Amount of substance

In a nutshell

Atoms are far too small to count, so chemists count them by weighing. The mole is the link between a mass, a gas volume or a solution volume you can measure and the number of particles that actually react.

This subtopic is the calculation toolkit for the whole course: convert what you measured into an amount in mol, use the balanced equation's ratio, then convert back. It also covers formulae from experimental data, percentage yield and atom economy.

Assumed knowledge: Atomic structure and isotopes for relative atomic mass, and Compounds, formulae and equations for writing formulae and balancing equations.

Core content

The mole

Amount of substance, symbol n, is the quantity that counts particles. Its unit is the mole (mol).

OCR writes amounts using the formula of the substance, so you will see "amount of NaCl" or "amount of O₂". Always name the particle you are counting.

• Mole: one mole is the amount of substance containing 6.02 × 10²³ particles (atoms, molecules, ions or electrons).
• Avogadro constant, N_A: the number of particles per mole, 6.02 × 10²³ mol^-1 (given on the data sheet).
• Molar mass, M: the mass per mole of a substance, units g mol^-1. Numerically it equals the relative formula mass: M(H₂O) = 18.0 g mol^-1.
• Molar gas volume: the gas volume per mole, units dm³ mol^-1. At room temperature and pressure (RTP) it is 24.0 dm³ mol^-1 for any gas (given on the data sheet).

OCR also accepts the older definition of the mole: the amount of substance containing as many particles as there are atoms in exactly 12 g of carbon-12. Either earns the mark.

Still don't get it? · what a mole really is

Imagine a bank that needs to count a huge pile of identical coins. Nobody counts them one at a time. They weigh 100 coins once, then weigh the whole pile, and work out the number from the mass. Weighing replaces counting.

Chemists do the same trick with atoms, which are far smaller and far more numerous.

1. Pick a fixed counting number: 6.02 × 10²³. Call that many particles "one mole", the way a dozen means 12.
2. That number was chosen so that one mole of a substance has a mass in grams equal to its relative formula mass. One mole of carbon atoms is 12.0 g; one mole of water molecules is 18.0 g.
3. So weighing out 18.0 g of water gives you 6.02 × 10²³ molecules without counting any of them.
4. Weigh out 9.0 g and you have half as many: 0.500 mol.

Back to the exam wording: the mole is the unit of amount of substance, one mole contains 6.02 × 10²³ particles (the Avogadro constant, N_A), and the molar mass is the mass of one mole in g mol⁻¹. That is why $n = \frac{m}{M}$.

Every route into and out of moles. Whatever you measure, turn it into an amount in mol first.

Unit conversions you will need:

• 1 dm³ = 1000 cm³, and 1 m³ = 1000 dm³ = 10⁶ cm³
• 1 kPa = 1000 Pa
• temperature in K = temperature in °C + 273
• 1 kg = 1000 g, and 1 tonne = 10⁶ g (given on the data sheet)

Significant figures. Give the final answer to the same number of significant figures as the least precise data in the question. Keep extra figures in your calculator during the working and round only at the end.

Determination of formulae

• Empirical formula: the simplest whole number ratio of atoms of each element present in a compound.
• Molecular formula: the number and type of atoms of each element in a molecule.

Formal definitions are not required, but you must use the terms correctly. Ethene has molecular formula C₂H₄ and empirical formula CH₂.

Empirical formula from masses or percentages:

1. Write down the mass (or percentage by mass) of each element.
2. Divide each by that element's A_r to get the amount in mol.
3. Divide every answer by the smallest one.
4. If a value is not close to a whole number (for example 1.33 or 1.5), multiply all of them by the same small whole number (3 or 2) until they are.

The same method works for elemental analysis data, which is just a percentage composition by mass.

Molecular formula from the empirical formula: divide the M_r by the mass of one empirical formula unit. The answer is a whole number; multiply every subscript by it.

Hydrated salts and water of crystallisation

Many salts crystallise with water molecules locked into the crystal.

• Hydrated: a crystalline compound containing water molecules (water of crystallisation).
• Anhydrous: a substance that contains no water molecules.
• Water of crystallisation: water molecules that form an essential part of the crystalline structure of a compound.

The water is shown after a dot in the formula: CuSO₄·5H₂O is hydrated copper(II) sulfate; CuSO₄ is anhydrous copper(II) sulfate.

