Notes · ChemistryUK · A-Levels

Amount of substance

The mole is the chemist's counting unit, the bridge between the masses we can weigh and the numbers of particles that actually react. This chapter defines the mole and the Avogadro constant, uses n = m/M, n = cV and the ideal gas equation to convert between mass, gas volume and solution concentration, and applies these to empirical formulae, reacting masses, atom economy, percentage yield and titrations.

5 sections~16 min reading time3 competenciesLevel Foundation 1 · Standard 4

T·0222 / 18

Exam profile

AO1 · Define the mole, the Avogadro constant, empirical and molecular formula, atom economy and percentage yieldAO2 · Carry out mole, gas, solution, titration, yield and atom-economy calculations to appropriate significant figuresAO3 · Process titration and gas data, identify the limiting reagent and evaluate the accuracy of a result

Operators:statecalculatedeterminededuceexplain

basic level

AS-Level requires n = m/M, n = cV, the ideal gas equation, empirical/molecular formulae, reacting masses, atom economy, percentage yield and the acid-base titration required practical.

higher level

The full A-Level applies the same calculations synoptically throughout - in energetics, equilibria, rates, electrode potentials and redox titrations - so fluency and correct significant figures here pay off across the whole course.

Depth

Text

Contents · 5 sections

Amount of substance

Relative masses and the mole#

FoundationAQA 7405 3.1.2.1DfE GCE Chemistry - amount of substance

The mole as the central hub

Fig. 1Every quantitative method converts to moles first: mass, gas volume, concentration and particle number all connect through n.

Atoms are far too small and too numerous to count individually, so chemists count them in moles. One mole is the amount of substance that contains as many particles as there are atoms in exactly

12\,\text{g}

of carbon-12; that number is the Avogadro constant,

N_A = 6.02 \times 10^{23}\ \text{mol}^{-1}

. The particle may be an atom, a molecule, an ion or an electron, so you must always say 'moles of what' - a mole of

\text{O}

atoms and a mole of

\text{O}_2

molecules contain different numbers of oxygen atoms.

The mass of one mole of a substance in grams is numerically equal to its relative atomic or molecular mass; this is the molar mass

M

, with units

\text{g mol}^{-1}

. The central equation of the whole topic is therefore

n = \dfrac{m}{M}

, connecting the number of moles

n

, the mass

m

in grams and the molar mass

M

. Rearranged,

m = nM

gives a mass from a number of moles, and

M = m/n

gives a molar mass from a measured mass and amount.

The number of individual particles is found by multiplying the number of moles by the Avogadro constant:

N = n \times N_A

. So

0.25\ \text{mol}

of water contains

0.25 \times 6.02 \times 10^{23} = 1.505 \times 10^{23}

molecules, and because each water molecule contains two hydrogen atoms, it contains

3.01 \times 10^{23}

hydrogen atoms. Keeping careful track of what is being counted - molecules, formula units or the atoms within them - is where marks are won or lost.

Because the mole ties mass to particle number, it is the hub through which every quantitative method in the course passes: a mass is converted to moles, the balanced equation gives the mole ratio to the substance of interest, and the moles are converted back to a mass, a gas volume or a concentration. Building the habit of 'convert to moles first' makes even unfamiliar calculations routine.

n = \dfrac{m}{M}

Moles from mass

Moles = mass in grams divided by molar mass in g mol^-1.

N = n \times N_A, \qquad N_A = 6.02 \times 10^{23}\ \text{mol}^{-1}

Number of particles

Multiply the number of moles by the Avogadro constant to get the number of individual particles.

Worked example

Moles and particles from a mass

How many moles, and how many molecules, are there in 4.00 g of methane, CH4 (Mr = 16.0)?

Moles from mass
$n = m/M = 4.00 / 16.0 = 0.250\ \text{mol}$ .
$n = \dfrac{4.00}{16.0} = 0.250\ \text{mol}$
Molecules
$N = n N_A = 0.250 \times 6.02 \times 10^{23} = 1.505 \times 10^{23}$ molecules.

