(e) Chemical formulae, equations and calculations Archives

1:25 write word equations and balanced chemical equations (including state symbols): for reactions studied in this specification and for unfamiliar reactions where suitable information is provided

Example:

Sodium (Na) reacts with water (H₂O) to produce a solution of sodium hydroxide (NaOH) and hydrogen gas (H₂).

Word equation:

sodium + water –> sodium hydroxide + hydrogen

Writing the chemical equation

A chemical equation represents what happens in terms of atoms in a chemical reaction.

Step 1: To write a chemical equation we need to know the chemical formulae of the substances.

Na + H₂O –> NaOH + H₂

Step 2: The next step is to balance the equation: write a large number before each compound so the number of atoms of each element on the left hand side (reactants) matches the number on the right (products). This large number is the amount of each compound or element.

During this balancing stage the actual formulas for each compound must not be changed. Only the number of each compound changes.

2Na + 2H₂O –> 2NaOH + H₂

If asked for an equation, the chemical equation must be given.

State symbols are used to show what physical state the reactants and products are in.

State symbols	Physical state
(s)	Solid
(l)	Liquid
(g)	Gas
(aq)	Aqueous solution (dissolved in water)

Example:

A solid piece of sodium (Na) reacts with water (H₂O) to produce a solution of sodium hydroxide (NaOH) and hydrogen gas (H₂).

2Na(s) + 2H₂O(l) –> 2NaOH(aq) + H₂(g)

Hydr0Gen2023-09-18T11:05:10+00:00Categories: (e) Chemical formulae, equations and calculations, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:25, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Alkanes & Crude Oil, Topic: Fundamentals, Topic: Ions, Topic: Key Reactions, Topic: Moles|

1:26 calculate relative formula masses (including relative molecular masses) (Mᵣ) from relative atomic masses (Aᵣ)

Relative formula mass (M_r) is mass of a molecule or compound (on a scale compared to carbon-12).

It is calculated by adding up the relative atomic masses (A_r) of all the atoms present in the formula.

Example:

The relative formula mass (M_r) for water (H₂O) is 18.

Water = H₂O

Atoms present = (2 x H) + (1 x O)

M_r = (2 x 1) + (1 x 16) = 18

Hydr0Gen2022-11-30T14:12:46+00:00Categories: (e) Chemical formulae, equations and calculations, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:26, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Calculations, Topic: Fundamentals, Topic: Key Calculations, Topic: Moles|

1:27 know that the mole (mol) is the unit for the amount of a substance

In Chemistry, the mole is a measure of amount of substance (n).

The abbreviation for mole is mol.

The mass of 1 mole of a substance is the relative formula mass (M_r) of the substance in grams.

Example:

The M_r of water is 18.

Therefore the mass of 1 mol of water equals 18 g.

Hydr0Gen2018-04-25T21:37:27+00:00Categories: (e) Chemical formulae, equations and calculations, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:27, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Key Calculations, Topic: Moles|

1:28 understand how to carry out calculations involving amount of substance, relative atomic mass (Aᵣ) and relative formula mass (Mᵣ)

The following formula allows for the interconversion between a mass in grams and a number of moles for a given substance:

$Amount(mol) = \frac{mass(g)}{M_r}$

Example 1:

Calculate the amount, in moles, of 8.8 g of carbon dioxide (CO₂).

Step 1: Calculate the relative formula mass (M_r) of carbon dioxide (CO₂).

$Carbon\,dioxide = CO_2$

$Atoms\,present = (1 \times C) + (2 \times O)$

$M_r = (1 \times 12) + (2 \times 16) = 44$

Step 2: Use the formula to calculate the amount in moles.

$Amount = \frac{mass(g)}{M_r}$

$Amount = \frac{8.8}{44}$

$Amount = 0.2 mol$

Example 2:

Calculate the mass of 2 mol of copper(II) sulfate (CuSO₄).

Step 1: Calculate the relative formula mass (M_r) of copper(II) sulfate (CuSO₄).

$copper(II)\,sulfate = CuSO_4$

$Atoms\,present = (1 \times Cu) + (1 \times S) + (4 \times O)$

$M_r = (1 \times 63.5) + (1 \times 32) + (4 \times 16) = 159.5$

Step 2: Rearrange the formula to calculate the mass.

