C4 - Quantitative chemistry
Independent learning
Introduction
We can use chemical formulas to work out what we can how much of a product we can expect to make during a chemical reaction and work out how efficient the process is. The key to understanding this is to understand what chemists mean when they are talking about moles (no not small furry burrowing animals). Read on to find out more:
Specification links:
Task 1: Know
Task 1a: Use look, cover, write, check and quizlet to learn the answers to the core questions and the keywords for this topic.
Learn
Task 2: Relative masses and moles
Task 2c: Calculate the Mr of the following:
Water (H2O)
Calcium Carbonate (CaCO3)
Lithium Oxide (Li2O)
Hydrochloric Acid (HCl)
Ammonia (NH4)
Water (H2O) 1+1+16=18
Calcium Carbonate (CaCO3) 40+12+16+16+16=100
Lithium Oxide (Li2O) 7+7+16=30
Hydrochloric Acid (HCl) 1+35.5=36.5
Ammonia (NH4) 14+1+1+1+1=18
Task 2d:
Give the formula for calculating percentage by mass
Calculate the % of hydrogen in water - write this out as an example calculation.
Task 2e Complete the exam question
Task 2f: Check your answer
CO2 so there are 2 oxygen atoms in each molecule
Ar of O = 16
Mr of CO2 = 12 + (2x16) = 44
Formula: number x Ar / Mr
Insert: 2 x 16 / 44
Answer: 72.7%
Set 3 move on to task 7
Set 1&2 - Task 2g: Watch the video
Task 2h: Make notes that
Define mole and give Avogadro’s number
Explain how to calculate the number of moles when given mass and relative formula mass
Task 2i: Calculate the number of moles in:
54g of Water (H2O)
149g of Potassium Chloride (KCl)
30g of Lithium Oxide (Li2O)
198g of Copper Chloride (CuCl2)
495g of White Arsenic (As2O3)
54g of Water (H2O) = 3 moles
149g of Potassium Chloride (KCl) = 2 moles
30g of Lithium Oxide (Li2O) = 1 mole
198g of Copper Chloride (CuCl2) = 1.5 moles
495g of White Arsenic (As2O3) = 2.5 moles
Task 2j: Calculate the mass of:
50 moles of calcium carbonate CaCO3
0.05 moles of hydrogen H2
0.6 moles of phosphorus P4
50 moles of calcium carbonate CaCO3 = 5000g or 5kg
0.05 moles of hydrogen H2 = 0.1g
0.6 moles of phosphorus P4 = 74.4g
Task 2k: Complete and self mark the exam questions
Task 3: Equations and calculations - Set 1 and 2 only
Task 3d: Calculate the following:
1. How much iron sulphide (FeS) do you get from sulphur reacting with 56g of iron?
(Mr: Fe = 56, S = 32)
Fe + S → FeS
2. Hydrogen reacts with oxygen to form water. What mass of water can be made from 8g of hydrogen?
(Mr: H = 1, O = 16)
2H2 + O2 → 2H2O
3. Carbon burns in air to form carbon dioxide. What mass of carbon is needed to make 22g of carbon dioxide?
(Mr: C = 12, O = 16)
C + O2 → CO2
4. Use the chemical equation to find the mass of carbon which reacts with 223g of lead oxide.
(Mr: Pb = 207, C = 12, O = 16)
2PbO + C → 2Pb + CO2
5. When sulphur burns, it forms sulphur dioxide. Calculate the mass of sulphur dioxide which would be obtained from 160 tonnes of sulphur.
(Mr: S = 32, O = 16)
S + O2 → SO2
6. To make ammonium nitrate fertiliser, ammonia(NH3) is reacted with nitric acid.(HNO3) Calculate the mass of nitric acid needed to make 400 tonnes of ammonium nitrate (NH4NO3).
(Mr: N = 14, H = 1, O = 16)
NH3 + HNO3 → NH4NO3
7. Calcium carbonate decomposes on heating to give calcium oxide and carbon dioxide.
(Mr: Ca = 40, C = 12, O = 16)
a. What mass of CaO can be made from 25g of CaCO3?
b. What mass of CaCO3 would be needed to form 11g of CO2?
CaCO3 → CaO + CO2
8. The reaction between magnesium carbonate and hydrochloric acid produces magnesium chloride, water and carbon dioxide.
(Mr: Mg = 24, C = 12, O = 16, Cl = 35.5)
What mass of magnesium chloride would you expect to get from 8.4g of magnesium carbonate?
MgCO3 + 2HCl → MgCl2 + H2O + CO2
9. The balanced symbol equation for the reaction between ammonia and sulphuric acid is
How many tonnes of sulphuric acid are needed to exactly neutralise 340 tonnes of ammonia?
(Mr: N = 14, H = 1, O = 16, S = 32)
2NH3 + H2SO4 → (NH4)2SO4
10. Uranium metal is produced from uranium fluoride using magnesium.
What mass of uranium would be produced from 34.9g of uranium fluoride?
