1.2Unit 1. Structure, Bonding and Introduction to Organic Chemistry
Amount of Substance — IAL Edexcel Modular (Chemistry) Revision Notes & Topical Questions
5 min read 13 parts Lesson 3 of 20 2 quizzes
Lesson summary
1.2 Amount of Substance — IAL EDEXCEL MODULAR (CHEMISTRY) notes: Calculate amounts in moles from mass, solution and gas data, and carry out reacting mass, reacting volume, limiting reagent, percentage yield and atom economy calculations.Amount of substance, n = the number of mole…
- Conversions
- Mass concentration
- Molar concentration
- Converting between the two
- ppm in solution
- ppm of an atmospheric gas
Section 1 of 13Amount of substance equations
- Amount of substance, n the number of moles of particles, a blanket term for most mole calculations
| Situation | Equation | Units |
|---|---|---|
| Any substance from mass | n = m ÷ M | mol, g, g mol-1 |
| Solutions | n = c × V | mol, mol dm-3, dm3 |
| Gases at r.t.p. | n = V ÷ 24.0 | mol, dm3 |
| Number of particles | n = N ÷ L | L = 6.02 × 1023 mol-1 |
| Ideal gas | n = pV ÷ RT | mol, Pa, m3, K |
- Conversions
- 1 dm3 = 1000 cm3 and 1 m3 = 1000 dm3, so cm3 is divided by 1000 to give dm3
Section 2 of 13Concentration of solutions
1Mass concentration
Mass concentration mass of solute in g ÷ volume of solution in dm3
- The units are usually g dm-3; g cm-3, kg m-3 and mg ml-1 are also used.
2Molar concentration
Molar concentration moles of solute ÷ volume of solution in dm3
- The units are mol dm-3.
- Moles of solute are found from n = m ÷ M.
3Converting between the two
- Converting between the two mass concentration in g dm-3 = molar concentration × M
4Dilution
- Dilution the moles of solute stay constant, so c1V1 = c2V2
Section 3 of 13Parts per million (ppm)
Parts per million (ppm)
ppm in solution
- 1 ppm is 1 mg of solute in 1 dm3 (1 kg) of water
- ppm = (mass of solute ÷ mass of solution) × 106.
- It is used for very low concentrations such as pollutants in water.
ppm of an atmospheric gas
- ppm of an atmospheric gas compared by volume, not mass
- ppm = (volume of gas ÷ volume of air) × 106.
- Both volumes must be in the same units; 1 ppm means 1 cm3 of gas in 106 cm3 of air.
- The identity of the gas does not affect the calculation.
Section 4 of 13Reacting mass calculations
Reacting mass calculations
Write the balanced equation.
Calculate moles of the known substance from n
Calculate moles of the known substance from n m ÷ M.
Use the mole ratio in the equation to find moles of the unknown substance.
Convert moles to mass with m
Convert moles to mass with m n × M.
Give the answer to a suitable number of significant figures with units.
- Units of mass
- any mass unit may be used provided it is consistent, because reacting masses stay in proportion
- (g, kg or tonnes)
- any mass unit may be used provided it is consistent, because reacting masses stay in proportion
Section 5 of 13Deducing an equation from reacting masses
Deducing an equation from reacting masses
Convert the mass of each reactant and product to moles by dividing by its molar mass.
Divide all the mole values by the smallest to give a ratio.
Use the whole-number ratio as the coefficients and simplify.
Section 6 of 13Reacting volumes of gases
Molar gas volume
- Molar gas volume all gases occupy the same volume under the same conditions
- At room temperature and pressure (r.t.p., 293 K and 101.3 kPa) one mole of gas occupies 24.0 dm3.
- At standard temperature and pressure (s.t.p., 273 K and 101.3 kPa) one mole of gas occupies 22.4 dm3.
Moles from volume
Moles from volume n = volume in dm3 ÷ molar gas volume
Volume from moles
V = n × molar gas volume
Gas volume ratios
- Gas volume ratios equal volumes of gases contain equal numbers of molecules, so gases react in the same ratio as their moles
- N2 + 3H2 → 2NH3 uses 1 volume of nitrogen with 3 volumes of hydrogen to give 2 volumes of ammonia.
Section 7 of 13The ideal gas equation
pV
pV nRT
What the symbols mean: p = pressure (Pa) · V = volume (m3) · n = amount of gas (mol) · R = 8.31 J mol−1 K−1 · T = temperature (K = °C + 273)
- p is pressure in Pa, V is volume in m3, n is moles, R = 8.31 J mol-1 K-1 and T is temperature in K.