Heating drives the water off as vapour:

$\text{MgSO}_4\cdot x\text{H}_2\text{O(s)} \longrightarrow \text{MgSO}_4\text{(s)} + x\text{H}_2\text{O(g)}$

To find x from experimental masses:

1. Mass of water lost = mass of hydrated salt − mass of anhydrous salt.
2. Amount of anhydrous salt = its mass ÷ its molar mass.
3. Amount of water = mass of water ÷ 18.0.
4. x = amount of water ÷ amount of anhydrous salt, rounded to a whole number.

If you are given a percentage composition instead, treat it exactly like an empirical formula: the anhydrous salt and H₂O are the two "units" in the ratio.

Calculation of reacting masses, gas volumes and mole concentrations

Concentration of a solution can be given two ways:

• in mol dm^-3: amount of solute in mol per dm³ of solution, $c = \dfrac{n}{V}$
• in g dm^-3: mass of solute in g per dm³ of solution

To convert: concentration in g dm^-3 = concentration in mol dm^-3 × M.

When a solution is diluted, the amount of solute stays the same; only the volume changes. For example, 25.0 cm³ of 0.200 mol dm^-3 solution made up to 250.0 cm³ becomes 0.0200 mol dm^-3 (ten times the volume, one tenth the concentration).

An ionic solute can release more than one of an ion. In 0.10 mol dm^-3 CaCl₂(aq), [Cl^-] is 0.20 mol dm^-3, because each CaCl₂ gives two Cl^- ions.

The ideal gas equation links the amount of any gas to its pressure, volume and temperature:

$pV = nRT$

R = 8.314 J mol^-1 K^-1 (given on the data sheet). Everything must be in SI units before you substitute:

Rearranged, $n = \dfrac{pV}{RT}$ and $V = \dfrac{nRT}{p}$. Combined with $n = \dfrac{m}{M}$ it gives the molar mass of a volatile liquid: $M = \dfrac{mRT}{pV}$.

Use 24.0 dm³ mol^-1 only at RTP. At any other temperature or pressure you need pV = nRT.

Stoichiometric relationships. The balancing numbers in an equation are mole ratios. In

$\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \longrightarrow 2\text{NH}_3\text{(g)}$

1 mol of N₂ reacts with 3 mol of H₂ to make 2 mol of NH₃. Every reacting-quantity calculation follows three steps:

1. Convert the quantity you are given into an amount in mol.
2. Use the mole ratio from the balanced equation to find the amount of the other substance.
3. Convert that amount into what the question asks for (mass, gas volume or concentration).

For gases, equal volumes at the same temperature and pressure contain equal numbers of molecules. So gas volumes react in the same ratio as the balancing numbers, and you can work in cm³ without converting to mol at all.

Limiting reagent. When amounts of two reactants are given, find which one runs out first. The limiting reagent decides how much product forms; the other reactant is in excess and some is left over.

Still don't get it? · finding the limiting reagent

Think about making cheese sandwiches. Each sandwich needs 2 slices of bread and 1 slice of cheese. You have 10 slices of bread and 8 slices of cheese. How many sandwiches?

1. The bread is enough for 10 ÷ 2 = 5 sandwiches.
2. The cheese is enough for 8 ÷ 1 = 8 sandwiches.
3. You stop when the bread runs out, so you make 5 sandwiches and 3 slices of cheese are left over.
4. Notice that having more slices of cheese did not make cheese the thing that ran out. What matters is how much each ingredient is needed per sandwich.

Chemistry is identical. The balanced equation is the recipe, and the amounts in mol are the ingredients. Divide each reactant's amount by its balancing number: the smallest result is the limiting reagent. Base every product calculation on the limiting reagent; the other reactant is in excess. Never compare masses directly, only amounts in mol.

Percentage yields and atom economy

Percentage yield measures how much product you actually got compared with the maximum possible.

$\text{percentage yield} = \frac{\text{actual amount of product}}{\text{theoretical amount of product}} \times 100$

The theoretical amount comes from a reacting-mass calculation using the limiting reagent. You may use masses or amounts in mol, as long as top and bottom are the same quantity.

Yields are below 100% because of:

• the reaction not going to completion (reversible or incomplete reactions)
• side reactions forming other products
• product lost during transfer between containers
• product lost during purification

Atom economy measures how much of the reactants' mass ends up in the desired product.