(Mr: U = 235, F = 19)
UF6 + 3Mg → 3MgF2 + U
Answers
88g
72g
6g
6g
320 tonnes
315 tonnes
a = 14g; b = 25g
9.5g
980 tonnes
23.5g
Task 4: Masses to balanced equations - Set 1 and 2 only
Task 4c: When copper metal reacts with oxygen, copper oxide, CuO, is formed.
6.35g of copper reacted with 1.60g of oxygen gas O2, to form 7.95g of copper oxide
a). Calculate the number of moles of each reactant and product
b) Show the balanced symbol equation for the reaction
Formula: Cu + O2 → CuO
Mass: 6.35g + 1.60g → 79.5g
Mr: 63.5 16x2 = 32 63.5 + 16 = 79.5
Moles : 6.35/63.5 = 0.1 1.60/32 = 0.05 7.95/79.5 = 0.1
Ratio 0. 1 : 0.05 : 0.1
0.1/0.05 = 2 0.05/0.05 = 1 0.1/0.05 = 2
Whole number ratio: 2 : 1 : 2
Balanced formula: 2Cu + O2 → 2CuO
Task 4d: If I react 44.6g of lead oxide (PbO) with 1.2g of carbon, I can obtain 41.4g of lead and ?g of carbon dioxide.
a. Work out how much carbon dioxide is produced.
b. Write the balanced symbol equation for this reaction.
Ar Pb = 207 Ar O = 16 Ar C = 12
Formula: PbO + C → Pb + CO2
a. Remember conservation of mass so:
44.6 +1.2 = 41.4 + mass of CO2
45.8 = 41.4 + mass CO2
45.8 - 41.4 = mass CO2 = 4.4g
b.
Masses: 44.6 + 1.2 = 41.4 + 4.6
Mr: 207 + 16 = 223 12 207 12+(2x16)= 44
Moles: 44.6/223= 0.2 1.2/12 = 0.1, 41.4/207= 0.2, 4.4/44 = 0.1
Ratio: 0.2 : 0.1 : 0.2 : 0.1
Whole number ratio: 2 : 1 : 2 : 1
Balanced formula: 2PbO + C → 2Pb + CO2
Task 4e: Aluminium reacts with iron (III)oxide, Fe2O3, to give iron metal and aluminium oxide, Al2O3
a. Write a balanced symbol equation for this reaction.
In an experiment,32.0g of iron(III)oxide was reacted with 16.2g of aluminium.
b. Which of the two reactants was in the limiting reactant? Show your working.
c. Calculate the theoretical maximum mass of iron that could be collected at the end of this experiment.
a. 2Al + Fe2O3 → 2Fe + Al2O3
b. Al Fe2O3
Mass: 16.2g 32g
Mr: 27 (2x56)+(3x16)=160
Moles: 16.2/27 = 0.6 32/160 = 0.2
Actual ratio: 3 1
Ideal ratio: 2 1
so Iron oxide will run out first and is therefore the limiting reactant
c.
Balanced formula tells us that the ratio of Fe2O3 : Fe is 1 : 2.
So if we have 0.2 moles of Fe2O3 we should get 2 x 02. = 0.4 moles of Fe.
mass = moles x Mr = 0.4 x 56 = 22.4g of Fe
Set 2 go to task 7 now
Task 5: Percentage yield - Set 1 only
Task 5b: Make notes that:
Define:
theoretical yield
experimental yield
Give the equation for calculating percentage yield
Explain using a worked example how to calculate the % yield
List the reasons why reactions don’t have 100% yield
List 3 reasons why higher yield reactions are better for the environment
Task 5c: N2 + 3H2 → 2NH3
If 7g of nitrogen reacts with excess hydrogen and 1.8g ammonia is collected, what is the % yield?
Step 1: Calculate theoretical yield:
Mr of N2 is: 2 x 14 = 28
7g of N2 is 7 / 28 = 0.25 moles
From formula the ratio N2 : 2NH3 = 1 : 2
Theoretically we would expect 0.25 x 2 = 0.5 moles of NH3
Mr of NH3 is : 14 + (3x1) = 17
Mass of 0.5 moles of NH3 is: 17 x 0.5 = 8.5g
Step 2: Calculate percentage yield:
Formula: % yield = (actual/ theoretical) x 100
Insert: % = (1.8/8.5) x100
Answer: 21.2% (in this case it is because this is a reversible reaction)
Task 5d: Sodium hydrogencarbonate, NaHCO3, can be converted into sodium carbonate, Na2CO3, by heating. This is a thermal decomposition reaction in which water vapour and carbon dioxide are also produced.
