Quantitative Aspects of Chemical Change
The mole and Avogadro's number
An equation for a chemical reaction provides the identities of reactants and products and the ratio in which the reactants combine. For example:
Fe + S → FeS
This equation shows that one atom of iron (Fe) combines with one atom of sulphur (S) to form iron(II) sulphide (FeS). The equation, however, does not tell us the amount (how many atoms or how much mass) of each substance involved. In chemistry we use the mole and Avogadro's number to quantify the number of particles (atoms, molecules, ions) and to relate particle numbers to mass.
Mole
Definition: The mole (abbreviation mol) is the SI unit for amount of substance. A mole is a counting unit for particles (atoms, molecules, ions, formula units) analogous to a dozen but much larger.
Avogadro's number
Definition: The number of particles in one mole is called Avogadro's number and equals 6.022 × 1023 particles. This number is usually written as NA = 6.022 × 1023.
Examples: 12.0 g of carbon contains one mole of carbon atoms (6.022 × 1023 atoms). 63.5 g of copper is one mole of copper atoms.
Molar mass
Definition: The molar mass (M) of a substance is the mass of one mole of that substance. The unit is grams per mole (g · mol-1).
On the periodic table the relative atomic mass (atomic weight) of an element has two useful interpretations:
- It gives the mass (in atomic mass units) of an average atom of the element;
- Numerically, the same value (but expressed in grams) is the mass of one mole of that element - the molar mass (g · mol-1).
| Element | Relative atomic mass (u) | Molar mass (g · mol-1) | Mass of one mole (g) |
|---|---|---|---|
| Magnesium | 24.3 | 24.3 | 24.3 |
| Lithium | 6.94 | 6.94 | 6.94 |
| Oxygen | 16.0 | 16.0 | 16.0 |
| Nitrogen | 14.0 | 14.0 | 14.0 |
| Iron | 55.8 | 55.8 | 55.8 |
Calculations using moles and mass
The central relation connecting mass, moles and molar mass is:
n = m / M
where n is the number of moles, m is the mass in grams and M is the molar mass in g · mol-1.
A convenient mnemonic is a triangle with m over n and M, meaning n = m ÷ M, m = n × M and M = m ÷ n. Always write units next to values used in calculations.
QUESTION Calculate the number of moles of iron (Fe) in an 11.7 g sample.
Sol
The molar mass of iron is 55.8 g · mol-1. Using n = m / M, n = 11.7 g ÷ 55.8 g · mol-1 = 0.21 mol (approximately).
QUESTION You have a sample that contains 5 moles of zinc. (a) What is the mass of the zinc? (b) How many atoms of zinc are in the sample?
Sol
The molar mass of zinc is 65.4 g · mol-1. The mass is m = n × M = 5 mol × 65.4 g · mol-1 = 327 g.
The number of atoms is 5 mol × 6.022 × 1023 atoms · mol-1 = 3.011 × 1024 atoms.
QUESTION Calculate the number of moles of copper in a 127 g sample.
Sol
Use n = m / M. M(Cu) = 63.5 g · mol-1, n = 127 g ÷ 63.5 g · mol-1 = 2.00 mol.
QUESTION Calculate the number of atoms in an 81 g sample of aluminium.
Sol
M(Al) = 27.0 g · mol-1. Number of moles is n = 81 g ÷ 27.0 g · mol-1 = 3.00 mol.
Number of atoms = 3.00 mol × 6.022 × 1023 = 1.8066 × 1024 atoms.
Exercises (selected)
- How many atoms are there in:
- 1 mole of a substance?
- 2 moles of calcium?
- 5 moles of phosphorus?
- 24.3 g of magnesium?
- 24.0 g of carbon?
- Complete the table of element, relative atomic mass, sample mass and number of moles:
| Element | Relative atomic mass (u) | Sample mass (g) | Number of moles in the sample |
|---|---|---|---|
| Hydrogen | 1.01 | 1.01 | |
| Magnesium | 24.3 | 24.3 | |
| Carbon | 12.0 | 24.0 | |
| Chlorine | 35.45 | 70.9 | |
| Nitrogen | 14.0 | 42.0 |
Compounds and molar mass of molecules
The same mole and molar mass concepts apply to compounds. For a covalent molecule, add the molar masses of each atom to find the molar mass of the molecule. For ionic solids we use formula mass (the mass of a formula unit). The mole count applies to whole molecules or formula units.
