About Chemical Engineering
Chemical engineering questions in a recruitment paper are the mole, the strength of a solution, the dilution of one solution into another, a balanced reaction, the gas laws and a heat duty - each one a single relation applied to the numbers given. This topic covers all of those areas on one page, so that the subject can be practised as a whole rather than in fragments.
What you need to understand
- The mole is the chemist unit of amount: one mole of a substance weighs its molar mass in grams. Water is 18 g per mole, sodium chloride 58.5 and calcium carbonate 100.
- Mass and moles are turned into one another with n = m / M, where m is the mass in grams and M the molar mass. The answer is in moles, so a mass left in kilograms is a factor of a thousand wrong.
- Molarity is moles of solute per litre of solution: M = n / V, with the volume in litres. A volume given in millilitres has to be divided by 1000 first.
- Diluting a solution adds solvent but no solute, so the number of moles is unchanged: M1 V1 = M2 V2. Adding water lowers the molarity in proportion to the volume added.
- A balanced equation gives the ratio in which substances react. The numbers in front of the formulas are the mole ratio, and they apply to moles, never to grams.
- Stoichiometry is worked in three steps: mass to moles, moles through the ratio from the equation, and moles back to mass. Trying to do it in one step is where most of the errors come from.
- The limiting reactant is the one that runs out first. If two reactants are given, work out how much product each could give, and the smaller figure is what the reaction can actually produce.
- The ideal gas law is p V = n R T, with p in kilopascals, V in litres, n in moles and T in kelvin. R is 8.314, and the units have to match it.
- Temperature in the gas law is absolute. Zero degrees Celsius is 273 K, and a temperature left in Celsius gives an answer that is simply wrong, not slightly out.
- Heat duty is Q = m c dT, where c is the specific heat capacity. Water is about 4.18 kJ per kilogram per kelvin.
- A latent heat is the heat needed to change phase at constant temperature, Q = m L. It is added to any sensible heat, never used instead of it.
- A percentage yield compares what was obtained with what the equation says was possible. The possible amount comes from the stoichiometry, not from whichever reactant was named first.
- Concentration can be quoted as a mass per volume, such as grams per litre, as well as in moles. Read which one the question wants before answering.
- A material balance says that what goes in either comes out or accumulates. It is the check that catches a lost stream in a longer problem.
How to work through these questions
- Write down the given quantities with their units, and convert to grams, litres, moles and kelvin before doing anything else with them.
- Balance the equation if it is not already balanced. A ratio taken from an unbalanced equation is wrong however careful the arithmetic afterwards.
- For a dilution, decide which of the two states is the concentrated one and label the unknown. M1 V1 = M2 V2 then gives it in one line.
- Check the size of the answer as you go: a molarity of 50 is not a solution, and a mass of 0.002 g is not a laboratory batch.
- Carry the units through the working. If the units of the answer are not the units that were asked for, the wrong relation has been used.
- Wherever a temperature appears, put it in kelvin at once. That single step removes the commonest fault in the gas and heat questions.
Mistakes that cost marks
- Applying the mole ratio from the equation to a mass instead of to a mole count. The ratio links moles and nothing else.
- Leaving the volume in millilitres in M = n / V, which makes the molarity a thousand times too large.
- Forgetting the 1000 when a volume in millilitres is used in the gas law, where litres are required.
- Using a Celsius temperature in the gas law or in a heat calculation, both of which need the absolute scale.
- Assuming that the first reactant named is the limiting one. It is the one that runs out, which has to be worked out.
- Adding the solvent volume to the solution volume instead of using M1 V1 = M2 V2. Volumes are not always additive, and it is the moles that are conserved.
- Mixing the units of R. 8.314 belongs with kilopascals and litres, and any other pair of units needs another value.
- Quoting a yield above 100 per cent, which almost always means the possible amount came from the wrong reactant or the product was weighed wet.
Worked example
A 2 M stock solution of hydrochloric acid is used to make 500 mL of a 0.4 M solution. What volume of the stock is needed, and how many grams of calcium carbonate of molar mass 100 would 0.2 mol of that acid dissolve, given CaCO3 + 2 HCl -> CaCl2 + H2O + CO2?
- The dilution is a question about moles: the concentrated and the dilute solution contain the same number of moles, so M1 V1 = M2 V2.
- Put in what is known: 2 x V1 = 0.4 x 500, so V1 = 200 / 2 = 100 mL.
- Check it against the sense of the question: making a solution five times weaker from 500 mL needs one fifth of the volume of stock, which is 100 mL. That is what the equation gave.
- For the second part, the equation says two moles of acid are used for every mole of calcium carbonate, so 0.2 mol of acid reacts with 0.1 mol of the carbonate.
- Mass is moles times molar mass: 0.1 x 100 = 10 g. The ratio was applied to moles and the multiplication by the molar mass came afterwards; doing it in the other order gives a wrong answer.