Result: 0.250 mol of methane, containing 1.51 x 10^23 molecules.

Typical mistakes

Confusing moles of molecules with moles of the atoms inside them (e.g. O2 vs O).
Using the relative formula mass of the wrong species, or leaving the molar mass in the wrong units.

Active revision

Calculate the number of moles, and the number of oxygen atoms, in $8.8\,\text{g}$ of carbon dioxide ( $M_r = 44.0$ ).

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Chemistry 7405 specification (AQA)

The ideal gas equation and molar volume#

StandardAQA 7405 3.1.2.3DfE GCE Chemistry - the ideal gas equation

Volume against temperature at constant pressure

Fig. 2At fixed pressure and amount, V is proportional to T (in kelvin) - a straight line through absolute zero.

The behaviour of a gas is captured by the ideal gas equation

pV = nRT

, where

p

is the pressure in pascals (

\text{Pa}

V

the volume in cubic metres (

\text{m}^3

n

the amount in moles,

R

the gas constant (

8.31\ \text{J K}^{-1}\text{mol}^{-1}

) and

T

the temperature in kelvin (

\text{K}

). The single biggest source of error is units: pressure must be in

\text{Pa}

(multiply

\text{kPa}

1000

), volume in

\text{m}^3

(

1\ \text{m}^3 = 1000\ \text{dm}^3 = 10^6\ \text{cm}^3

), and temperature in

\text{K}

(add

273

to a Celsius value).

The equation follows from combining the simple gas laws. At fixed amount and temperature,

pV

is constant (Boyle's law); at fixed amount and pressure,

V \propto T

(Charles's law, a straight line through absolute zero); and at fixed temperature and pressure,

V \propto n

(Avogadro's law). Rearranging lets you find whichever quantity is unknown, most often the volume

V = nRT/p

or the amount

n = pV/RT

, and, combined with

n = m/M

, the molar mass of a volatile substance.

An 'ideal' gas is a model in which the particles have negligible volume of their own and no forces between them, and all collisions are perfectly elastic. Real gases approach this behaviour best at high temperature and low pressure, where the molecules are far apart and moving fast; they deviate most at low temperature and high pressure, where the molecules are close enough for intermolecular forces and their own volume to matter. Being able to state these assumptions and conditions is examinable.

A useful shortcut for room conditions is the molar gas volume: at room temperature and pressure (rtp, taken as

293\text{-}298\ \text{K}

and

100\ \text{kPa}

) one mole of any gas occupies about

24.0\ \text{dm}^3

, so

n = V/24.0

with

V

\text{dm}^3

. This is quick for rtp problems, but the full

pV = nRT

must be used whenever the conditions are stated and differ from rtp.

pV = nRT

Ideal gas equation

p in Pa, V in m^3, n in mol, R = 8.31 J K^-1 mol^-1, T in K. Convert every quantity to these SI units first.

M = \dfrac{mRT}{pV}

Molar mass of a gas

Combining pV = nRT with n = m/M lets you find the molar mass of a volatile substance from a measured mass and volume.

Worked example

Volume of a gas from pV = nRT

Calculate the volume, in dm^3, occupied by 0.500 mol of an ideal gas at 298 K and a pressure of 100 kPa. (R = 8.31 J K^-1 mol^-1.)

Convert to SI
$p = 100\ \text{kPa} = 1.00 \times 10^{5}\ \text{Pa}$ ; $T = 298\ \text{K}$ ; $n = 0.500\ \text{mol}$ .
Rearrange and substitute
$V = \dfrac{nRT}{p} = \dfrac{0.500 \times 8.31 \times 298}{1.00 \times 10^{5}}$ .
$V = \dfrac{0.500 \times 8.31 \times 298}{1.00 \times 10^{5}} = 1.238 \times 10^{-2}\ \text{m}^3$
Convert to dm³
$1.238 \times 10^{-2}\ \text{m}^3 \times 1000 = 12.4\ \text{dm}^3$ .

Result: The gas occupies 12.4 dm^3 (3 s.f.), close to half the molar volume, as expected for 0.5 mol.