Example of mole ratios in a balanced chemical equation:
N2 + 3H2 → 2NH3
This means 1 mole of nitrogen molecules reacts with 3 moles of hydrogen molecules to produce 2 moles of ammonia molecules.
QUESTION Calculate the molar mass of H2SO4.
Sol
M(H) = 1.01 g · mol-1, M(S) = 32.1 g · mol-1, M(O) = 16.0 g · mol-1.
M(H2SO4) = (2 × 1.01) + 32.1 + (4 × 16.0) = 98.12 g · mol-1.
QUESTION Calculate the number of moles in 1 kg of MgCl2.
Sol
Convert mass: 1 kg = 1000 g.
M(MgCl2) = 24.3 + 2 × 35.45 = 95.2 g · mol-1.
n = 1000 g ÷ 95.2 g · mol-1 = 10.50 mol (approximately).
Group tasks and understanding
Suggested classroom discussion items:
- What are the units of the mole?
- Given a 46 g sample of NO2 determine:
- moles of NO2,
- moles of nitrogen atoms,
- moles of oxygen atoms,
- number of molecules of NO2.
- Explain the difference between a mole and a molecule.
- Write Avogadro's number without scientific notation and estimate how long it would take to count to it at two numbers per second.
Exercises on molar mass and mole calculations
- Give the molar mass of:
- hydrogen gas (H2),
- nitrogen gas (N2),
- bromine gas (Br2).
- Calculate the number of moles in:
- 21.6 g of boron (B),
- 54.9 g of manganese (Mn),
- 100.3 g of mercury (Hg),
- 50 g of barium (Ba),
- 40 g of lead (Pb).
Composition and formula determination
Four common types of composition problems:
- Given the formula, calculate percentage by mass of each element.
- Given percentage composition, determine empirical or molecular formula.
- Combustion or reaction analysis: deduce formula of a reactant from products.
- Determine number of moles of water of crystallisation from mass loss on heating.
QUESTION Calculate the percentage by mass of each element in sulphuric acid, H2SO4.
Sol
M(H2SO4) = 98.12 g · mol-1 as calculated earlier.
Mass contributed by hydrogen = 2 × 1.01 = 2.02 g · mol-1. Percentage = 2.02 ÷ 98.12 × 100% = 2.06% (approximately).
Mass contributed by sulphur = 32.1 g · mol-1. Percentage = 32.1 ÷ 98.12 × 100% = 32.7% (approximately).
Mass contributed by oxygen = 4 × 16.0 = 64.0 g · mol-1. Percentage = 64.0 ÷ 98.12 × 100% = 65.2% (approximately).
QUESTION A compound contains 52.2% carbon, 13.0% hydrogen and 34.8% oxygen by mass. Determine its empirical formula.
Sol
Assume 100 g sample: C = 52.2 g, H = 13.0 g, O = 34.8 g.
Number of moles: n(C) = 52.2 ÷ 12.0 = 4.35 mol; n(H) = 13.0 ÷ 1.01 ≈ 12.87 mol; n(O) = 34.8 ÷ 16.0 = 2.175 mol.
Divide by smallest (2.175): C = 4.35 ÷ 2.175 = 2; H = 12.87 ÷ 2.175 ≈ 6; O = 2.175 ÷ 2.175 = 1.
Empirical formula: C2H6O.
QUESTION 207 g of lead combines with oxygen to form 239 g of a lead oxide. Determine the formula of the lead oxide (Pb = 207.2 u, O = 16.0 u).
Sol
Mass of oxygen combined = 239 g - 207 g = 32 g.
Moles of Pb = 207 g ÷ 207.2 g · mol-1 ≈ 0.999 mol ≈ 1.00 mol.
Moles of O = 32 g ÷ 16.0 g · mol-1 = 2.00 mol.
Mole ratio Pb : O = 1 : 2, so formula is PbO2.
QUESTION A sample of acetic acid has percentage composition 39.9% C, 6.7% H and 53.4% O. (1) Determine the empirical formula. (2) Determine the molecular formula if the molar mass is 60.06 g · mol-1.
Sol
Assume 100 g sample: C = 39.9 g, H = 6.7 g, O = 53.4 g.
Moles: n(C) = 39.9 ÷ 12.0 = 3.325 mol; n(H) = 6.7 ÷ 1.01 ≈ 6.634 mol; n(O) = 53.4 ÷ 16.0 = 3.3375 mol.
