Chemistry Formulas, Reactions and Compounds

Symbols and formulas

Writing a formula from valencies (criss-cross rule)

A^x and B^y → A_yB_x

In words: Swap the valencies and write them as subscripts.

Valency is the combining power of an atom. To write the formula of a compound, write the symbols side by side, swap the valencies, and write each as a small number after the other symbol. If both numbers can be divided by a common factor, reduce them. Leave out a subscript of 1.

What the letters mean

A, B

the two elements (or groups such as SO_{4})

x, y

their valencies

A_yB_x

the formula of the compound

Worked examples

1. Example 1. Calcium has valency 2 and chlorine has valency 1. How many chlorine atoms are there in the formula of calcium chloride?

   1. Ca has valency 2 and Cl has valency 1
   2. Swap: Ca takes 1, Cl takes 2
   3. Formula: CaCl₂

   Answer: 2 chlorine atoms (CaCl₂)

2. Example 2. Aluminium has valency 3 and oxygen has valency 2. How many aluminium atoms are in aluminium oxide?

   1. Al has valency 3 and O has valency 2
   2. Swap: Al takes 2, O takes 3
   3. Formula: Al₂O₃

   Answer: 2 aluminium atoms (Al₂O₃)

Watch out: Reduce the numbers when you can: magnesium (valency 2) and oxygen (valency 2) give Mg₂O₂ after swapping, which reduces to MgO.

Chemical reactions

Photosynthesis

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

In words: Six carbon dioxide and six water make one glucose and six oxygen.

Green plants make their own food using sunlight, which chlorophyll absorbs. Carbon dioxide from the air and water from the soil are turned into glucose, and oxygen is released. Sunlight and chlorophyll are needed but are not written in the equation.

The substances

CO₂

carbon dioxide, taken in from the air

H₂O

water, taken in through the roots

C₆H₁₂O₆

glucose, the food the plant makes

O₂

oxygen, released into the air

Worked examples

1. Example 1. How many moles of carbon dioxide are needed to make 1 mole of glucose?

   1. The equation shows 6CO₂ for each C₆H₁₂O₆

   Answer: 6 moles

Chemical reactions

Respiration

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

In words: One glucose and six oxygen make six carbon dioxide and six water, releasing energy.

Living cells use oxygen to break glucose down and release energy. It is the reverse of photosynthesis in terms of substances. The energy released is not written as a substance in the equation.

The substances

C₆H₁₂O₆

glucose, the fuel

O₂

oxygen, breathed in

CO₂

carbon dioxide, breathed out

H₂O

water

Worked examples

1. Example 1. How many moles of carbon dioxide form when 2 moles of glucose are used up?

   1. 1 mol C₆H₁₂O₆ gives 6 mol CO₂
   2. 2 × 6

   Answer: 12 moles

Chemical reactions

Rusting of iron

4Fe + 3O₂ → 2Fe₂O₃

In words: Four iron and three oxygen make two iron oxide.

Iron slowly joins with oxygen to make a reddish-brown substance. Real rust also needs water and is a hydrated form of iron(III) oxide, Fe₂O₃·xH₂O; the equation shows the main change. Painting and galvanising keep air and water away from the iron.

The substances

Fe

iron

O₂

oxygen from the air

Fe₂O₃

iron(III) oxide, the main part of rust

Worked examples

1. Example 1. How many grams of Fe₂O₃ form when 112 g of iron rust completely? (Fe = 56, Fe₂O₃ = 160 g/mol)

   1. Moles of Fe = 112 ÷ 56 = 2
   2. 4 mol Fe give 2 mol Fe₂O₃, so 2 mol Fe give 1 mol
   3. Mass = 1 × 160

   Answer: 160 g

Chemical reactions

Burning of magnesium

2Mg + O₂ → 2MgO

In words: Two magnesium and one oxygen make two magnesium oxide.

