There are several different ways of calculating moles and you need to be confident with all of them. You’ll usually have to use more than one approach to calculating moles in any given question.
I’ve worked through some very straightforward examples to help you practice with each of the equations and ideas. The examples only use a single approach at a time. We’ll look at more complex examples later on.
- From a number of particles
By definition, there are 6.0 x 1023 particles in 1 mole of substance. This is Avogadro’s number (NA). The equation that links number of particles and moles is: n = N / NA, where:
n = number of moles (mol)
N = number of particles
NA = Avogadro’s number
| Q1: How many moles are there in 1.2 x 1024 molecules of N2? n =N / NA = 1.2 x 1024 / 6.0 x 1023 = 2.0 mol |
| Q2: How many molecules are there in 5.0 mol of O2? Re-arrange the equation to make N the subject. N = n x NA= 5.0 x 6.0 x 1023 = 3.0 x 1024 |
| Q3: How many atoms are there in 4.0 mol of CO2? Re-arrange the equation to make N the subject. N = n x NA= 4.0 x 6.0 x 1023 = 2.4 x 1024 There are 3 atoms in each molecule of CO2 so there will be 3 x 2.4 x 1024 = 7.2 x 1024 atoms in total |
- From the mass of a substance
The Relative Formula Mass (Mr) is calculated by adding up the Relative Atomic Masses (Ar) of each atom in the formula unit. To do this, you will need to understand how to count atoms of each type in the formula unit. The equation that links mass and moles is: n = m / Mr, where:
n = number of moles (mol)
m = mass of substance (g)
Mr = Relative Formula Mass
Remember: You will sometimes need to convert the mass unit.
| Q4: How many moles are there in 2200 mg of CO2? Convert the 2200 mg into g by dividing by 1000 –> m = 2.2 g Work out Mr from the formula –>Mr = 44 n = m / Mr = 2.2 / 44 = 0.050 mol |
| Q5: What is the mass of 0.1 mol of O2? Work out Mr from the formula –> Mr = 32 Re-arrange the equation to make m the subject m = n x Mr = 0.1 x 32 = 3.2 g |
| Q6: What is the Relative Formula Mass if 10 mol of a substance has a mass of 0.71 kg? Convert the 0.71 kg into g by multiplying by 1000 –> m = 710 g Re-arrange the equation to make Mr the subject Mr = m / n = 710 / 10 = 71 |
- From the concentration and volume of a solution
In an aqueous solution the number of moles is proportional to the concentration and volume of the solution. The equation that links concentration, volume and moles is: n = cv, where:
n = number of moles (mol)
c = concentration (mol dm-3)
v = volume (dm3)
Remember: You will sometimes need to convert the volume unit.
| Q7: How many moles are there in 100 cm3 of 0.20 mol dm-3 HCl? Convert the 100 cm3 into dm3 first by dividing by 1000 –> v = 0.1 dm3 n = c x v = 0.2 x 0.1 = 0.020 mol |
| Q8: What is the concentration if there is 0.10 mol HNO3 in 500 cm3? Convert the 500 cm3 into dm3 by dividing by 1000 –> v = 0.5 dm3 Re-arrange the equation to make c the subject c = n / v = 0.10 / 0.5 = 0.2 mol dm-3 |
| Q9: What is the volume in cm3 if there is 0.40 mol of 0.5 mol dm-3 HNO3? Re-arrange the equation to make v the subject v = n / c = 0.4 / 0.5 = 0.8 dm3 Convert the 0.8 dm3 into cm3 by multiplying by 1000 v = 800 cm3 |
| Q10: How many moles in a 10 cm3 sample if there are 0.150 mol in the 50 cm3 total solution? This question is different and many students find this sort of step quite difficult. There isn’t really an equation to use but instead we need to think carefully about proportions. We start by comparing the volumes: 50 / 10 = 5 The volume of the sample is 5x smaller than the volume of the total solution. We know that the number of moles is proportional to volume. This means the number of the moles in the sample must be 5x smaller too. 0.150 / 5 = 0.030 mol |
| Q11: How many moles in the 87.5 cm3 total solution if there are 0.60 mol in a 12.5 cm3 sample? We start by comparing the volumes: 87.5 / 12.5 = 7 The volume of the total solution is 7x larger than the volume of the sample. This means that the number of moles in the total solution be 7x larger too. 0.60 x 7 = 4.2 mol |
| Q12: 100 cm3 of distilled water is added to 400 cm3 of 0.50 mol dm-3 HCl. What is the new concentration of the solution? We need to use the idea that dilution does not affect the total number of moles of substance in the solution. Convert the the 400 cm3 into dm3 by dividing by 1000 –> v = 0.4 dm3 n = c x v = 0.50 x 0.4 = 0.020 mol This means that there is 0.020 mol of HCl at the start and the same number of moles of at the end. Convert the the new total volume 500 cm3 (i.e., 100 + 400) into dm3 by dividing by 1000 –> v = 0.5 dm3 Re-arrange the equation to make c the subject c = n / v = 0.020 / 0.5 = 0.40 mol dm-3 |
| Q13: What volume of distilled water must be added to 200 cm3 of 0.80 mol/dm3 HCl to make a solution with a concentration of 0.50 mol/dm3? Convert the 200 cm3 into dm3 by dividing by 1000 –> v = 0.2 dm3 n = c x v = 0.8 x 0.2 = 0.16 mol Re-arrange the equation to make v the subject v = n / c = 0.16 / 0.5 = 0.32 dm3 This is, of course, equivalent to 320 cm3. The original volume is 200 cm3 so 120 cm3 of distilled water must be added to make the a solution with the required concentration. |
- From the pressure, volume and temperature of a gas
If you have a gas, the number of moles is proportional to the pressure and volume, and inversely proportional to the temperature of the solution. The equation that links pressure, volume, temperature of a gas and moles is: pV = nRT, where:
n = number of moles (mol)
p = pressure (Pa)
v = volume (m3)
R = molar gas constant = 8.31 J K-1 mol-1
T = Temperature (K)
Remember you will sometimes need to convert the pressure, volume and temperature units.