$mass = amount \times M_r$

$mass = 2 \times 159.5$

$mass = 319g$

Hydr0Gen2018-04-25T21:37:27+00:00Categories: (e) Chemical formulae, equations and calculations, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:28, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Key Calculations, Topic: Moles|

1:29 calculate reacting masses using experimental data and chemical equations

Example: When calcium carbonate (CaCO₃) is heated calcium oxide is produced. You can use reacting mass calculations to calculate the mass of calcium oxide produced when heating 25 g of calcium carbonate.

CaCO₃ –> CaO + CO₂

Step 1: Calculate the amount, in moles, of 25 g of calcium carbonate (CaCO₃)

$M_r(CaCO_3)=(1 \times 40) + (1\times12) + (3 \times 16)$

$M_r(CaCO_3)= 100$

$Amount = \frac{mass(g)}{M_r}$

$Amount = \frac{25}{100}$

$Amount = 0.25 mol$

Step 2: Deduce the amount, in moles, of CaO produced from 0.25 mol of CaCO₃.

This step involves using the ratio of CaCO₃ to CaO from the chemical equation.

CaCO₃ –> CaO + CO₂

From the equation you can see that the ratio of CaCO₃ to CaO is 1:1.

Therefore if you have 0.25 mol of CaCO₃ this will produced 0.25 mol of CaO.

Step 3: Calculate the mass of 0.25 mol of CaO.

$M_r(CaO)=(1 \times 40) + (1 \times 16)$

$M_r(CaO)= 56$

$Mass(CaO)=amount \times M_r$

$Mass(CaO)= 0.25 \times 56$

$Mass(CaO)= 14g$

A simple format for laying out this method can be used.

Example: What mass of ammonia (NH₃) is formed when 7 g of nitrogen (N₂) is combined with hydrogen (H₂).

Hydr0Gen2018-04-25T21:37:27+00:00Categories: (e) Chemical formulae, equations and calculations, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:29, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Key Calculations, Topic: Moles|

1:30 calculate percentage yield

Yield is how much product you get from a chemical reaction.

The theoretical yield is the amount of product that you would expect to get. This is calculated using reacting mass calculations.

In most chemical reactions, however, you rarely achieved your theoretical yield.

For example, in the following reaction:

CaCO₃ –> CaO + CO₂

You might expect to achieve a theoretical yield of 56 g of CaO from 100 g of CaCO₃.

However, what if the actual yield is only 48 g of CaO.

By using the following formula, the % yield can be calculated:

$yield= \frac{actual\,amount\,of\,product}{theoretical\,amount\,of\,product}$

$yield= \frac{48}{56}$

$yield=0.86$

$\% yield=86\%$

Hydr0Gen2023-01-10T11:13:31+00:00Categories: (e) Chemical formulae, equations and calculations, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:30, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Equilibria, Topic: Key Calculations, Topic: Reversible Reactions (D)|

1:31 understand how the formulae of simple compounds can be obtained experimentally, including metal oxides, water and salts containing water of crystallisation

Finding the formula of a metal oxide experimentally

The formulae of metal oxides can be found experimentally by reacting a metal with oxygen and recording the mass changes.

Example: When magnesium is burned in air, it reacts with oxygen (O₂) to form magnesium oxide (MgO).

Method:
• Weigh a crucible and lid
• Place the magnesium ribbon in the crucible, replace the lid, and reweigh
• Calculate the mass of magnesium
(mass of crucible + lid + Magnesium – mass of crucible + lid)
• Heat the crucible with lid on until the magnesium burns
(lid prevents magnesium oxide escaping therefore ensuring accurate results)
• Lift the lid from time to time (this allows air to enter)
• Stop heating when there is no sign of further reaction
(this ensures all Mg has reacted)
• Allow to cool and reweigh
• Repeat the heating , cooling and reweigh until two consecutive masses are the same
(this ensures all Mg has reacted and therefore the results will be accurate)
• Calculate the mass of magnesium oxide formed (mass of crucible + lid + Magnesium oxide – mass of crucible + lid)

Finding the formula of a salt containing water of crystallisation

When some substances crystallise from solution, water becomes chemically bound up with the salt. This is called water of crystallisation and the salt is said to be hydrated. For example, hydrated copper sulfate has the formula CuSO₄.5H₂O which formula indicates that for every CuSO₄ in a crystal there are five water (H₂O) molecules.