A student started with 16.8 g of sodium hydrogencarbonate and collected 9.20g of sodium carbonate
Write a balanced symbol equation for the reaction
Calculate the % yield
2NaHCO3 → Na2CO3 + H2O + CO2
86.8%
Task 5e: Complete the progress quiz
Task 6: Atom economy - Set 1 only
Task 6c:
Task 6d:
Answer
Useful product = 2 x PbO, 2 x Mr PbO = 2 x (207 + 16) = 446
Reactants = 2 PbS + 3O2
2 x Mr PbS
= 2 x (207 + 32) = 478
3 x Mr O2
= 3 x (16 + 16) = 96
So % atom economy
= 446 / (478 + 96) x 100
= 77.7%
Answer
a. Mr of C2H5Cl (useful product)
= (2x12)+(5x1)+35.5
= 64.5
Mr of C2H5OH (reactant 1)
= (2x12)+(5x1)+16+1
= 46
Mr of HCl (reactant 2)
= 1 + 35.5
= 36.5
Mr of reactants = 46 + 36.5 = 82.5
% yield = (useful / reactants) x 100
% yield = 64.5/(82.5) x 100 = 78.2%
b. Reaction 1 as it has an economy of 100% so there are no unwanted products to dispose of making it better for the environment
Task 6e : Answer the exam questions and self mark
Task 7: Expressing concentrations
Task 7b: Make notes that:
Define solute, solvent, solution and concentration
State the equation for calculating concentration
Explain how to convert from cm3 to dm3
State how to calculate mass if you are given concentration and volume
Task 7c: Calculate the concentrations of the following solutions in g/dm3.
10.0 g of sodium chloride dissolved in 2.0 dm3 of water.
2.5g of glucose dissolved in 0.5 dm3 of water.
36g of potassium chloride dissolved in 1.8 dm3 of water.
Which solution is more concentrated:
Solution A - 25g of sodium chloride dissolved in 2.0 dm3 of water
Solution B - 40g of sodium chloride dissolved in 3.5 dm3 of water?
10g/2dm3 = 5g/dm3
2.5g/0.5dm3 = 5g/dm3
36g/1.8dm3 = 20g/dm3
4)
Solution A 25g/2dm3 = 12.5 g/dm3
Solution B 40g/3.5dm3 = 11.43 g/dm3
So solution A is more concentrated.
Task 7d: Convert the following to dm3:
250cm3
125cm3
1.5cm3
50cm3
1000cm3
You remove 50cm3 of solution from a total volume of 0.75 dm3. How much solution do you have left over? Give your answer in dm3.
0.25 dm3
0.125 dm3
0.0015 dm3
0.05 dm3
1 dm3
50 cm3 = 0.05dm3
0.75 dm3 - 0.05 dm3 = 0.7 dm3
Task 7e: Calculate the concentrations of the following solutions in g/dm3:
1.5g of potassium iodide in 150cm3 of solution.
5g of potassium sulfate in 50cm3 of solution.
2.3g of lithium chloride in 500cm3 of solution.
Which is more dilute:
Solution A - 1.2g of potassium chloride in 50cm3 of water.
Solution B - 25g of potassium chloride in 1 dm3 of water.
150cm3 = 0.15dm3
1.5g/0.15dm3 = 10g/dm3
50cm3 = 0.05dm3
5g/0.05dm3 = 100g/dm3
500cm3 = 0.5dm32.3g/0.5dm3 = 4.6g/dm3
Solution A - 50cm3 = 0.05dm3 1.2g/0.05dm3 = 24g/dm3
Solution B - 25g/1dm3 = 25g/dm3
Solution A is more dilute.
Task 7f: Calculate the mass of solute needed to make the following solutions:
0.5dm3 of a 2g/dm3 solution of silver nitrate.
0.25dm3 of a 1.5g/dm3 solution of sodium chloride.
0.01dm3 of a 0.4g/dm3 solution of sodium carbonate.
500cm3 of a 5g/dm3 solution of magnesium sulfate.
25cm3 of a 93.6g/dm3 solution of sodium chloride.
Mass = 0.5dm3 x 2g/dm3 = 1g
Mass = 0.25dm3 x 1.5g/dm3 = 0.375g
Mass = 0.01dm3 x 0.4g/dm3 = 0.004g
500cm3 = 0.5dm3Mass = 0.5dm3 x 5g/dm3 = 2.5g
25cm3 = 0.025dm3
Mass = 0.025dm3 x 93.6g/dm3
= 2.34g
Set 2&3 go to task 10
Task 8: Titrations - Set 1 only
Task 8b:
Give the equation for calculating concentration in mol/dm3
Define titration
Describe how to carry out a titration using the video on slide 17.
Explain how to work out the concentration of the unknown solution using the results from a titration
Describe how to convert from mol/dm3 to g/dm3
Task 9: Molar gas volume - set 1 only
Task 9b: Make notes that:
Define the molar volume of gas
Answer the practise questions on slide 13 and self mark using slide 14
Describe how to calculate the relative formula mass of a gas using mass and volume
Answer the practise questions on slide 17 and 18 and self mark
Describe how to calculate molecular volume and copy the example calculation on slide 24
Answer the practise questions on slide 25, 27, 29 and 31 and self mark using slides 26, 28, 30 and 32
Task 10: Summary
Task 10a: Complete the seneca learning unit for your set.
Task 10b: Make a mind map or single page revision summary of this topic and add it to your revision folder.