| Q14: Calculate the pressure in kPa when 0.05 mol of an ideal gas is confined in a 500 cm3 at 27 oC Convert the 500 cm3 into m3 by dividing by 106 –> V =500 x 10-6 cm3 Convert the 27 oC into K by adding 273 –> T = 300 K Re-arrange the equation to make p the subject p = nRT/V = (0.05 x 8.31 x 300) / (500 x 10-6) = 2.49 x 105 Pa This is, of course, equivalent to 249 kPa |
| Q15: Calculate the volume in dm3 when 0.10 mol of an ideal gas is held at a pressure of 0.2 kPa at 330 oC Convert the 0.2 kPa in Pa by multiplying by 1000 –> p = 0.2 x 103 Pa Convert the 330 oC into K by adding 273 –> T = 603 K Re-arrange the equation to make p the subject V = nRT / p = (0.10 x 8.31 x 603) / (0.2 x 103 Pa) = 2.51 m3 This is, of course, equivalent to 2510 dm3 |
| Q16: Calculate the temperature in oC when 0.50 mol of an ideal gas is held at a pressure of 10 kPa in a volume of 400 dm3 Convert the 10 kPa in Pa by multiplying by 1000 –> p = 10 x 103 Pa Convert the 400 dm3 into m3 by dividing by 1000 –> V =0.4 m3 Re-arrange the equation to make p the subject T = pV / nR= (10 x 103 x 0.4) / (0.50 x 8.31) = 963 K This is, of course, equivalent to 690 oC |
| Q17: Calculate the moles of gas when an ideal gas is held at a pressure of 500 kPa in a volume of 7000 cm3 at 100 oC, Convert the 500 kPa in Pa by multiplying by 1000 –> p = 500 x 103 Pa Convert the 7000 cm3 into m3 by dividing by 106 –> V =7000 x 10-6 m3 Convert the 100 oC into K by adding 273 –> T = 373 K Re-arrange the equation to make n the subject n = pV / RT= (500 x 103 x 7000 X 10-6) / (8.31 x 373) = 1.13 mol |
- From a balanced symbol equation
If you have a balanced symbol equation, the changes in moles of each substance are in the same ratio as the equation. The amount of reactants will decrease and the amount of the products will increase.
| Q18: Calculate the moles of O2 used up and the moles of CO2 and H2O formed when 0.20 mol of C3H8 combusts. Equation: 1 C3H8 + 5 O2 –> 3 CO2 + 4 H2O This equation tells us that the moles of C3H8, O2, CO2 and H2O change in a 1:5:3:4 ratio. If the moles of C3H8 decrease by 0.20 mol (= 1 / 1 x 0.20) The moles of O2 will decrease by 1.0 mol (= 5 / 1 x 0.20) The moles of CO2 will increase by 0.60 mol (= 3 / 1 x 0.20) The moles of H2O will increase by 0.80 mol (= 4 / 1 x 0.20) If you can’t immediately see where these numbers have come from, look at how the calculation in the brackets relate to the numbers in the balanced symbol equation. |
| Q19: Calculate the moles of MgO, NO2 and O2 formed when 0.50 mol of Mg(NO3)2 undergoes thermal decomposition Equation: 2 Mg(NO3)2 –> 2 MgO + 4 NO2 + 1 O2 This equation tells us that moles of Mg(NO3)2, MgO, NO2 and O2 change in a 2:2:4:1 ratio. If the moles of Mg(NO3)2 decrease by 0.50 mol (= 2 / 2 x 0.50) The moles of MgO will increase by 0.50 mol (= 2 / 2 x 0.50) The moles of NO2 will increase by 1.0 mol (= 4 / 2 x 0.50) The moles of O2 will increased by 0.25 mol (= 1 / 2 x 0.50) |
| Q20: C2H6 burns completely in O2 to form 1.20 mol H2O. Calculate the changes in the moles of all substances. Equation: 2 C2H6 + 7 O2 –> 4 CO2 + 6 H2O This equation tells us that the moles of C2H6, O2, CO2 and H2O change in a 2:7:4:6 ratio. If the moles of H2O increase by 1.2 mol (= 6 / 6 x 1.20) The moles of CO2 will increase by 0.80 mol (= 4 / 6 x 1.20) The moles of O2 will decrease by 1.4 mol (= 7 / 6 x 1.20) The moles of C3H8 will decrease by 0.40 mol (= 2 / 6 x 1.20) |
The changes in moles of each substance are always in the same ratio as in the balanced symbol equation. This is a really powerful idea in Chemistry and means that if we know the initial amounts of all the substances, and the change of one of the substances, we can work out the final amounts of all the substances. Big Idea #3 will introduce the wonder that is an ICE table!