When you heat a salt that contains water of crystallisation, the water is driven off leaving the anhydrous (without water) salt behind. If the hydrated copper sulfate (CuSO₄.5H₂O) are strongly heated in a crucible then they will break down and the water lost, leaving behind anhydrous copper sulfate (CuSO₄). The method followed is similar to that for metal oxides, as shown above. The difference of mass before and after heating is the mass of the water lost. These mass numbers can be used to obtain the formula of the salt.

Hydr0Gen2023-01-10T11:36:41+00:00Categories: (e) Chemical formulae, equations and calculations, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:31, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Acids & Salts, Topic: Acids & Salts (D), Topic: Key Calculations, Topic: Specification experiments|

1:32 know what is meant by the terms empirical formula and molecular formula

The empirical formula shows the simplest whole-number ratio between atoms/ions in a compound.

The molecular formula shows the actual number of atoms of each type of element in a molecule.

For example, for ethane:

Molecular formula = C₂H₆

Empirical formula = CH₃

Here are some more examples:

Name	Molecular formula	Empirical Formula
pentene	C₅H₁₀	CH₂
butene	C₄H₈	CH₂
glucose	C₆H₁₂O₆	CH₂O
hydrogen peroxide	H₂O₂	HO
propane	C₃H₈	C₃H₈

Notice from the table that several different molecules can have the same empirical formula, which means that it is not possible to deduce the molecular formula from the empirical formula without some additional information. Also notice that sometimes it is not possible to simplify a molecular formula into simpler whole-number ratio, in which case the empirical formula is equal to the molecular formula.

Hydr0Gen2023-01-10T08:29:26+00:00Categories: (e) Chemical formulae, equations and calculations, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:32, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Calculations, Topic: Crude oil to plastics (D), Topic: Key Calculations|

1:33 calculate empirical and molecular formulae from experimental data

Calculating the Empirical Formula

Example: What is the empirical formula of magnesium chloride if 0.96 g of magnesium combines with 2.84 g of chlorine?

Step 1: Put the symbols for each element involved at the top of the page.

Step 2: Underneath, write down the masses of each element combining.

Step 3: Divide by their relative atomic mass (A_r).

Step 4: Divide all the numbers by the smallest of these numbers to give a whole number ratio.

Step 5: Use this to give the empirical formula.

(If your ratio is 1:1.5 then multiple each number by 2

If your ratio is 1:1.33 then x3. If your ratio is 1:1.25 x4)

Calculating the Molecular Formula

If you know the empirical formula and the relative formula mass you can work out the molecular formula of a compound.

Example: A compound has the empirical formula CH₂, and its relative formula mass is 56. Calculate the molecular formula.

Step 1: Calculate the relative mass of the empirical formula.

Step 2: Find out the number of times the relative mass of the empirical formula goes into the M_r of the compound.

Step 3: This tells how many times bigger the molecule formula is compared to the empirical formula.

Empirical formula calculations involving water of crystallisation

When you heat a salt that contains water of crystallisation, the water is driven off leaving the anhydrous (without water) salt behind. For example, barium chloride crystals contain water of crystallisation, and therefore would have the formula BaCl₂.nH₂O where the symbol ‘n’ indicates the number of molecules of water of crystallisation. This value can be calculated using the following method.

Example: If you heated hydrated barium chloride (BaCl₂.nH₂O) in a crucible you might end up with the following results.

Step 1: Calculate the mass of the anhydrous barium chloride (BaCl₂) and the water (H₂O) driven off.

Step 2: Use the empirical formula method to find the value of n in the formula.

Hydr0Gen2023-01-10T08:30:04+00:00Categories: (e) Chemical formulae, equations and calculations, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:33, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Calculations, Topic: Crude oil to plastics (D), Topic: Key Calculations|

1:34 (Triple only) understand how to carry out calculations involving amount of substance, volume and concentration (in mol/dm³) of solution

Concentration is a measurement of the amount of substance per unit volume.

In Chemistry, concentration is measured in mol/dm³ (read as moles per cubic decimetre).

The following formula allows for the interconversion between a concentration (in mol/dm³), the amount (in moles) of a substance in a solution, and the volume of the solution (in dm³).

$amount(mol) = concentration(mol/dm^3) \times volume(dm^3)$

$concentration(mol/dm^3) = \frac{amount(mol)}{volume(dm^3)}$

Note: 1 dm³ = 1000 cm³

For example, to convert 250 cm³ into dm³:

$250cm^3 = \frac{250}{1000}$

$250cm^3 = 0.25dm^3$

Example 1: 0.03 mol of sodium carbonate (Na₂CO₃) is dissolved in 300 cm³ of water. Calculate the concentration of the solution.