Divide by smallest (3.325 ≈ 3.3375): relative ratio ≈ 1 : 2 : 1 → empirical formula CH2O.
Molar mass of empirical formula = 12.0 + 2×1.01 + 16.0 = 30.02 g · mol-1. Given molar mass 60.06 g · mol-1 is twice this, so molecular formula is C2H4O2 (commonly written CH3COOH).
Waters of crystallisation
QUESTION Aluminium trichloride crystals are AlCl3·nH2O. A sample had mass 5.0 g before heating and 2.8 g after heating (all water lost). Determine n.
Sol
Mass of water lost = 5.0 g - 2.8 g = 2.2 g. Mass of anhydrous AlCl3 = 2.8 g.
Moles AlCl3 = 2.8 g ÷ 133.35 g · mol-1 = 0.0210 mol (approx).
Moles H2O = 2.2 g ÷ 18.02 g · mol-1 = 0.1221 mol (approx).
Ratio H2O : AlCl3 = 0.1221 ÷ 0.0210 ≈ 5.81 ≈ 6. Therefore n = 6 and formula is AlCl3·6H2O.
Experimental determination of waters of crystallisation is commonly done by heating a hydrated salt, weighing before and after heating, and converting masses to moles to find an integer ratio.
Amount of substance: gases and molar volume
Molar volume of gases: One mole of any ideal gas occupies 22.4 dm3 at standard temperature and pressure (S.T.P.), where S.T.P. is defined as T = 273.15 K and pressure ≈ 1 atm. Thus one mole of gas at S.T.P. has volume 22.4 dm3.
Concentration of solutions
Definition: Concentration (molarity) C is the amount of solute (moles n) dissolved per unit volume V of solution (in dm3):
C = n / V
Units: mol · dm-3 (often written mol L-1). Volume must be expressed in dm3 (1 dm3 = 1 litre = 1000 cm3).
QUESTION If 3.5 g of sodium hydroxide (NaOH) is dissolved in 2.5 dm3 of water, what is the concentration in mol · dm-3?
Sol
Molar mass M(NaOH) = 40.01 g · mol-1. Moles n = 3.5 g ÷ 40.01 g · mol-1 = 0.0875 mol.
Concentration C = n ÷ V = 0.0875 mol ÷ 2.5 dm3 = 0.035 mol · dm-3.
QUESTION What mass of KMnO4 is needed to prepare 1.00 dm3 of a 0.20 mol · dm-3 solution?
Sol
Moles required n = C × V = 0.20 mol · dm-3 × 1.00 dm3 = 0.20 mol.
M(KMnO4) ≈ 158.0 g · mol-1. Mass m = n × M = 0.20 mol × 158.0 g · mol-1 = 31.6 g.
QUESTION How much NaCl (in g) is needed to prepare 500 cm3 of 0.01 mol · dm-3 solution?
Sol
Convert volume: 500 cm3 = 0.5 dm3.
Moles n = C × V = 0.01 mol · dm-3 × 0.5 dm3 = 0.005 mol.
M(NaCl) ≈ 58.45 g · mol-1. Mass m = 0.005 mol × 58.45 g · mol-1 = 0.292 g (≈ 0.29 g).
Exercises on concentration
- 5.95 g of potassium bromide is dissolved in 400 cm3 of water. Calculate its concentration.
- 100 g of sodium chloride is dissolved in 450 cm3 of water.
- How many moles of NaCl are present?
- What is the volume in dm3?
- Calculate the concentration in mol · dm-3.
- What is the molarity of the solution formed by dissolving 80 g of NaOH in 500 cm3 of water?
- What mass of HCl is needed to make up 1000 cm3 of a 1.0 mol · dm-3 solution?
- How many moles of H2SO4 are there in 250 cm3 of a 0.8 mol · dm-3 solution? What mass of acid is in this solution?
Stoichiometry: calculations with balanced equations
Stoichiometry is the calculation of quantities of reactants and products in chemical reactions using the mole concept and balanced chemical equations. The coefficients in a balanced equation give mole ratios between substances.
QUESTION What volume of oxygen at S.T.P. is needed for the complete combustion of 2.0 dm3 of propane (C3H8)? Combustion produces CO2 and H2O.
Sol
Balanced equation: C3H8 + 5O2 → 3CO2 + 4H2O.
Mole (volume) ratio O2 : C3H8 = 5 : 1. Therefore 1 volume of propane requires 5 volumes of oxygen.
Volume of O2 required = 2.0 dm3 × 5 = 10.0 dm3 at S.T.P.