Typical mistakes

Leaving pressure in kPa or volume in cm^3 or dm^3 instead of converting to Pa and m^3.
Using temperature in degrees Celsius rather than kelvin.

Active revision

A flask of volume $250\ \text{cm}^3$ contains a gas at $101\ \text{kPa}$ and $25\,^{\circ}\text{C}$ . Calculate the amount of gas, in moles.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

Empirical and molecular formulae#

StandardAQA 7405 3.1.2.2DfE GCE Chemistry - empirical and molecular formulae

Empirical formula method

Fig. 3For CH2O: dividing moles by the smallest gives the 1:2:1 ratio.

The empirical formula is the simplest whole-number ratio of the atoms of each element in a compound; the molecular formula gives the actual number of atoms of each element in one molecule. For water the two are the same (

\text{H}_2\text{O}

); for hydrogen peroxide the empirical formula is

\text{HO}

but the molecular formula is

\text{H}_2\text{O}_2

. Ionic and giant covalent substances are described only by an empirical formula, because they have no discrete molecules.

To find an empirical formula from experimental data, convert the mass (or percentage by mass) of each element to moles by dividing by its relative atomic mass, then divide every result by the smallest to get the simplest ratio. If the ratio is not whole numbers, multiply through by a small integer (e.g.

\times 2

to turn

1.5

into

3

). This is exactly the method used to interpret combustion analysis or the mass gained when a metal is oxidised.

To go from the empirical formula to the molecular formula you need the relative molecular mass, usually from a mass spectrum. Find the mass of one empirical unit, divide

M_r

by it to get the whole-number multiplier, and multiply every subscript by that number. If the empirical formula is

\text{CH}_2\text{O}

(empirical mass

30

) and

M_r = 180

, the multiplier is

180/30 = 6

, so the molecular formula is

\text{C}_6\text{H}_{12}\text{O}_6

The method is quantitative and unforgiving of rounding: keep at least three significant figures through the mole calculations, and only round the final ratio. A frequent trap is treating a percentage as if it were already a number of moles, or forgetting that a hydrated salt contains water of crystallisation, whose moles must be found from the mass loss on heating and folded into the formula (for example determining the

x

\text{CuSO}_4 \cdot x\,\text{H}_2\text{O}

\text{multiplier} = \dfrac{M_r}{\text{empirical formula mass}}

Empirical to molecular

Multiply every subscript of the empirical formula by this whole number to obtain the molecular formula.

Worked example

From composition to molecular formula

A compound is 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass, with Mr = 180. Find the molecular formula.

Moles of each element
C: $40.0/12.0 = 3.33$ ; H: $6.7/1.0 = 6.70$ ; O: $53.3/16.0 = 3.33$ .
Simplest ratio
Divide by the smallest (3.33): C 1, H 2.01 ~ 2, O 1, so the empirical formula is CH2O.
Scale to molecular
Empirical mass $= 12 + 2 + 16 = 30$ ; multiplier $= 180/30 = 6$ .
$\dfrac{180}{30} = 6 \;\Rightarrow\; \text{C}_6\text{H}_{12}\text{O}_6$

Result: The molecular formula is C6H12O6 (glucose).

Typical mistakes

Rounding the mole values too early and losing the true ratio.
Reporting the empirical formula when the molecular formula is asked for (or vice versa).

Active revision

A hydrocarbon contains $85.7\%$ carbon by mass and has $M_r = 56.0$ . Determine its empirical and molecular formulae.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

Equations, reacting masses, atom economy and yield#

StandardAQA 7405 3.1.2.5DfE GCE Chemistry - balanced equations and calculations

A balanced equation is a mole recipe: its coefficients give the ratio in which substances react and form. To find a reacting mass, convert the known mass to moles with

n = m/M

, multiply by the mole ratio from the equation to get the moles of the substance you want, then convert back to a mass with

m = nM

. When two reactants are given, first identify the limiting reagent - the one that runs out first - because it alone determines how much product can form; the other is in excess.