$\text{atom economy} = \frac{\text{sum of molar masses of the desired products}}{\text{sum of molar masses of all products}} \times 100$

Because mass is conserved, the sum of molar masses of all products equals that of all reactants. Use the balancing numbers when you add the molar masses up.

A reaction can have a high percentage yield and still a low atom economy, if it makes a lot of waste by-product.

Atom economy and sustainability

Developing processes with a high atom economy benefits sustainability because:

• fewer waste products are made, so less waste needs disposal
• raw materials are used more efficiently, so finite resources last longer
• less pollution and environmental damage from waste
• processes are usually cheaper to run, with less waste to separate and treat

If a low atom economy process is unavoidable, selling or using the by-product improves its overall efficiency. Addition reactions, where every atom ends up in one product, have 100% atom economy.

PAG1: moles determination

OCR's PAG1 practicals use measurement of mass and measurement of volume of gas to find an amount in mol (OCR's suggested activity is determining the composition of copper(II) carbonate). The two core techniques are below.

Heating to constant mass (finding x in a hydrated salt):

1. Weigh an empty, dry crucible (and lid).
2. Add the hydrated salt and reweigh.
3. Heat gently at first, then more strongly, with the lid slightly ajar.
4. Allow to cool, then reweigh.
5. Repeat heating, cooling and weighing until two readings agree (constant mass). This shows all the water of crystallisation has been removed.

In this run the crucible weighed 23.00 g. Readings 3 and 4 agree at 24.20 g, so heating can stop: the hydrated salt was 2.46 g and the anhydrous salt 1.20 g (used in Worked example 6).

What goes wrong, and which way it pushes the answer:

• Not heating to constant mass: some water stays in the solid, so the mass of water lost is too small and x comes out too low.
• Solid spitting out of the crucible: extra mass is lost, so x comes out too high. The lid reduces this.
• Overheating: some salts decompose further, losing more than water, so x comes out too high.

Measuring a volume of gas: collect the gas from a reaction in a gas syringe, or over water in an upturned measuring cylinder or burette. Record the room temperature and pressure, because gas volume depends on both.

Sources of error in gas collection:

• gas escapes before the bung is put in (fix: keep one reactant separate, for example in a small tube inside the flask, until the bung is in)
• some gases, such as CO₂, are slightly soluble in water, so collecting over water gives a volume that is too low
• the gas syringe plunger sticks

Measuring volumes of solutions: use a volumetric pipette to deliver a fixed volume accurately, a burette for a variable volume, and a volumetric flask to make up a solution to an exact volume. A measuring cylinder is much less precise. (Making a standard solution and titration are covered in Acids, 2.1.4.)

Worked examples

Each example is a different calculation type from this subtopic. A_r values are from the data sheet; every step is one line, with units carried through.

1. Mass, amount and number of particles

Calculate the number of chloride ions in 2.22 g of calcium chloride, CaCl₂.

$M(\text{CaCl}_2) = 40.1 + (2 \times 35.5) = 111.1 \text{ g mol}^{-1}$ $n(\text{CaCl}_2) = \frac{m}{M} = \frac{2.22}{111.1} = 0.01998 \text{ mol}$ $n(\text{Cl}^-) = 2 \times 0.01998 = 0.03996 \text{ mol}$ $N(\text{Cl}^-) = n \times N_A = 0.03996 \times 6.02 \times 10^{23} = 2.41 \times 10^{22}$

The trap is the third line: each CaCl₂ contains two Cl^- ions. Stopping at 0.01998 × N_A counts formula units, not ions. Answer to 3 significant figures, matching 2.22 g.

2. Concentration in mol dm^-3 and g dm^-3

3.18 g of sodium carbonate, Na₂CO₃, is dissolved in water and made up to 250.0 cm³. Calculate its concentration in mol dm^-3 and in g dm^-3.

$M(\text{Na}_2\text{CO}_3) = (2 \times 23.0) + 12.0 + (3 \times 16.0) = 106.0 \text{ g mol}^{-1}$ $n = \frac{3.18}{106.0} = 0.0300 \text{ mol}$ $V = \frac{250.0}{1000} = 0.2500 \text{ dm}^3$ $c = \frac{n}{V} = \frac{0.0300}{0.2500} = 0.120 \text{ mol dm}^{-3}$ $\text{mass concentration} = 0.120 \times 106.0 = 12.7 \text{ g dm}^{-3}$

Check: 3.18 g ÷ 0.2500 dm³ = 12.7 g dm^-3. Dividing by 250 instead of 0.2500 gives an answer 1000 times too small.

