A-level Chemistry/WJEC/Module 1/Calculations

Balanced Equations and Associated Calculations

Central to chemical calculations is the balanced chemical equation. Here is an example:

2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O

Two molecules of ethane (C₂H₆) react with 7 molecules of oxygen (O₂) to form 4 molecules of carbon dioxide (CO₂) and 6 molecules of water (H₂O).

A balanced chemical equation is required to convert information about one chemical into information about a second chemical.

Examples:

How many molecules of oxygen would react with 200 molecules of ethane?

A very useful idea is the extent (ξ) of the reaction. This is the amount of a chemical divided by the number of molecules for that chemical which react.

For ethane, 200 molecules is an extent of 200 / 2 = 100 molecules.

For oxygen, if ξ = 100 molecules then it means that 100 x 7 = 700 molecules of oxygen can react.

How many molecules of carbon dioxide form if 300 molecules of ethane react with 350 molecules of oxygen?

For ethane, 300 molecules is an extent of 300 / 2 = 150 molecules.

For oxygen, 350 molecules is an extent of 350 / 7 = 50 molecules.

ξ for oxygen is lower, which means that during the reaction, oxygen will run out first and the reaction will stop. We call oxygen the limiting reagent and ethane is said to be in excess.

For carbon dioxide, if the reaction is limited to ξ = 50 molecules then it means that 50 x 4 = 200 molecules of carbon dioxide will form.

For ethane, if the reaction is limited to ξ = 50 molecules then it means that 50 x 2 = 100 molecules of ethane can react. The reaction began with 300 molecules of ethane, so 200 molecules of ethane will be left unreacted.

How many molecules of water form if 500 molecules of ethane react with 2100 molecules of oxygen?

For ethane, 500 molecules is an extent of 500 / 2 = 250 molecules.

For oxygen, 2100 molecules is an extent of 2100 / 7 = 300 molecules.

This time, ξ for ethane is lower, which means that during the reaction, ethane will run out first and the reaction will stop. In this scenario, ethane is the limiting reagent and oxygen is said to be in excess.

For water, if the reaction is limited to ξ = 250 molecules then it means that 250 x 6 = 1500 molecules of water will form.

For oxygen, if the reaction is limited to ξ = 250 molecules then it means that 250 x 7 = 1750 molecules of oxygen can react. The reaction began with 2100 molecules of oxygen, so 350 molecules of oxygen will be left unreacted.

For practical chemistry, we need to work with much larger numbers of molecules. Chemists originally defined a "much larger number" to be the number of atoms in 1 g of hydrogen, the lightest element. This number is called a mole and is now defined as 6.02 x 10²³.

The Mole

One "mole" is a number of particles. The number chosen for 1 mole is a very large, so that the mass of 1 mole of atoms, or molecules, is a convenient mass for laboratory use. e.g. 1 mole of ¹²C atoms have a mass of 12.0 g. Until 2019, the value of the mole was defined using ¹²C atoms as the standard.

The number of particles in 1 mole is the Avogadro Constant, which is given the symbol N_A (or sometimes L). N_A = 6.02 x 10²³ mol^-1 - this figure is in your Data Booklet. Since 2019 the value of N_A has simply been defined as 6.022 140 76 x 10²³ mol^-1. The first 3 significant figures are adequate for A-level calculations.

So, a mole of particles is N_A particles which is 6.02 x 10²³ particles.

If a mole of ¹²C atoms have a mass of 12.0 g then a single ¹²C atom has a mass of 12.0 g ÷ 6.02 x 10²³ = 1.99 x 10^-23 g.

What is left if 0.500 moles of ethane react with 2.800 moles of oxygen?

2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O

For ethane, 0.500 moles is an extent of 0.500 / 2 = 0.250 moles.

For oxygen, 2.800 moles is an extent of 2.800 / 7 = 0.400 moles.

ξ for ethane is lower, which means that during the reaction, ethane will run out first and the reaction will stop. In this scenario, ethane is the limiting reagent and oxygen is said to be in excess.

For carbon dioxide, if the reaction is limited to ξ = 0.250 moles then it means that 0.250 x 4 = 1.000 moles of carbon dioxide will form.

For water, if the reaction is limited to ξ = 0.250 moles then it means that 0.250 x 6 = 1.500 moles of water will form.

