  Page 80

1 a. Mass of H2S= 2(Mass of H)+ Mass of S

=2+32

34 g

1. Mass of SO2= Mass of S + 2 (Mass of O)

= 32+32

64 g

1. C₂H4 = 2(Mass of C)+ 4(Mass of H)

= 2X12 + 4X1

=24+4

=28 g

1. NaOH = Mass of Na + Mass of O + Mass of H

= 23 +16 +1

= 40 g

1. Na2CO3 = 2 Mass of Na + Mass of C + 3 Mass of O

= 2X23 + 12 + 3X16

=46 + 12 + 48

=106 g

1. Al₂(SO₄)₃= 2 Mass of Al + 3 Mass of S + 12 Mass of O

= 27*2 + 3*32 + 12*16

=54 + 96 + 192

=342 g

1. NaAl(OH)₄= Mass of Na + Mass of Al + 4 Mass of O + 4 Mass of H

=23 + 27+ 4*16 + 4*1

=118 g

2 a Moles = MAss /Mr

27 g of silver will have 27/108 = 0.25 moles

1. 0.056 g of iron will have 0.056/56 = 0.001 moles
2. 6.2 g of phosphorus will have 6.2/124 = 0.05 moles

3. CaCO3 = CaO + CO2

According to the balanced chemical equations, 1 moles of calcium carbonate will give 1 moles of Calcium oxide

= 100 g of CaCO3 or 100 tonnes of CaCO3 will give 56 g or 56 tonnes of CaO

= 1500 tonnes of calcium carbonate will give

= (56/100)* = 840 tonnes of Calcium oxide

But the yield is only 804 tonnes

So percentage yield = (804/840)*100

= 95.7%

4.a. Answer.

C₂H₄(g) + H₂O(g) ↔ C₂H₅OH(g)

1. According to the balanced chemical equation, 1 moles of ethene gives 1 moles of ethanol.

So 28 g of ethene gives 46 g of ethanol

1 g of ethene gives 46/28 of ethanol

So 14 g of ethene gives = (46/28)*14

=  23 g of ethanol

But the yield of ethanol is 17.25 g

So percentage yield = 75%

5. a. According to the balanced chemical equation, 1 moles of S will produce 1 moles of SO2 and 1 moles of SO2 produces one moles of H2SO4. So from 1 moles of S we will get 1 mole of sulphuric acid.

4. According to the balanced chemical equations, 1 moles of S will produce 1 moles of sulphuric acid. 32 kg of S will produce 98 Kg of H2SO4 so 64 kg of sulphur will produce 196 Kg of H2SO4.

188. From the answer above, the H2SO4 produced should be 196 Kg but 188.16 kg is produced so the percentage yield is 96%.

d.The yield is less than 100% due to the following reasons

1. Reaction was not completed or reverse reaction started
2. Some of the reactants or products might have lost from the reaction mixture.
3. Reactant might not be pure.

6. a. Answer.

Balanced chemical equation of potassium hydroxide and hydrochloric Acid.

KOH(aq) + HCl(aq) → KCl(aq) + H₂O(l)

1. (i)Moles of Hydrochloric Acid in 20 cm3 of 0.50 mol/dm3

= 20*0.50/1000

0.01 moles

(ii) 0.01 moles of hydrochloric acid will react with 0.01 moles of potassium hydroxide

1. (i) Moles of Potassium hydroxide = 0.01 moles

Volume of potassium hydroxide = 12.5 cm3 = 0.0125 dm3

Concentration of potassium hydroxide = 0.01/0.0125

= 0.8 mol/dm³

1. Answer.

0.8 mol of KOH = 0.8 * 56 = 44.8 g/dm³

^7.Answer. Equation for thermal decomposition of magnesium carbonate :-

MgCO₃ (s)  = MgO (s) + CO_{2 (g)}

 1.Answer.

1.1 g of carbon dioxide = 1.1/44 g = 0.025 moles

1 mole of each gas occupies 24 dm3 of volume at room temperature and pressure

So 0.025 moles of gas will occupy = 0.6 dm3

1. Answer.

According to the balanced chemical equation,

If 1 mole of carbon dioxide is formed than 1 moles of magnesium carbonate decomposed.