3. Gas volume at RTP

Calculate the mass of oxygen gas, O₂, that occupies 600 cm³ at RTP. (Molar gas volume = 24.0 dm³ mol^-1 at RTP.)

$V = \frac{600}{1000} = 0.600 \text{ dm}^3$ $n(\text{O}_2) = \frac{V}{24.0} = \frac{0.600}{24.0} = 0.0250 \text{ mol}$ $m = n \times M = 0.0250 \times 32.0 = 0.800 \text{ g}$

Oxygen gas is O₂, so M = 32.0 g mol^-1, not 16.0.

4. Ideal gas equation: molar mass of a volatile liquid

0.215 g of a volatile liquid is vaporised. The vapour occupies 75.7 cm³ at 101 kPa and 95 °C. Calculate the molar mass of the liquid. (R = 8.314 J mol^-1 K^-1)

$p = 101 \text{ kPa} = 1.01 \times 10^{5} \text{ Pa}$ $V = 75.7 \text{ cm}^3 = 75.7 \times 10^{-6} \text{ m}^3 = 7.57 \times 10^{-5} \text{ m}^3$ $T = 95 + 273 = 368 \text{ K}$ $n = \frac{pV}{RT} = \frac{(1.01 \times 10^{5}) \times (7.57 \times 10^{-5})}{8.314 \times 368} = 2.499 \times 10^{-3} \text{ mol}$ $M = \frac{m}{n} = \frac{0.215}{2.499 \times 10^{-3}} = 86.0 \text{ g mol}^{-1}$

The first three lines are where most marks go: kPa to Pa, cm³ to m³, °C to K. A molar mass of 86.0 g mol^-1 fits hexane, C₆H₁₄.

5. Empirical and molecular formula

(a) Empirical formula needing a multiplier. An oxide of iron contains 72.4% iron by mass. Find its empirical formula.

$\%\,\text{O} = 100 - 72.4 = 27.6\%$ $n(\text{Fe}) = \frac{72.4}{55.8} = 1.297 \qquad n(\text{O}) = \frac{27.6}{16.0} = 1.725$ $\text{divide by smallest:} \quad \text{Fe} = \frac{1.297}{1.297} = 1 \qquad \text{O} = \frac{1.725}{1.297} = 1.33$ $\times 3: \quad \text{Fe} = 3 \qquad \text{O} = 3.99 \approx 4$ $\text{empirical formula} = \text{Fe}_3\text{O}_4$

Rounding 1.33 down to 1 would give FeO, which is wrong. A ratio ending in .33 or .67 means multiply by 3; ending in .5 means multiply by 2.

(b) Molecular formula. A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass, and has M_r = 56.0. Find its molecular formula.

$n(\text{C}) = \frac{85.7}{12.0} = 7.142 \qquad n(\text{H}) = \frac{14.3}{1.0} = 14.3$ $\text{divide by smallest:} \quad \text{C} = 1 \qquad \text{H} = \frac{14.3}{7.142} = 2.00$ $\text{empirical formula} = \text{CH}_2, \quad \text{mass of unit} = 12.0 + 2.0 = 14.0$ $\frac{M_r}{\text{empirical unit mass}} = \frac{56.0}{14.0} = 4$ $\text{molecular formula} = (\text{CH}_2)_4 = \text{C}_4\text{H}_8$

6. Formula of a hydrated salt from experimental results

A student heated 2.46 g of hydrated magnesium sulfate, MgSO₄·xH₂O, to constant mass. 1.20 g of anhydrous MgSO₄ remained. Find x.

$m(\text{H}_2\text{O}) = 2.46 - 1.20 = 1.26 \text{ g}$ $M(\text{MgSO}_4) = 24.3 + 32.1 + (4 \times 16.0) = 120.4 \text{ g mol}^{-1}$ $n(\text{MgSO}_4) = \frac{1.20}{120.4} = 9.967 \times 10^{-3} \text{ mol}$ $n(\text{H}_2\text{O}) = \frac{1.26}{18.0} = 0.0700 \text{ mol}$ $x = \frac{0.0700}{9.967 \times 10^{-3}} = 7.02 \approx 7$

The formula is MgSO₄·7H₂O. Divide the amount of water by the amount of the anhydrous salt (1.20 g), not the hydrated salt (2.46 g).

