Edexcel IGCSE Chemistry E Chemical Formulae Equations and Calculations Questions
Practise balancing equations and using formulae, Mr and mole ratios for reacting masses, yields and composition.
- Syllabus
- First assessment 2019
- Course
- Chemistry 4CH1
Practise balancing equations and using formulae, Mr and mole ratios for reacting masses, yields and composition.
This question is about magnesium metal and its compounds.
Magnesium burns in oxygen gas to form solid magnesium oxide.
Complete the chemical equation for the reaction by balancing the equation and adding the state symbols.
2Mg(s)+O2(g)→2MgO(s)
Marking guidance:
- 1 mark for balancing.
- 1 mark for state symbols.
- Allow multiples or fractions, but O2 must also be changed to score the balancing mark.
A student uses this method to make a solution of magnesium chloride.
Step 1. measure 25 cm3 of dilute hydrochloric acid into a beaker
Step 2 add magnesium powder a little at a time
Step 3 keep adding magnesium powder until it is in excess
Step 4 remove the excess magnesium powder by filtration
This is the equation for the reaction.
Calculate the minimum mass of magnesium needed to react with 25.0 cm3 of 2.00 mol/dm3 hydrochloric acid.
[for Mg,Ar=24 ]
mass = g
- amount of HCl = (25 x 2) / 1000 = 0.050 mol
- amount of magnesium = 0.050 / 2 = 0.025 mol
- mass of magnesium = 0.025 x 24 = 0.60 g / 0.6 g
Allow:
- ECF from amount of HCl divided by 2
- ECF from amount of magnesium multiplied by 24
Marking guidance:
- Answer of 0.60 g without working scores 3.
- Answer of 1.20 g with working scores 2.
Crude oil is a mixture of hydrocarbons.
One of the hydrocarbons in the refinery gas fraction is an alkane with the structural formula CH3CH2CH2CH3
Calculate the relative molecular mass (Mr) of this alkane.
(Mr = 4 x 12 + 10 x 1 =) 58
The diagram shows the positions of some elements in part of the Periodic Table.
Na Al S Cl
K Xe
In
A teacher adds a small piece of sodium to a glass trough containing water and universal indicator.
The universal indicator changes colour.
The equation for the reaction is
The reaction with sodium produces 0.036 g of hydrogen gas.
One mole of hydrogen gas contains 6.0×1023 molecules.
Calculate the number of molecules of hydrogen gas produced in the reaction with sodium.
Give your answer to two significant figures.
number of molecules of hydrogen gas =
- moles of hydrogen = 0.036 / 2 = 0.018 mol
- number of molecules = 0.018 x 6.0 x 10^23 = 1.08 x 10^22
- to two significant figures = 1.1 x 10^22
Allow:
- error carried forward from the mole calculation multiplied by 6.0 x 10^23
- error carried forward from the number of molecules, but the answer must be to 2 significant figures
Marking guidance:
- Correct answer with no working scores 3.
- 2.16 x 10^22 scores 1.
- 2.2 x 10^22 scores 2.
This question is about magnesium metal and its compounds.
Magnesium burns in oxygen gas to form solid magnesium oxide.
Complete the chemical equation for the reaction by balancing the equation and adding the state symbols.
2Mg(s)+O2(g)→2MgO(s)
Marking guidance:
- 1 mark for balancing.
- 1 mark for state symbols.
- Allow multiples or fractions, but O2 must also be changed to score the balancing mark.
A sample of iron(III) oxide reacts with excess carbon monoxide.
This is the equation for the reaction.
Calculate the minimum mass of iron(III) oxide needed to produce a theoretical yield of 28 g of iron.
[for Fe2O3Mr=160 ]
minimum mass of iron(III) oxide =
M1 (amount of Fe= ) 28÷56 OR 0.5(0)(mol)
M2 (amount of Fe2O3= ) 0.5÷2 OR 0.25( mol)
M3 (mass of Fe2O3=0.25×160= ) 40( g)
ALLOW ecf throughout
correct answer without working scores 3
A sample of iron(III) oxide reacts with excess carbon monoxide.
This is the equation for the reaction.
The actual yield of iron from this sample is 21 g .
Calculate the percentage yield of iron.
- 21 / 28 x 100
- 75%
Allow:
- ECF for any correct calculation involving 21 as the numerator
- Correct answer without working scores 2.
- 0.75 scores 1.
Sodium sulfate can be prepared by the reaction between sodium hydroxide solution and sodium hydrogensulfate ( NaHSO4 ) solution.
This is the equation for the reaction.
Sodium hydrogensulfate solution is acidic.
A student adds 25.0 cm3 of sodium hydroxide solution to a conical flask and adds two drops of indicator.
The student does a titration.
Crystals of hydrated sodium sulfate decompose when heated.
This is the equation for the decomposition.