If 0.025 moles of carbon dioxide is formed than 0.025 moles of magnesium carbonate decomposed

0.025 moles = 0.025*84

= 2.1 g of magnesium carbonate decomposed

1. Magnesium carbonate also reacts with dilute hydrochloric acid:

MgCO₃ (s) + 2HCl (aq) → MgCl₂ (aq) + H₂O (l) + CO₂ (g)

1. i. Answer.

Since the reaction has the gaseous products so the best way to tell that the reaction is completed is to see the bubbled of the gas. When the bubbles of the gas stops the reaction is completed.

1. Answer.

Moles of hydrochloric Acid = Concentration * Volume

= 0.2*25/1000

= 0.005 mole

Moles of magnesium carbonate = 8.4/84

=0.1 mole

0.1 mol of magnesium carbonate will require 0.2 moles of hydrochloric acid but the moles of acid is 0.05 moles so acid is limiting and magnesium chloride is in excess.

iii. Answer.

Since acid is limiting it will be completely used up in the reaction.

According to the balanced chemical equation,

The acid and carbon dioxide ratio is 1:2

For 0.005 moles of acid carbon dioxide produced will be 0.0025 moles

And each moles of a gas at r.t.p occupes 24 dm3 of gas

So 0.0025 moles of gas will occupy = 0.0025*24

= 0.06 dm³ (or 60 cm³)

8.a. Answer.

A_r of Br = (50 ×79) + (50×81) = 80

100

⁸⁰Br

1. Answer.

A_rof Fe = (5.8 × 54) + (91.8 × 56) + (2.1 × 57) + (0.3 × 58)

100

= 5591.1

100

= 55.9

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Practice Questions AQA GCSE Chemistry C4 Chemical Calculations Kerboodle Answers  Page 81

01.1.Balanced chemical equation of the reaction between magnesium and oxygen is :-

2Mg(s) + O₂(g) → 2MgO(s)

01.2 Answer.

At the start crucible has only magnesium. During the reaction, oxygen from the air reacts with magnesium and magnesium oxide is formed which increased the mass of the crucible.

01.3. Answer.

To make sure that the reaction is complete, the student can Reweigh the crucible. If the two masses are the same, the reaction is complete.

01.4. Answer.

Since the reaction involves gaseous products, the mass will decrease because carbon dioxide is produced in this reaction which will escape to the surroundings.

02.1. Answer.

1.76 g of carbon dioxide = 0.04 moles of carbon dioxide

0.90 g of water = 0.05 moles of water

So balancing ratio will be 4CO₂ and 5H₂O

02.2 Answer.

In the balancing number carbon atoms are 4 and hydrogen atoms are 10 in 4CO₂ and 5H₂O so the formula of the hydrocarbon will be :-

C₄H₁₀

 02.3. Answer.

0.22 g of carbon dioxide = 0.22/44  = 0.005 moles

Volume = 400 cm3 = 0.4 dm3

Concentration = 0.005/0.4

0.0125 mol / dm³

03.1. Answer.

Ammonia has a lower boiling point due to weak intermolecular forces between the molecule of ammonia.

 03.2. Answer.

Moles of nitrogen = 84 tonnes = 84000000/28

=3,000,000 moles of nitrogen

Moles of hydrogen = 30 tonnes = 30000000/2

=15,000,000

 According to the balanced chemical equation the ratio of Nitrogen to hydrogen is 1:3. So for 3,000,000 moles of nitrogen we require 9,000,000 moles of hydrogen and we have 15,000,000 moles so nitrogen is limiting and hydrogen is in excess.

03.3. Answer.

Ratio of nitrogen to ammonia is 1:2

So for 3,000,000 moles of nitrogen we will get 6,000,000 moles of ammonia

=6,000,000*17

= 102 tonnes

 03.4. Answer.

If 18 tonnes of ammonia is formed then percentage yield is :

(18/102)*100

17.6%.

03.5. Answer.

Since only one product is formed which is the desired product so the percentage atom economy will be 100% as all the reactants are used up to make only one single desired product.

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This page contains the detailed and easy notes for AQA GCSE Chemistry Quantitative Chemistry for revision and understanding Quantitative Chemistry.

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AQA GCSE Paper 1: 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

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:

Wikipedia

Wikimedia

Commons

Flickr

Pixabay

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.

C4-Analytical  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