Answer: 100 mL of stock, which would dissolve 10 g of calcium carbonate
Practice questions with answers
A few Chemical Engineering questions with the full solution shown, so you can see
how the method is applied before you attempt the timed set.
Question 1
How many moles are there in 70 g of a substance whose molar mass is 28 g/mol?
Answer: Option D — with explanation
A mole is a fixed number of particles, and its mass in grams is the molar mass, so the number of moles is the mass divided by the molar mass.
70 / 28 = 5/2 mol. Multiplying the mass by the molar mass instead of dividing gives a number with the wrong units as well as the wrong size.
Common mistakes
- 2/5 is not the answer: 2/5
- 5 is not the answer: 5
- 1960 is not the answer: 1960
Question 2
What is the molarity of a solution containing 3 mol of solute in 500 mL of solution? Give the answer in mol per litre.
Answer: Option B — with explanation
Molarity is the number of moles of solute in one litre of solution, so the volume has to be in litres before the division.
3 mol in 500 mL is 1/2 L, and 3 / 1/2 = 6 mol per litre. Leaving the volume in millilitres makes the answer a thousand times too large, which is the commonest slip with this question.
Common mistakes
- 12 is not the answer: 12
- 3/500 is not the answer: 3/500
- 1500 is not the answer: 1500
Question 3
25 mL of a 1 mol per litre solution is diluted with water to a total volume of 125 mL. What is the new concentration, in mol per litre?
Answer: Option C — with explanation
Diluting adds water, not solute, so the number of moles stays the same and only the volume changes. That is why C1 V1 = C2 V2.
1 x 25 = 25 millimoles of solute, and spreading those over 125 mL gives 1/5 mol per litre. The concentration must fall when a solution is diluted, so any answer above 1 has the ratio the wrong way up.
Common mistakes
- 2/5 is not the answer: 2/5
- 1 is not the answer: 1
- 5 is not the answer: 5
Question 4
The reaction takes 1 mol of Fe2O3, of molar mass 160 g/mol, and gives 2 mol of Fe, of molar mass 56 g/mol. Starting from 320 g of Fe2O3, what mass of Fe is formed? Give the answer in grams.
Answer: Option B — with explanation
Work in moles, not in grams. First find the moles of reactant, then use the reaction's own ratio to find the moles of product, and only then turn those moles back into a mass.
320 g / 160 g per mol = 2 mol of Fe2O3, which gives 4 mol of Fe, and that weighs 4 x 56 = 224 g. Applying the ratio to the masses instead of to the moles is the mistake the other answers are built on: mass is not conserved molecule for molecule, because the products and the reactants have different molar masses.
Common mistakes
- 280 is not the answer: 280
- 56 is not the answer: 56
- 112 is not the answer: 112
Question 5
4 mol of an ideal gas is held at 200 degrees Celsius and 100 kPa. What volume does it occupy, in litres? Take R as 8.314 J per mol per kelvin and give the answer to two decimal places.
-
A
157.30
-
B
157300.88
-
C
0.16
-
D
66.51
Answer: Option A — with explanation
The ideal gas law is PV = nRT, so the volume is nRT over P - but T has to be in kelvin, and the pressure in pascals if R is 8.314.
200 degrees Celsius is 473 K. The volume in cubic metres is 4 x 8.314 x 473 / 100000 = 0.16, and a cubic metre is a thousand litres, so the answer is 157.30 litres. Leaving the temperature in Celsius is the mistake that gives a volume that is too small.
Common mistakes
- 157300.88 is not the answer: 157300.88
- 0.16 is not the answer: 0.16
- 66.51 is not the answer: 66.51
Frequently asked questions
Why does the mole ratio from the equation not apply to grams?
Because different substances have different molar masses, so equal masses are not equal amounts. The equation balances atoms, and atoms are counted in moles. Convert to moles, apply the ratio, and convert back.
How do I know whether a question wants molarity or a mass?
From the wording and the units: per litre means molarity, grams means mass. If the question asks for a concentration after a dilution, the answer is a molarity and M1 V1 = M2 V2 is enough on its own.
What is the limiting reactant?
The one used up first. Work out how much product each reactant could give on its own, and the smaller figure is what the reaction can produce. Everything after that is worked from the limiting reactant.
Do I have to convert to moles for a dilution question?
No. M1 V1 = M2 V2 already accounts for the moles, provided the two concentrations are quoted in the same units and the two volumes in the same units.
Which value of R should I use?
8.314 with kilopascals and litres, which is the usual pairing in a question of this kind. Convert the pressure and the volume to those units first, and convert the temperature to kelvin as well.
Why has my yield come out above 100 per cent?
Almost always because the theoretical amount was worked out from the wrong reactant, or because the product was weighed while still wet. Check the limiting reactant first, then the state of the product.