For oxygen, if the reaction is limited to ξ = 0.250 moles then it means that 0.250 x 7 = 1.750 moles of oxygen can react. The reaction began with 2.800 moles of oxygen, so 1.050 moles of oxygen will be left unreacted.

How do we figure out the number of moles of a chemical? There are three main methods:

1. comparing the mass of a chemical sample to the mass of 1 mole of that chemical
2. comparing the volume of a pure gas to the volume of 1 mole of gas, and
3. measuring volumes of solutions where the number of moles dissolved in a standard volume is known.

Relative Masses and Molar Masses

1 mole of the isotope ³⁵Cl is 35.0 g. We say that the relative isotopic mass (I_r) of ³⁵Cl is 35.0.

Chlorine naturally occurs as a mixture of two isotopes, ³⁵Cl and ³⁷Cl. 76 % of chlorine atoms are ³⁵Cl. The average mass of chlorine atom is 76 % of 35 + 24 % of 37 = 26.60 + 8.88 = 35.48. This is chlorine's relative atomic mass (A_r), which we round up to 35.5.

No single chlorine atom has a mass of 35.5, the atoms have an I_r of either 35.0 or 37.0. This is similar to the average UK family having 1.5 children - some families have 1 child, some have 2 or more. On average, there are 1.5 children per family but obviously there is no such thing as 0.5 of a child.

Chlorine atoms do not usually exist on their own. Chlorine atoms usually form diatomic molecules, Cl₂ which have a relative molecular mass (M_r) of 71.0. In the same way, chloroform (CHCl₃) has an M_r of 119.5 (12.0 + 1.01 + 3 x 35.5).

1 mole of sodium chloride, NaCl, has a mass of 58.5 g. We can say that its relative formula mass (M_r) is 58.5. Because NaCl is not made of molecules, we refer to its relative formula mass but the symbol (M_r) is the same as for relative molecular mass.

A useful alternative term is molar mass (M). This has units of g mol^-1 ("grams per mole") or kg mol^-1 on the Physics specification. Molar mass can refer to isotopes, atoms, molecules or formulae, so you do need to be specific:

• The molar mass of the chlorine-35 isotope is 35.0 g mol^-1
• The molar mass of chlorine atoms is 35.5 g mol^-1
• The molar mass of chlorine molecules is 71.0 g mol^-1

In Biology, masses of molecules are often measured in daltons (Da) or kilodaltons (kDa). For all practical purposes, 1 Da = 1 g mol^-1. (It is actually 0.999 999 999 65 g mol^-1). Glucose, for example, has a relative molecular mass of 180.156, a molar mass of 180.156 g mol^-1 and a mean mass of 180.156 Da (2.99171 x 10^-22 g).

Empirical and Molecular Formulae

Definitions:

• Empirical formula: The simplest whole number ratio of atom types present in a molecule of a compound.
• Molecular formula: The actual number of atoms of each element present in a molecule of a compound.

To determine a molecular formula you need the given mass of the substance and the empirical formula mass. Divide the given mass by the empirical formula mass.

Multiply the given answer by the ratios in the empirical formula and you will have the molecular formula.

A typical question might ask about a compound with 39.97 % carbon, 6.73 % hydrogen and 53.30 % oxygen and M_r = 180.12

There are two ways to find the empirical and molecular formulae:

If we assume a 100 g sample, we would have:

• 39.97 g carbon ÷ 12.0 = 3.331 mol ÷ 3.331 mol = 1
• 6.73 g hydrogen ÷ 1.01 = 6.663 mol ÷ 3.331 mol = 2
• 53.30 g oxygen ÷ 16.0 = 3.3310 mol ÷ 3.331 mol = 1

Divide each mass by the A_r for that element, and then divide the amounts by the smallest amount to find the ratio.

Empirical formula: CH₂O

If CH₂O were the molecular formula, then M_r would be 30.02

M_r is actually 180.12.

180.12 / 30.02 = 6, so the molecular formula is C₆H₁₂O₆

Alternatively:

If we assume a 180.12 g sample (1 mole), we would have:

• 39.97 % carbon x 180.12 g = 71.99 g ÷ 12.0 = 6.000 mol
• 6.73 % hydrogen x 180.12 g = 12.12 g ÷ 1.01 = 12.000 mol
• 53.30 % oxygen x 180.12 g = 96.00 g ÷ 16.0 = 6.000 mol

Find the mass of each element in 1 mole, then divide each mass by the A_r for that element.