Step 1: Convert the volume of water from cm³ into dm³.

$volume = 300cm^3$

$volume = \frac{300}{1000}dm^3$

$volume = 0.300dm^3$

Step 2: Use the molar concentration formula to calculate of the concentration of the solution.

$concentration = \frac{amount}{volume}$

$concentration = \frac{0.03}{0.300}$

$concentration = 0.1 mol/dm^3$

Example 2: Calculate the amount, in moles, of 25cm³ of hydrochloric acid (HCl) with a concentration of 2 mol/dm³.

Step 1: Convert the volume of HCl from cm³ into dm³.

$volume = 25cm^3$

$volume = \frac{25}{1000}dm^3$

$volume = 0.025dm^3$

Step 2: Rearrange the molar concentration formula to calculate of the amount, in moles of HCl.

$amount= concentration \times volume$

$amount = 2 \times 0.025$

$amount = 0.050 mol$

Titration calculations, example 1

The titration method that is used to prepare a soluble salt is also used to determine the concentration of an unknown solution.

For example, a titration problem will look like this:

A pupil carried out a titration to find the concentration of a solution of hydrochloric acid (HCl). She found that 25.0 cm³ of 0.100 mol/dm³ sodium hydroxide solution (NaOH) required 23.50 cm³ of dilute hydrochloric acid for neutralisation. Calculate the concentration, in mol/dm³ of the acid.

The chemical equation for this reaction is:

NaOH(aq) + HCl(aq) –> NaCl(aq) + H₂O(l)

volume 25.0cm³ 23.50cm³

concentration 0.100mol/dm³

It can be useful, as shown above, to write the values of the volumes & concentration underneath the equation.

Step 1: Calculate the amount, in moles, of the solution that you know the values for both volume and concentration. In this case sodium hydroxide (NaOH).

First convert the volume of NaOH from cm³ into dm³.

$volume = 25cm^3$

$volume = \frac{25}{1000}dm^3$

$volume = 0.025dm^3$

Then rearrange the molar concentration formula to calculate of the amount, in moles of NaOH.

$amount= concentration \times volume$

$amount = 0.100 \times 0.025$

$amount = 0.0025 mol$

Step 2: Deduce the amount, in moles of the solution with the unknown concentration. In this case hydrochloric acid (HCl).

From the chemical equation the ratio of NaOH to HCl is 1:1

Therefore if you have 0.0025 mol of NaOH, this will react with 0.0025 mol of HCl.

Step 3: Calculate the concentration of the hydrochloric acid (HCl).

Convert the volume of HCl from cm³ into dm³.

$volume = 23.5cm^3$

$volume = \frac{23.5}{1000}dm^3$

$volume = 0.0235dm^3$

Use the molar concentration formula to calculate of the concentration of the HCl.

$concentration = \frac{amount}{volume}$

$concentration = \frac{0.0025}{0.0235}$

$concentration = 0.106 mol/dm^3$

Titration calculations, example 2

25.0 cm³ of sodium carbonate solution (Na₂CO₃) of unknown concentration was neutralised by 30.0 cm³ of 0.100 mol/dm³ nitric acid (HNO₃).

Na₂CO₃(aq) + 2HNO₃(aq) –> 2NaNO₃(aq) + CO₂(g) + H₂O(l)

Calculate the concentration, in mol/dm³ of the sodium carbonate solution.

Step 1: Calculate the amount, in moles, of nitric acid (HNO₃).

$volume = 30cm^3$

$volume = \frac{30}{1000}dm^3$

$volume = 0.030dm^3$

$amount= concentration \times volume$

$amount = 0.100 \times 0.030$

$amount = 0.0030 mol$

Step 2: Deduce the amount, in moles of sodium carbonate (Na₂CO₃).

Using the equation:

Na₂CO₃(aq) + 2HNO₃(aq) –> 2NaNO₃(aq) + CO₂(g) + H₂O(l)

Ratio Na₂CO₃:HNO₃ = 1:2

$amount=\frac{0.0030}{2} = 0.0015mol$

Step 3: Calculate the concentration of sodium carbonate (Na₂CO₃).