A student uses this apparatus to find the value of x .
hydrated sodium sulfate
The student heats the crystals until the decomposition is complete.
The table shows the student's results.
Use the results to calculate the value of x .
[for Na2SO4,Mr=142 for H2O,Mr=18 ]
value of x=
Method 1:
- mass of sodium sulfate = 2.84 g
- mass of water = 3.60 g
- moles of sodium sulfate = 0.02 and moles of water = 0.20
- x = 10
Method 2:
- mass of sodium sulfate = 2.84 g
- mass of hydrated salt = 6.44 g
- Mr of hydrated salt = 6.44 / 0.02 = 322
- xH2O = 322 - 142 = 180, so x = 10
Marking guidance:
- Answer of 10 without working scores 4.
- Allow ECF for x if dividing by the smallest number, but only if x is given as a whole number.
Ocimene is an organic compound that gives some plants their particular smell.
The molecular formula of ocimene is C10H16
Using ocimene as an example, explain what is meant by the term empirical formula.
M1 simplest (whole number) ratio of atoms present (in a compound)
M2 empirical formula (of ocimene/ C10H16 ) is C5H8
ALLOW elements for atoms
ALLOW C : H ratio 5:8
The table shows the structures of six organic compounds.
Another compound, G, has this percentage composition by mass.
Show by calculation that the empirical formula of compound G is C2H4Cl
M1 37.8 / 12, 6.3 / 1, 55.9 / 35.5
M2 3.15, 6.3, 1.57
M3 divide by smallest to get 2:4:1
OR
M1 Mr of C_2 H_4 Cl = 63.5
M2 24/63.5 x 100 and 4/63.5 x 100 and 35.5/63.5 x100
M3 37.8% and 6.3% and 55.9%
M2 subsumes M1
M3 must be calculated
The relative formula mass ( Mr ) of G is 127
Determine the molecular formula of G .
M1 127÷63.5=2
M2 Molecular formula =C4H8Cl2
Answer of C4H8Cl2 without working scores 2
This question is about magnesium metal and its compounds.
A student uses this method to make a solution of magnesium chloride.
Step 1. measure 25 cm3 of dilute hydrochloric acid into a beaker
Step 2 add magnesium powder a little at a time
Step 3 keep adding magnesium powder until it is in excess
Step 4 remove the excess magnesium powder by filtration
This is the equation for the reaction.
Calculate the minimum mass of magnesium needed to react with 25.0 cm3 of 2.00 mol/dm3 hydrochloric acid.
[for Mg,Ar=24 ]
mass = g
- amount of HCl = (25 x 2) / 1000 = 0.050 mol
- amount of magnesium = 0.050 / 2 = 0.025 mol
- mass of magnesium = 0.025 x 24 = 0.60 g / 0.6 g
Allow:
- ECF from amount of HCl divided by 2
- ECF from amount of magnesium multiplied by 24
Marking guidance:
- Answer of 0.60 g without working scores 3.
- Answer of 1.20 g with working scores 2.
This question is about octane (C8H18) which is produced in the gasoline fraction during fractional distillation of crude oil.
A car is driven at constant speed for 4.00 km .
The exhaust gases are collected and their volume at room temperature and pressure (rtp) is 5.02×105 cm3.
The exhaust gases include carbon dioxide and oxides of nitrogen.
The carbon dioxide is removed from the exhaust gases. The volume of the remaining gases at rtp is 2.96×105 cm3.
Show that the car produces less than 100 g of carbon dioxide per km.
[molar volume of carbon dioxide at rtp =24000 cm3 ]
- volume of carbon dioxide = 206000 cm3 / 2.06 x 10^5 cm3 / 206 dm3
- volume of carbon dioxide per km = 51500 cm3 / 5.15 x 10^4 cm3 / 51.5 dm3
- 51500 / 24000 = 2.15 mol
- Mr of carbon dioxide is 44
- mass of carbon dioxide per km = 94.4 g
Allow:
- 94-95 g
- error carried forward from an incorrect Mr
Marking guidance:
- Division by 4 can happen in the volume, mole, or mass steps.
- Correct answer of 94-95 g scores 5 marks.
A student uses this apparatus to determine the formula of magnesium oxide.
This is the student's method.
- find mass of empty crucible and lid
- place some magnesium ribbon in crucible
- find mass of crucible, lid and magnesium ribbon
- heat crucible, lifting lid occasionally
- after five minutes, find mass of crucible, lid and magnesium oxide
Explain why the student lifts the lid.
An explanation that links the following two points
M1 to allow air/oxygen to enter the crucible OWTTE
M2 so that oxygen can react with the magnesium
The mass of magnesium oxide formed is less than expected.
Describe what the student should do next to obtain a more accurate value for the mass of magnesium oxide.
- heat the crucible again and reweigh
- repeat until constant mass
Marking guidance:
- Heat and reweigh to constant mass scores 2.