Percentage yield measures how much product you actually obtained compared with the maximum the equation predicts:

\text{percentage yield} = \dfrac{\text{actual amount}}{\text{theoretical amount}} \times 100

. Yields are below

100\%

because reactions may not go to completion, may reach equilibrium, may have competing side reactions, or product may be lost during separation and purification. It is a measure of efficiency of conversion for the desired product.

Atom economy measures how much of the mass of all the reactants ends up in the desired product:

\text{atom economy} = \dfrac{M_r\,(\text{desired product})}{\sum M_r\,(\text{all products})} \times 100

. A reaction with a single product has

100\%

atom economy; one that also makes waste by-products has less. High atom economy is central to green chemistry because it means less waste, lower raw-material cost and less to dispose of - addition reactions score highly, substitution and elimination reactions lower.

The two measures answer different questions and are often confused. Percentage yield is about how well a particular reaction was carried out (it could be improved by better technique); atom economy is a fixed property of the chosen reaction route (it can only be improved by choosing a different reaction). A process can have a high yield but poor atom economy, or vice versa, so industrial chemists weigh both alongside cost, energy and the value of any by-products.

\%\ \text{yield} = \dfrac{\text{actual moles (or mass) of product}}{\text{theoretical moles (or mass) of product}} \times 100

Percentage yield

Compares the product obtained with the maximum the equation predicts.

\%\ \text{atom economy} = \dfrac{M_r\,(\text{desired product})}{\sum M_r\,(\text{all products})} \times 100

Atom economy

The fraction of reactant mass that becomes the useful product; higher means less waste.

Worked example

Reacting mass and percentage yield

Calcium carbonate decomposes on heating: CaCO3 -> CaO + CO2. 25.0 g of CaCO3 is heated and 12.6 g of CaO is obtained. Calculate the percentage yield. (Mr: CaCO3 = 100.1, CaO = 56.1.)

Moles of CaCO3
$n = 25.0/100.1 = 0.2498\ \text{mol}$ .
Theoretical moles and mass of CaO
The ratio is 1:1, so theoretical CaO $= 0.2498\ \text{mol} = 0.2498 \times 56.1 = 14.01\ \text{g}$ .
Percentage yield
$\%\ \text{yield} = \dfrac{12.6}{14.01} \times 100 = 89.9\%$ .
$\%\ \text{yield} = \dfrac{12.6}{14.01} \times 100 = 89.9\%$

Result: The percentage yield is 89.9% (3 s.f.).

Worked example

Atom economy of an extraction

Calcium oxide is produced by CaCO3 -> CaO + CO2. Calculate the atom economy with respect to CaO. (Mr: CaO = 56.1, CO2 = 44.0.)

Identify all products
The products are CaO (desired) and CO2 (waste).
Apply the formula
$\dfrac{56.1}{56.1 + 44.0} \times 100 = \dfrac{56.1}{100.1} \times 100 = 56.0\%$ .
$\dfrac{56.1}{56.1+44.0}\times100 = 56.0\%$

Result: The atom economy is 56.0%; nearly half the reactant mass is lost as carbon dioxide.

Typical mistakes

Assuming the reactant of smaller mass is limiting without converting to moles first.
Confusing atom economy with percentage yield, or including reactants in the atom-economy denominator (it uses total products, which by conservation of mass equals total reactants).

Active revision

Ethanol is made industrially by hydration of ethene: $\text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{OH}$ . State the atom economy and explain your answer.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

Solutions, concentration and titrations#

StandardAQA 7405 3.1.2.5DfE GCE Chemistry - required practical 1

Acid-base titration apparatus

Fig. 4The burette delivers the titrant into a pipetted volume in the conical flask; the white tile helps judge the colour change.

The concentration of a solution is the amount of solute per unit volume, in

\text{mol dm}^{-3}

, linked to amount and volume by

n = c \times V

, where

V

is the volume in

\text{dm}^3

. Because volumes are usually measured in

\text{cm}^3

, remember that

V\,(\text{dm}^3) = V\,(\text{cm}^3)/1000

. A standard solution is one of accurately known concentration, made by dissolving a weighed mass of a pure solid in a volumetric flask and making up to the mark.