Unit conversions
- Unit conversions kPa × 1000 gives Pa, cm3 × 10-6 gives m3, dm3 ÷ 1000 gives m3, and °C + 273 gives K
Ideal gas assumptions
- Ideal gas assumptions molecules move rapidly and randomly, have negligible volume and exert no intermolecular forces
- Collisions are elastic, so no kinetic energy is lost.
- Temperature reflects the mean kinetic energy of the molecules.
- Real gases deviate most at high pressure and low temperature.
Finding the Mr of a gas
- Finding the Mr of a gas measure mass, volume, pressure and temperature of a sample
- Calculate n = pV ÷ RT, then M = m ÷ n.
Section 8 of 13Limiting reagent and excess
- Limiting reagent = the reactant that is completely used up first
- It sets the maximum amount of product (the theoretical yield).
- The other reactant is in excess.
1
Calculate moles of each reactant.
2
Divide each by its coefficient in the balanced equation.
3
The smaller value identifies the limiting reagent.
4
Use the moles of limiting reagent to find the moles of product.
Section 9 of 13Percentage yield
Percentage yield
(actual yield ÷ theoretical yield) × 100
- The actual yield is the mass or moles of product obtained experimentally.
- The theoretical yield comes from a reacting mass calculation.
Reasons for a yield below 100%
- The reaction does not go to completion or is reversible.
- Side reactions take place at the same time.
- Product is lost during separation and purification.
A yield above 100% indicates an error
- The product may still be wet or contain impurities.
- The mass of the filter paper or container may have been included.
Section 10 of 13Atom economy
Atom economy
Atom economy
(Mr of desired product ÷ sum of Mr of all reactants) × 100
- It is found from the balanced equation alone and can also be worked out using masses.
Addition reactions have 100% atom economy
- Addition reactions have 100% atom economy all reactant atoms end up in the single product
- Substitution and elimination reactions produce by-products, which lowers the atom economy.
Importance
- Importance a high atom economy means less waste and a more sustainable process
Section 11 of 13More from the specification
1Concentration
- Concentration amount of solute per volume of solution, c = n/V in mol per dm3
2Moles from mass
- Moles from mass n = mass/molar mass, used with balanced equations for magnesium, aluminium, sodium or copper reactions
3Empirical formula
- Empirical formula found from masses of elements,
- such as magnesium and oxygen in magnesium oxide, by dividing by molar mass and finding the simplest ratio
4Determining a formula experimentally
- Determining a formula experimentally heat magnesium in a crucible, record mass gain as oxygen added, then calculate moles of each
5Gas and carbon dioxide volumes
- Gas and carbon dioxide volumes molar gas volume or the ideal gas equation pV = nRT converts moles of CO2 to volume
6Concentration
- Concentration amount of solute per volume of solution, c = n/V in mol per dm3
7Moles from mass
- Moles from mass n = mass/molar mass, used with balanced equations for magnesium, aluminium, sodium or copper reactions
8Empirical formula
- Empirical formula found from masses of elements,
- such as magnesium and oxygen in magnesium oxide, by dividing by molar mass and finding the simplest ratio
9Determining a formula experimentally
- Determining a formula experimentally heat magnesium in a crucible, record mass gain as oxygen added, then calculate moles of each
10Gas and carbon dioxide volumes
- Gas and carbon dioxide volumes molar gas volume or the ideal gas equation pV = nRT converts moles of CO2 to volume
11Oxygen
- A gas that relights a glowing splint.
12Magnesium oxide
- White powder made when magnesium burns.
13Carbon dioxide
- A gas that turns limewater cloudy.
14Kinetic theory
- The model that matter consists of particles in constant motion.
Section 12 of 13Key terms
| Term | Meaning |
|---|---|
| Mole | The amount of substance containing 6.02 × 10²³ particles (the Avogadro constant). |
| Molar mass (M) | The mass of one mole of a substance in g mol⁻¹. |
| Concentration | The amount of solute per unit volume of solution, usually mol dm⁻³. |
| Molar gas volume | The volume of one mole of any gas: 24.0 dm³ at r.t.p. |
| Limiting reagent | The reactant that is used up first and so determines the amount of product. |
| Percentage yield | Actual yield divided by theoretical yield, multiplied by 100. |
| Atom economy | The Mr of the desired product divided by the total Mr of reactants, multiplied by 100. |
| Parts per million | Mass or volume of a substance in one million parts of the mixture. |
Section 13 of 13Exam technique
1
Forgetting to convert cm³ to dm³ — divide by 1000 before using concentration or molar volume
2
Using volumes in dm³ or cm³ in pV = nRT — volume must be in m³ and pressure in Pa
3
Comparing raw moles to find the limiting reagent — divide each amount by its coefficient first
4
Confusing percentage yield with atom economy — yield is experimental, atom economy comes from the equation only
5
Dropping units or significant figures — answers should match the data given
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