7. Reacting quantities with a limiting reagent

0.486 g of magnesium is added to 50.0 cm³ of 0.500 mol dm^-3 hydrochloric acid. Calculate the volume of hydrogen, in cm³ at RTP, that forms.

$\text{Mg(s)} + 2\text{HCl(aq)} \longrightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)}$ $n(\text{Mg}) = \frac{0.486}{24.3} = 0.0200 \text{ mol}$ $n(\text{HCl}) = c \times V = 0.500 \times \frac{50.0}{1000} = 0.0250 \text{ mol}$ $0.0200 \text{ mol Mg needs } 2 \times 0.0200 = 0.0400 \text{ mol HCl; only } 0.0250 \text{ mol present}$ $\text{HCl is the limiting reagent; Mg is in excess}$ $n(\text{H}_2) = \frac{0.0250}{2} = 0.0125 \text{ mol}$ $V(\text{H}_2) = 0.0125 \times 24\,000 = 300 \text{ cm}^3$

Using the magnesium (the reagent in excess) would give 480 cm³, a volume the acid cannot produce.

8. Reacting gas volumes

50 cm³ of propane is burned in 300 cm³ of oxygen. All volumes are measured at RTP, where water is a liquid. Find the total volume of gas left at the end.

$\text{C}_3\text{H}_8\text{(g)} + 5\text{O}_2\text{(g)} \longrightarrow 3\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(l)}$ $\text{O}_2 \text{ used} = 5 \times 50 = 250 \text{ cm}^3, \text{ so O}_2 \text{ left} = 300 - 250 = 50 \text{ cm}^3$ $\text{CO}_2 \text{ formed} = 3 \times 50 = 150 \text{ cm}^3$ $\text{total gas} = 50 + 150 = 200 \text{ cm}^3$

No conversion to mol is needed: gas volumes at the same conditions react in the ratio of the balancing numbers. Do not count the liquid water, and do not forget the excess oxygen.

9. Percentage yield

25.0 g of calcium carbonate is heated and 12.6 g of calcium oxide is collected. Calculate the percentage yield.

$\text{CaCO}_3\text{(s)} \longrightarrow \text{CaO(s)} + \text{CO}_2\text{(g)}$ $n(\text{CaCO}_3) = \frac{25.0}{100.1} = 0.2498 \text{ mol}$ $1\,\text{CaCO}_3 : 1\,\text{CaO} \implies n(\text{CaO}) = 0.2498 \text{ mol}$ $\text{theoretical mass of CaO} = 0.2498 \times 56.1 = 14.01 \text{ g}$ $\text{percentage yield} = \frac{12.6}{14.01} \times 100 = 89.9\%$

The actual yield goes on top. If your answer is over 100%, you have the fraction upside down (or a wet product).

10. Atom economy

Titanium is extracted by the reaction below. Calculate the atom economy for making titanium.

$\text{TiCl}_4 + 4\text{Na} \longrightarrow \text{Ti} + 4\text{NaCl}$ $\text{sum of molar masses of all products} = 47.9 + 4 \times (23.0 + 35.5) = 47.9 + 234.0 = 281.9 \text{ g mol}^{-1}$ $\text{check (reactants)} = (47.9 + 4 \times 35.5) + (4 \times 23.0) = 189.9 + 92.0 = 281.9 \text{ g mol}^{-1}$ $\text{atom economy} = \frac{47.9}{281.9} \times 100 = 17.0\%$

The 4 in front of NaCl must be used. Leaving it out gives 47.9 ÷ 106.4 × 100 = 45.0%, which is wrong.