Molecular formula: C₆H₁₂O₆

The ratios in the molecular formula will simplify to CH₂O, the empirical formula.

Anhydrous and Hydrated Salts

Solid salts consist of a lattice of positive and negative ions, sometimes water molecules are incorporated into the lattice. The water in a lattice is called the water of crystallisation. When a salt contains water the salt is hydrated, if a salt does not contain water of crystallisation the salt is anhydrous.

A mole of a particular hydrated salt usually has the same number of moles of water of crystallisation, its formula shows how many water molecules are present for every molecule of salt e.g. copper sulfate, CuSO₄, has 5 moles of water for every mole of salt, therefore its hydrated formula is CuSO₄⋅5H₂O, (a dot ⋅ separates the salt's formula from the water of crystallisation).

Examples:

• CuSO₄⋅5H₂O
• Na₂CO₃⋅10H₂O
• MgSO₄⋅7H₂O

Dehydration

Many hydrated salts lose their water of crystallisation when heated. For example:

CuSO₄⋅5H₂O(s) → CuSO₄(s) + 5 H₂O(g)

M_r of CuSO₄ is 159.6 and M_r of H₂O is 18.0

An experiment might start with 12.48 g of CuSO₄⋅5H₂O(s) and find that after heating, only 7.98 g of solid are left.

7.98 g of CuSO₄ is 7.98 / 159.6 = 0.0500 mol

The 4.50 g of mass that is "lost" is the H₂O(g). 4.50 g of H₂O is 4.50 / 18.0 = 0.250 mol

The ratio of CuSO₄ to H₂O is 0.0500 mol to 0.250 mol which is 1 to 5.

Practically, these experiments are prone to errors if not all the water is evaporated or if extra mass is lost. Extra mass can be lost if the solid "spits" during heating and some solid jumps out, or if the anhydrous salt itself can decompose if it is heated for too long:

2 CuSO₄ → 2 CuO(s) + 2 SO₂(g) + O₂(g)

Concentrations of Solutions

Water of crystallisation is confusing when working out concentrations of salt solutions. If we want to make 250 cm³ of a 0.100 mol dm^-3 solution of sodium carbonate, then how much sodium carbonate do we need?

The easy bit is (or should be) n = c x V = 0.250 dm³ x 0.100 mol dm^-3 = 0.0250 mol.

What is the mass of 0.0250 mol of sodium carbonate? If we are using the decahydrate (Na₂CO₃⋅10H₂O) then its M_r is 286.2 and the mass is m = n x M_r = 7.16 g

If we are using the anhydrous salt (Na₂CO₃) then its M_r is 106.0 and the mass is m = n x M_r = 2.65 g

So 7.16 g of Na₂CO₃⋅10H₂O dissolved in 250 cm³ is an Na₂CO₃ solution (notice the water is no longer mentioned) with a concentration of 0.100 mol dm^-3.

Standard Solutions

A standard solution of a chemical has a known concentration.

concentration = quantity of solute / volume of solution

Various sizes of volumetric flask. In each case, the liquid level is made up so that the bottom of the meniscus is on the calibration line. The flasks are marked in "ml" (millilitres) which are identical to cm³. 5000 ml = 5000 cm³ = 5 dm³.

Usually, the solute is measured in moles, the volume is measured in dm³ and concentration is measured in mol dm^-3 ("moles per decimetre cubed").

c = n / V

It is important to be able to convert between dm³ (litres, L or l) and the common units used in the laboratory, cm³ (centimetres cubed, c.c., millilitres, mL or ml).

1 dm³ = 1000 cm³ 1 cm³ = 1/1000 dm³ (0.001 dm³ or 1 x 10^-3 dm³

Notice that concentration is based on the volume of the solution. Dissolving 1 mole (58.5 g) of sodium chloride in 1 dm³ of water does not make a 1 mol dm^-3 solution. This mixture would have a volume of 1.021 dm³ (1021 cm³) and a density of 1.037 g cm^-3. The concentration would actually be 0.979 mol dm^-3.

1 mol dm^-3 NaCl solution can be made by taking 58.5 g of NaCl and making the total volume of solution up to 1 dm³ by adding water. The volume is typically measured with a volumetric flask. This actually requires 0.980 dm³ (980 cm³) of water and the solution has a density of 1.036 g cm^-3. However, we do not need to know the volume of water, or the density, if we use a volumetric flask to measure the total volume of solution.

