Moles and Calculations
The Mole
The mole (mol) is the unit for the amount of substance in chemistry.
One mole of any substance contains the Avogadro constant (6.02 x 10^23) particles (atoms, molecules, ions or formula units).
The molar mass of a substance is the mass of one mole:
- For an element: molar mass = Ar in grams (e.g. one mole of iron = 56 g)
- For a compound: molar mass = Mr in grams (e.g. one mole of water = 18 g)
Moles, Mass and Relative Formula Mass
Rearrangements:
- mass = moles x molar mass
- molar mass = mass / moles
Example: How many moles are in 42 g of carbon? (Ar of C = 12)
moles = 42 / 12 = 3.5 mol
Example: What is the mass of 0.25 mol of zinc? (Ar of Zn = 65)
mass = 0.25 x 65 = 16.25 g
Reacting Masses
Balanced symbol equations show the molar ratios of reactants and products. These ratios can be used to calculate the mass of any reactant or product given the mass of another.
Method:
- Write the balanced equation.
- Calculate the moles of the given substance (moles = mass / Mr).
- Use the molar ratio from the equation to find moles of the desired substance.
- Calculate the mass (mass = moles x Mr).
Example: Calculate the mass of MgO produced when 6.0 g of Mg burns. 2Mg(s) + O2(g) → 2MgO(s) (Ar: Mg = 24, O = 16)
- Moles of Mg = 6.0 / 24 = 0.25 mol
- Ratio Mg:MgO = 2:2 = 1:1, so moles of MgO = 0.25 mol
- Mass of MgO = 0.25 x (24 + 16) = 0.25 x 40 = 10 g
Percentage Yield
In practice, the actual yield (amount of product obtained) is always less than the theoretical yield (maximum possible). Reasons include:
- The reaction may not go to completion (especially reversible reactions)
- Some product may be lost during separation and purification (e.g. filtration, evaporation)
- Side reactions may use up some of the reactants
Example: A student prepares copper(II) sulfate. Theoretical yield = 2.0 g; actual yield = 1.6 g.
Percentage yield = (1.6 / 2.0) x 100 = 80%
Exam Tip
Percentage yield can never exceed 100%. If you get a value above 100%, you have divided the wrong way round. Check that you are dividing actual yield by theoretical yield.
Empirical and Molecular Formulae
Molecular formula: gives the actual number of atoms of each element in one molecule or formula unit.
Empirical formula: gives the simplest whole-number ratio of atoms of each element in a compound.
Example: Ethanoic acid has molecular formula C2H4O2. Dividing all subscripts by 2 gives the empirical formula CH2O.
Calculating Empirical Formula
Given masses or percentages of each element:
- Write down the mass (or percentage) of each element.
- Divide each mass by the element’s Ar to get moles.
- Divide all mole values by the smallest value to get the ratio.
- If the ratio contains decimals (e.g. 1.5), multiply all values by a suitable whole number to get whole numbers.
Example: A compound contains 10 g H and 80 g O. (Ar: H = 1, O = 16)
| Element | Mass (g) | Ar | Moles | Ratio |
|---|---|---|---|---|
| H | 10 | 1 | 10 | 2 |
| O | 80 | 16 | 5 | 1 |
Empirical formula: H2O
Calculating Molecular Formula from Empirical Formula
- Find Mr of the empirical formula.
- Divide the actual Mr of the compound by the Mr of the empirical formula.
- Multiply subscripts in the empirical formula by this number.
Example: Empirical formula CH2O, actual Mr = 180. Mr of CH2O = 12 + 2 + 16 = 30. 180 / 30 = 6. Molecular formula: C6H12O6
Finding Formulae Experimentally
Metal Oxide by Combustion (e.g. Magnesium Oxide)
- Weigh the empty crucible and lid.
- Add magnesium ribbon; weigh again to find mass of Mg.
- Heat strongly with the lid slightly open (to let in air but prevent MgO smoke escaping).
- Continue until mass is constant.
- Calculate: mass of oxygen = mass of MgO - mass of Mg.
- Find moles of each, simplify the ratio, write the empirical formula.
Metal Oxide by Reduction (e.g. Copper(II) Oxide)
- Weigh empty boiling tube.
- Add copper(II) oxide; weigh to find mass of CuO.
- Pass methane over the heated CuO; the oxygen is removed as water.
- Weigh the remaining copper metal.
- Calculate: mass of oxygen removed = mass of CuO - mass of Cu.
- Find moles of each, simplify the ratio.
Higher Tier
Concentration of Solutions
Concentration is the amount of solute dissolved in a given volume of solution.
Rearrangements:
- moles = concentration x volume
- volume = moles / concentration
Converting volumes: 1 dm3 = 1000 cm3. To convert cm3 to dm3, divide by 1000.
Example: Calculate the concentration of NaOH solution when 80 g is dissolved in 500 cm3 of water. (Mr of NaOH = 40)
- Moles of NaOH = 80 / 40 = 2 mol
- Volume = 500 / 1000 = 0.5 dm3
- Concentration = 2 / 0.5 = 4 mol/dm3
Gas Volumes
Avogadro’s Law: at the same temperature and pressure, equal amounts (in moles) of gases occupy the same volume.
At room temperature and pressure (RTP) (20 °C, 1 atm), 1 mole of any gas occupies 24 dm3 (or 24,000 cm3).
Example: What volume of gas is occupied by 0.5 mol at RTP?
Volume = 0.5 x 24 = 12 dm3
Example: Using gas volumes in equations: in C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g), if 150 cm3 of propane reacts, the volume of O2 needed = 5 x 150 = 750 cm3, and total volume of gaseous products = (3 + 4) x 150 = 1050 cm3.
Exam Questions
Q: Calculate the number of moles in 4.4 g of carbon dioxide. (Mr = 44)
moles = 4.4 / 44 = 0.1 mol
Q: In the reaction 2H2 + O2 → 2H2O, calculate the mass of water produced from 4 g of hydrogen. (Ar: H = 1, O = 16)
Moles of H2 = 4 / 2 = 2 mol; ratio H2:H2O = 1:1, so moles of H2O = 2 mol; mass = 2 x 18 = 36 g
Q: A student produces 4.5 g of a salt. The theoretical yield was 6.0 g. Calculate the percentage yield.
Percentage yield = (4.5 / 6.0) x 100 = 75%
Q: A compound contains 40% C, 6.67% H and 53.33% O by mass. Calculate its empirical formula. (Ar: C = 12, H = 1, O = 16)
Moles: C = 40/12 = 3.33; H = 6.67/1 = 6.67; O = 53.33/16 = 3.33. Divide by 3.33: C:H:O = 1:2:1. Empirical formula: CH2O
Q: Calculate the volume at RTP of 2 mol of carbon dioxide gas.
Volume = 2 x 24 = 48 dm3
Exam Tip
For all multi-step calculations, show every step of working. Examiners award marks for each correct step, so even if you make an arithmetic error, you may still get marks for the correct method.