$volume = 25cm^3$

$volume = \frac{25}{1000}dm^3$

$volume = 0.025dm^3$

$concentration = \frac{amount}{volume}$

$concentration = \frac{0.0015}{0.025}$

$concentration = 0.06 mol/dm^3$

Converting mol/dm³ into g/dm³

Concentration can also be expressed in g/dm³ (grams per cubic decimetre).

Therefore mol/dm³ can be converted into g/dm³.

Example: Convert 0.06 mol/dm³ of sodium carbonate (Na₂CO₃) into g/dm³

Step 1: calculate the relative formula mass (M_r) of sodium carbonate (Na₂CO₃).

$M_r= (2 \times 23) + (1 \times 12) (3 \times 16) = 106$

Step 2: Recall the formula giving the relationship between mass, amount and formula mass.

$mass= amount(in\,moles) \times M_r$

therefore

$mass/dm^3= moles/dm^3 \times M_r$

$mass/dm^3= 0.06 \times 106$

$mass/dm^3= 6.36 g/dm^3$

Hydr0Gen2019-05-13T15:41:32+00:00Categories: (e) Chemical formulae, equations and calculations, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:34, Chemistry, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Key Calculations, Triple|

1:35 (Triple only) understand how to carry out calculations involving gas volumes and the molar volume of a gas (24dm³ and 24,000cm³ at room temperature and pressure (rtp))

The molar volume of a gas is the volume that one mole of any gas will occupy.

1 mole of gas, at room temperature and pressure (rtp), will always occupy 24 dm³ or 24,000 cm³.

Note: 1 dm³ = 1000 cm³

The following formulae allows for the interconversion between a volume in dm³ or cm³ and a number of moles for a given gas:

$Amount(mol) = \frac{volume(dm^3)}{24}$

$Amount(mol) = \frac{volume(cm^3)}{24000}$

Example 1:

Calculate the amount, in moles, of 12 dm³ of carbon dioxide (CO₂).

$Amount = \frac{volume(dm^3)}{24}$

$Amount = \frac{12}{24}$

$Amount = 0.5mol$

Example 2:

Calculate the volume at rtp in cubic centimetres (cm³), of 3 mol of oxygen, (O₂).

$Volume = amount \times 24000$

$Volume = 3 \times 24000$

$Volume = 72000cm^3$

Hydr0Gen2019-06-21T14:38:23+00:00Categories: (e) Chemical formulae, equations and calculations, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:35, Chemistry, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Key Calculations, Triple|

1:36 practical: know how to determine the formula of a metal oxide by combustion (e.g. magnesium oxide) or by reduction (e.g. copper(II) oxide)

To determine the formula of a metal oxide by combustion.

Example: When magnesium is burned in air, it reacts with oxygen (O₂) to form magnesium oxide (MgO).

2Mg + O₂ –> 2MgO

Method:

Weigh a crucible and lid
Place the magnesium ribbon in the crucible, replace the lid, and reweigh
Calculate the mass of magnesium
Heat the crucible with lid on until the magnesium burns (lid prevents magnesium oxide escaping therefore ensuring accurate results)
Lift the lid from time to time (this allows air to enter)
Stop heating when there is no sign of further reaction
Allow to cool and reweigh
Repeat the heating, cooling and reweigh until two consecutive masses are the same (this ensures all Mg has reacted and therefore the results will be accurate)
Calculate the mass of magnesium oxide formed (mass after heating – mass of crucible)

It is then possible to use this data to calculate the empirical formula of the metal oxide.

To determine the formula of a metal oxide by reduction

Example: When copper (II) oxide is heated in a stream of methane, the oxygen is removed from the copper (II) oxide, producing copper, carbon dioxide and water:

4CuO + CH₄ –> 4Cu + CO₂ + 2H₂O

By comparing the mass of copper produced with the mass of copper oxide used it is possible to determine the formula of the copper oxide.

As an alternative, hydrogen gas could be used instead of methane:

CuO + H₂ –> Cu + H₂O

It is then possible to use this data to calculate the empirical formula of the metal oxide.

There is an important safety point to note with both versions of this experiment. Both methane and hydrogen are explosive if ignited with oxygen. It is important that a good supply of the the gas is allowed to fill the tube before the gas it lit. This flushes out any oxygen from the tube, so the gas will only burn when it exits the tube and comes into contact with oxygen in the air.

Hydr0Gen2019-03-16T10:58:55+00:00Categories: (e) Chemical formulae, equations and calculations, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:36, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Calculations, Topic: Key Calculations, Topic: Specification experiments|