Common exam mistakes

• Leaving kPa, cm³ or °C in pV = nRT. The equation needs Pa, m³ and K. Using dm³ instead of m³ makes the answer 1000 times too big; forgetting the 273 wrecks the temperature.
• Using 24.0 dm³ mol^-1 when the gas is not at RTP. If a temperature and pressure other than RTP are given, use pV = nRT.
• Dividing by a volume in cm³ in c = n/V. Concentration in mol dm^-3 needs V in dm³: divide cm³ by 1000 first.
• Using 16.0 for oxygen gas, 35.5 for chlorine gas or 1.0 for hydrogen gas. These elements are diatomic: O₂ = 32.0, Cl₂ = 71.0, H₂ = 2.0 g mol^-1.
• Ignoring the mole ratio, or using it the wrong way round. If 2 mol of NaOH react with 1 mol of H₂SO₄, the amount of H₂SO₄ is half that of NaOH, not double.
• Counting formula units instead of ions or atoms. "Number of chloride ions in CaCl₂" needs the factor of 2; "number of atoms in a mole of H₂O" needs the factor of 3.
• Basing the product on the reagent in excess. Find the limiting reagent by comparing amounts in mol divided by their balancing numbers, never by comparing masses.
• Rounding too early in an empirical formula. Rounding 1.33 to 1 or 1.5 to 2 gives the wrong formula. Keep 3 or more figures and look for .33, .5 and .67.
• Giving a ratio instead of a formula. The answer is Fe₃O₄, not "3 : 4".
• Hydrated salt: dividing by the moles of the hydrated salt instead of the anhydrous salt, or giving x as a decimal. x is a whole number.
• Saying "heat it for longer" without the method. The point to make is reheat, cool and reweigh until the mass is constant, which shows all the water has gone.
• Percentage yield upside down, or confusing it with atom economy. Yield uses an experimental result; atom economy uses only the balanced equation.
• Atom economy without the balancing numbers, or without multiplying by 100.
• Too many or too few significant figures. Match the least precise data; 0.0250 has 3 significant figures, not 4 and not 2.

Key definitions

• Amount of substance (n): the quantity whose unit is the mole; it counts the number of particles of a substance.
• Mole (mol): the amount of substance containing 6.02 × 10²³ particles. (Also accepted: the amount of substance containing as many particles as there are atoms in exactly 12 g of carbon-12.)
• Avogadro constant (N_A): the number of particles per mole, 6.02 × 10²³ mol^-1.
• Molar mass (M): the mass per mole of a substance, in g mol^-1.
• Molar gas volume: the gas volume per mole, in dm³ mol^-1; 24.0 dm³ mol^-1 at RTP.
• Empirical formula: the simplest whole number ratio of atoms of each element present in a compound.
• Molecular formula: the number and type of atoms of each element in a molecule.
• Hydrated: a crystalline compound containing water molecules (water of crystallisation).
• Anhydrous: a substance that contains no water molecules.
• Water of crystallisation: water molecules that form an essential part of the crystalline structure of a compound.
• Concentration: the amount of solute, in mol, per dm³ of solution (mol dm^-3); or the mass of solute per dm³ of solution (g dm^-3).
• Limiting reagent: the reactant that is completely used up and so determines the amount of product formed.
• Percentage yield = (actual amount of product ÷ theoretical amount of product) × 100.
• Atom economy = (sum of molar masses of the desired products ÷ sum of molar masses of all products) × 100.

Specification

• I can explain and use the terms amount of substance, mole, Avogadro constant, molar mass and molar gas volume, with their units. (2.1.3 a)
• I can use the terms empirical formula and molecular formula. (2.1.3 b)
• I can calculate an empirical formula from composition by mass or percentage composition by mass, including elemental analysis data, and a molecular formula using the relative molecular mass. (2.1.3 c)
• I can use the terms anhydrous, hydrated and water of crystallisation, and calculate the formula of a hydrated salt from percentage composition, mass composition or experimental results. (2.1.3 d)
• I can carry out calculations using amount of substance in mol involving mass, gas volume, and solution volume and concentration, giving concentrations in mol dm^-3 and g dm^-3. (2.1.3 e)
• I can use the ideal gas equation pV = nRT with every quantity in SI units. (2.1.3 f)
• I can use stoichiometric relationships from balanced equations in calculations, including limiting reagents and reacting gas volumes. (2.1.3 g)
• I can calculate the percentage yield of a reaction, or a related quantity, and the atom economy of a reaction. (2.1.3 h)
• I can describe the techniques for measuring mass, volumes of solutions and gas volumes, including heating to constant mass and collecting a gas (PAG1). (2.1.3 i)
• I can explain the benefits for sustainability of developing chemical processes with a high atom economy. (2.1.3 j)

Related notes

• Atomic structure and isotopes
• Compounds, formulae and equations
• Acids: standard solutions, titrations and PAG2
• Redox
• Enthalpy changes: uses amounts in mol to calculate enthalpy changes