Most importantly, if we measure 25 cm³ (for example) of this solution, we know that this 25 cm³ sample contains 1.00 mol dm^-3 x 0.025 dm³ = 0.025 mol of NaCl.

A burette filled with a coloured solution showing the meniscus at the top of the the liquid. In this case the burette reading is 20.05 cm³. Remember to read the bottom of the meniscus and notice that the burette scale is from top to bottom. Burette readings are expected to be recorded to the nearest 0.05 cm³ (even if the reading is 0.00 cm³, write "0.00" not "0").

The Ideal Gas Equation

pV=nRT

p is the pressure of the gas in Pa (Nm⁻²)

V is the volume of the gas in m³

Shortcut: p is often given in kPa and V is often given in dm³.

22.4 dm³ at 101 kPa can be written as 0.0224 m³ at 101 000 Pa, but notice that 22.4 x 101 = 0.0224 x 101 000.

So if p is given in kPa and V is given in dm³ then you need not convert p to Pa and V to m³ - just multiply the values you're given.

n is the amount (number of moles)

R is the gas constant (8.31 J mol⁻¹ K⁻¹)

T is the temperature in kelvin (equivalent to 273 + θ, the temperature in degrees Celsius)

Example: How many moles of gas occupy 250 cm³ at 15 °C and 200 kPa?

Using pV = nRT, n = pV / RT

n = 200 000 Pa x 2.5 x 10^-4 m³ / (8.31 J mol^-1 K^-1 x 288 K)

n = 2.089 x 10^-2 mol

(Using the V_m method (see below) the value comes out as 2.095 x 10^-2 mol. Using unrounded values for R, p and T, the answer should be 2.087 x 10^-2 mol i.e. the pV = nRT method is more precise if p is given in kPa or Pa.)

Molar Gas Volumes, V_m

The Data Booklet gives two values for V_m, 22.4 dm³ mol⁻¹ at 273 K and 1 atm, and 24.5 dm³ mol⁻¹ at 298 K and 1 atm.

These are the volumes of 1 mole of gas under these conditions.

To find the actual number of moles, divide the volume of gas by the appropriate value of V_m. For example, 49.0 dm³ of gas at 298 K and 1 atm is 49.0 / 24.5 = 2.00 mol.

V_m can be corrected for different temperatures and pressures. It is best to start with the 22.4 dm³ mol⁻¹ figure at 273 K and 1 atm. The 24.5 dm³ mol⁻¹ figure at 298 K and 1 atm is good only for those exact conditions because it has a larger rounding error (-0.01 % for 22.4 dm³ mol⁻¹ vs +0.19 % for 24.5 dm³ mol⁻¹).

At temperature T and pressure p, V_m = 22.4 dm³ mol⁻¹ x T / (273 x p)

T and p must be in kelvin and atmospheres respectively.

For example, at 100 °C and 2 atm, V_m = 22.4 dm³ mol⁻¹ x 373 K / (273 x 2 atm) = 15.3 dm³ mol⁻¹

Use this approach if the pressure is given in atm, or is stated to be 101 kPa (i.e. 1 atm). For pressure given in kPa, or Pa, the pV = nRT approach is more precise.

n = pV / RT (p is in Pa, V in m³, T in K, R = 8.31 J mol⁻¹ K⁻¹)

n = V / V_m = 273 K x pV / (22.4 dm³ mol⁻¹ x T) (p is in atm, V in dm³, T in K)

Example: How many moles of gas occupy 250 cm³ at 45 °C and 1.2 atm?

n = 273 K x pV / (22.4 dm³ mol⁻¹ x T)

T = 45 + 273 = 318 K

V = 0.250 dm³

n = 273 K x 1.2 atm x 0.250 dm³ / (22.4 dm³ mol⁻¹ x 318 K)

n = 0.01150 mol = 1.150 x 10^-2 mol

Using pV = nRT, n = pV / RT

n = 121 200 Pa x 2.5 x 10^-4 m³ / (8.31 J mol^-1 K^-1 x 318 K)

n = 1.147 x 10^-2 mol

(Using unrounded values for R, p and T, the answer should be 1.149 x 10^-2 mol i.e. the V_m method is more precise if p is given in atm.)