QUESTION What mass of iron(II) sulphide (FeS) is formed when 5.6 g of iron reacts completely with sulphur?
Sol
Balanced equation: Fe + S → FeS. Mole ratio Fe : FeS = 1 : 1.
M(Fe) = 55.8 g · mol-1. Moles Fe = 5.6 g ÷ 55.8 g · mol-1 = 0.100 mol.
Thus moles FeS produced = 0.100 mol. M(FeS) = 55.8 + 32.1 = 87.9 g · mol-1. Mass = 0.100 mol × 87.9 g · mol-1 = 8.79 g.
Theoretical yield and percentage yield
Theoretical yield is the maximum amount of product expected from a reaction based on stoichiometric calculations. The actual yield is what is obtained experimentally. Percentage yield is:
% yield = (actual yield ÷ theoretical yield) × 100
QUESTION Sulphuric acid reacts with ammonia to produce ammonium sulphate: H2SO4 + 2NH3 → (NH4)2SO4. If 2.0 kg of H2SO4 is used and 2.2 kg of fertiliser is obtained, what is the theoretical yield and the % yield?
Sol
Convert 2.0 kg to grams: 2000 g of H2SO4. M(H2SO4) = 98.12 g · mol-1. Moles H2SO4 = 2000 g ÷ 98.12 g · mol-1 ≈ 20.383 mol.
Mole ratio H2SO4 : (NH4)2SO4 = 1 : 1, so moles product = 20.383 mol.
M((NH4)2SO4) = 114.04 g · mol-1. Theoretical mass = 20.383 mol × 114.04 g · mol-1 ≈ 2324 g = 2.324 kg.
Percentage yield = (actual 2.200 kg ÷ theoretical 2.324 kg) × 100 ≈ 94.6%.
QUESTION Barium chloride and sulphuric acid react according to: BaCl2 + H2SO4 → BaSO4 + 2HCl. If you have 2.0 g of BaCl2, (1) what mass of H2SO4 is required to react completely? (2) What mass of HCl is produced?
Sol
M(BaCl2) = 208.2 g · mol-1. Moles BaCl2 = 2.0 g ÷ 208.2 g · mol-1 = 0.00961 mol (approx).
Mole ratio BaCl2 : H2SO4 = 1 : 1, so moles H2SO4 required = 0.00961 mol. Mass H2SO4 = 0.00961 mol × 98.12 g · mol-1 ≈ 0.94 g.
Mole ratio to HCl: 1 mol BaCl2 produces 2 mol HCl; moles HCl = 2 × 0.00961 = 0.01922 mol. M(HCl) = 36.46 g · mol-1. Mass HCl = 0.01922 mol × 36.46 g · mol-1 ≈ 0.70 g.
Stoichiometry exercises (selected)
- Diborane combustion: B2H6 + 3O2 → 2HBO2 + 2H2O. If 2.37 g B2H6 reacts, how many grams of water are produced?
- Sodium azide decomposition: 2NaN3 → 2Na + 3N2. If 23.4 g NaN3 decomposes, how many moles of N2 are produced and what volume would this gas occupy at S.T.P.?
- Photosynthesis: (a) Write and balance the equation for photosynthesis producing glucose (C6H12O6) and O2. (b) If 3.0 mol CO2 are used, what mass of glucose is produced?
Summary
- The mole (mol) is the SI unit for amount of substance.
- Avogadro's number NA = 6.022 × 1023 gives the number of particles in one mole.
- Molar mass (M) is the mass of one mole of a substance (g · mol-1) and numerically equals the relative atomic or molecular mass.
- Relationship between mass (m), moles (n) and molar mass (M): n = m / M.
- Coefficients in a balanced chemical equation give mole ratios of reactants and products.
- Empirical formula gives the simplest whole-number ratio of atoms in a compound; molecular formula gives the actual number of atoms in a molecule.
- Percentage composition by mass of elements in a compound can be calculated from the formula; conversely, percentage composition can be used to determine empirical formula.
- Waters of crystallisation can be determined by mass loss on heating; convert masses to moles and find integer ratios.
- One mole of an ideal gas occupies 22.4 dm3 at S.T.P.
- Concentration (molarity) C = n / V, with units mol · dm-3.
- Stoichiometry uses balanced equations and the mole concept to calculate quantities of reactants and products; theoretical yield is the calculated maximum amount of product and percentage yield compares actual to theoretical yield.