A titration finds an unknown concentration by reacting exactly measured volumes of two solutions. A pipette delivers a fixed volume of one solution into a conical flask, a burette adds the other drop by drop, and an indicator (or pH meter) signals the end point where the reaction is just complete. The titre - the burette reading - is repeated until concordant results (within

0.10\ \text{cm}^3

) are obtained, and only these are averaged, so that random errors are minimised.

The calculation follows the 'convert to moles' habit exactly. From the solution of known concentration, find

n = cV

. Use the balanced equation's mole ratio to find the moles of the other reactant. Then divide by its volume to find its concentration,

c = n/V

. For an acid-base titration such as

\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}

the ratio is

1:1

, but for

\text{H}_2\text{SO}_4 + 2\text{NaOH}

it is

1:2

, and using the wrong ratio is a classic error.

Accuracy depends on good technique and on understanding the apparatus uncertainties. Reading a burette to

\pm 0.05\ \text{cm}^3

at each end gives a

\pm 0.10\ \text{cm}^3

uncertainty on the titre; the percentage uncertainty is

\dfrac{0.10}{\text{titre}} \times 100

, which is why titres near

25\ \text{cm}^3

are more precise than small ones. Rinsing the burette with the solution it will hold, and the pipette with its solution, prevents dilution errors; rinsing the conical flask with water (not the solution) is correct because only the moles delivered matter.

n = c \times V

Amount from concentration

Amount in mol = concentration in mol dm^-3 x volume in dm^3 (so divide a cm^3 volume by 1000).

\%\ \text{uncertainty} = \dfrac{0.10}{\text{titre}} \times 100

Burette percentage uncertainty

Two readings at +/-0.05 cm^3 give a +/-0.10 cm^3 titre uncertainty; larger titres are proportionally more precise.

Worked example

Concentration of an alkali by titration

In a titration, 25.0 cm^3 of sodium hydroxide solution is exactly neutralised by 23.20 cm^3 of 0.100 mol dm^-3 hydrochloric acid. Calculate the concentration of the sodium hydroxide. (HCl + NaOH -> NaCl + H2O.)

Moles of HCl
$n(\text{HCl}) = c V = 0.100 \times 23.20/1000 = 2.320 \times 10^{-3}\ \text{mol}$ .
Moles of NaOH
The ratio is 1:1, so $n(\text{NaOH}) = 2.320 \times 10^{-3}\ \text{mol}$ .
Concentration of NaOH
$c = n/V = \dfrac{2.320 \times 10^{-3}}{25.0/1000} = 0.0928\ \text{mol dm}^{-3}$ .
$c = \dfrac{2.320 \times 10^{-3}}{0.0250} = 0.0928\ \text{mol dm}^{-3}$

Result: The sodium hydroxide is 0.0928 mol dm^-3 (3 s.f.).

Typical mistakes

Using a 1:1 ratio when the equation (e.g. H2SO4 + 2NaOH) demands 1:2.
Forgetting to convert cm^3 to dm^3 (dividing by 1000) before or after using n = cV.

Active revision

$25.0\ \text{cm}^3$ of sulfuric acid is exactly neutralised by $22.40\ \text{cm}^3$ of $0.150\ \text{mol dm}^{-3}$ sodium hydroxide. Calculate the concentration of the acid.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

References & sources

Sources

Department for Education

GCE AS and A level subject content for the sciences

AQA

AQA A-level Chemistry 7405 specification

Amount of substance

5 sections~16 min reading time3 competenciesLevel Foundation 1 · Standard 4

Amount of substance

Key points

Moles and particles from a mass

Key points

Volume of a gas from pV = nRT

Key points

From composition to molecular formula

Key points

Reacting mass and percentage yield

Atom economy of an extraction

Key points

Concentration of an alkali by titration

Sources

Amount of substance

Key points

Moles and particles from a mass

Key points

Volume of a gas from pV = nRT

Key points

From composition to molecular formula

Key points

Reacting mass and percentage yield

Atom economy of an extraction

Key points

Concentration of an alkali by titration

Sources