Errors

It is important to know how certain a measurement is; What is the potential error in the measurement?

A 25 cm³ volumetric pipette is marked ± 0.06 cm³. This is the absolute uncertainty, u.

If the uncertainty is not recorded, we can assume that the absolute uncertainty is:

• one-half of the divisions drawn on a scale
• one-half of the last decimal place shown on a digital display

For a ruler which has a 1 mm scale, the absolute uncertainty is ± 0.5 mm. For a digital thermometer showing 24.2 ᐤC, the absolute uncertainty is ± 0.05 ᐤC.

Ruler with 1 mm scales, marked every 1 cm

Digital thermometer

Be careful, many measurements are actually the difference between two measurements. If we used the rule to measure 297 mm, we rely on the 0 mm and the 297 mm marks being accurate. If we add the two absolute uncertainties we find that we have measured 297 mm ± 1 mm†. This is true of burettes and balances.

† To be exact, we shouldn't add the uncertainties, but square the uncertainties, add the squares, and then take the square root: √(0.5² + 0.5²) = 0.71 mm. There's no time for this at A-level: Just add the uncertainties!

Examples:

• Initial burette reading: 0.4 cm³. The scale is marked every 0.1 cm³ so this initial reading is 0.4 ± 0.05 cm³.
• Final burette reading: 24.7 cm³ ± 0.05 cm³.
• Titre: 24.7 - 0.4 = 24.3 ± 0.1 cm³

• Initial burette reading: 0.0 ± 0.05 cm³.
• Final burette reading: 24.3 cm³ ± 0.05 cm³.
• Titre: 24.3 - 0.0 = 24.3 ± 0.1 cm³

This is why it is often a waste of time to "zero" a burette. Just record the initial and final values.

• Mass of weighing boat: 1.210 g. The digital scale gives three decimal places so this reading is 1.210 ± 0.0005 g.
• Mass of weighing boat with sample: 2.210 ± 0.0005 g.
• Mass of weighing boat after sample transferred to beaker: 1.211 ± 0.0005 g.
• Mass of sample transferred to beaker: 2.210 - 1.211 = 0.999 ± 0.001 g

Notice how it is important to record the final zero if you have measured it. 0.0 cm³ is more precise than 0 cm³. 1.210 g is more precise than 1.21 g.

What if you use the 0.999 g of sample to make a 250 cm³ solution, and measure 24.3 cm³ from this solution using the burette? How many grams of sample are in the 24.3 cm³?

Well, 0.999 g x 24.3 cm³ ÷ 250 cm³ = 0.0971 g but how certain is this value?

We have three uncertainties to consider:

• Balance: 0.999 ± 0.001 g
• Volumetric flask: 250 ± 0.3 cm³
• Burette: 24.3 ± 0.1 cm³

The uncertainties add up, but how?

We can calculate the relative (percentage) uncertainty, u_r from the equation u_r = 100 % x absolute error ÷ the value measured.

• Balance: 100 % x 0.001 g ÷ 0.999 g = 0.100 %
• Volumetric flask: 100 % x 0.3 cm³ ÷ 250 cm³ = 0.120 %
• Burette: 100 % x 0.1 cm³ ÷ 24.3 cm³ = 0.411 %

Adding the percentage errors gives us ± 0.631 % (or 0.440 % using the square-root-of-the-sum-of-the-squares method you really don't need to know).

The 24.3 cm³ of solution contains 0.0971 g ± 0.631 %.

Significant figures

As a rule, give your answers to 3 significant figures unless told otherwise. This implies a percentage error between 0.5 % and 0.05 %.

0.999 g could be between 0.9995 and 0.9985 g so ± 0.0005 g or ± 0.05 %

0.100 g could be between 0.1005 and 0.0995 g so ± 0.0005 g or ± 0.5 %

• 2 significant figures implies a percentage error between 5 % and 0.5 %.
• 3 significant figures implies a percentage error between 0.5 % and 0.05 %.
• 4 significant figures implies a percentage error between 0.05 % and 0.005 %.

If you use several values in a calculation, the value with the lowest number of significant figures dictates the number of significant figures in the answer.

Example: 0.998 g ÷ 2.0 dm³ = 0.50 g dm^-3

Exception: When adding or subtracting numbers, the value with the lowest number of decimal places dictates the number of decimal places in the answer.

Example: 2.0 g - 0.998 g = 1.0 g