Magnesium burns with a dazzling white flame and leaves a white powder, magnesium oxide. It is an example of a combination reaction, in which two substances join to make one.

The substances

Mg

magnesium ribbon

O₂

oxygen from the air

MgO

magnesium oxide, the white ash

Worked examples

1. Example 1. How many grams of MgO form when 24 g of magnesium burns? (Mg = 24, MgO = 40 g/mol)

   1. Moles of Mg = 24 ÷ 24 = 1
   2. 2 mol Mg give 2 mol MgO, so 1 mol gives 1 mol
   3. Mass = 1 × 40

   Answer: 40 g

Chemical reactions

Neutralisation of an acid by a base

HCl + NaOH → NaCl + H₂O

In words: Hydrochloric acid and sodium hydroxide make salt and water.

An acid and a base react to make a salt and water, and the mixture becomes less acidic and less basic. This is called neutralisation. Indigestion tablets work this way on the extra acid in the stomach.

The substances

HCl

hydrochloric acid

NaOH

sodium hydroxide, a strong base

NaCl

sodium chloride, a salt

H₂O

water

Worked examples

1. Example 1. How many moles of NaOH are needed to neutralise 0.2 mole of HCl?

   1. The equation shows 1 HCl for each 1 NaOH, so the moles are equal

   Answer: 0.2 mole

Chemical reactions

Vinegar and baking soda

CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂

In words: Acetic acid and sodium hydrogencarbonate make sodium acetate, water and carbon dioxide.

When you pour vinegar on baking soda it fizzes, because carbon dioxide gas is released. An acid reacts with a hydrogencarbonate to make a salt, water and carbon dioxide.

The substances

CH₃COOH

acetic acid, the acid in vinegar

NaHCO₃

sodium hydrogencarbonate (baking soda)

CH₃COONa

sodium acetate, a salt

H₂O

water

CO₂

carbon dioxide, the gas that fizzes

Worked examples

1. Example 1. How many moles of CO₂ form from 0.5 mole of baking soda?

   1. 1 mol NaHCO₃ gives 1 mol CO₂

   Answer: 0.5 mole

Laws of chemical combination

Law of conservation of mass

mass of reactants = mass of products

In words: In a chemical reaction the total mass stays the same.

Atoms are not created or destroyed in a chemical reaction. They only join in new ways, so the total mass before the reaction equals the total mass after it. This is why every chemical equation must be balanced.

What the letters mean

reactants

the substances that react (left of the arrow)

products

the substances that form (right of the arrow)

Worked examples

1. Example 1. 12 g of carbon burns completely in 32 g of oxygen. What mass of carbon dioxide forms?

   1. Mass before = 12 + 32
   2. Mass after must be the same

   Answer: 44 g

2. Example 2. 2 g of hydrogen combines completely with 16 g of oxygen. What mass of water forms?

   1. 2 + 16

   Answer: 18 g

Laws of chemical combination

Law of constant proportions

In water, H : O = 1 : 8 by mass

In words: A compound always contains the same elements in the same ratio by mass.

Pure water always has hydrogen and oxygen in the mass ratio 1 : 8, whether it comes from a tap or a river. The ratio comes from the formula: H₂O has 2 g of hydrogen and 16 g of oxygen in every 18 g.

What the letters mean

H : O

ratio of the masses of hydrogen and oxygen in water

Worked examples

1. Example 1. How many grams of hydrogen are in 36 g of water?

   1. Hydrogen is 2 parts in 18 of water
   2. 36 × 2 ÷ 18

   Answer: 4 g

2. Example 2. How many grams of oxygen are in 9 g of water?

   1. Oxygen is 16 parts in 18 of water
   2. 9 × 16 ÷ 18

   Answer: 8 g

Mole concept

Moles from mass

n = m ÷ M

In words: Number of moles equals mass divided by molar mass.

A mole is a fixed number of particles, 6.022 × 10²³, just as a dozen is 12. The molar mass M is the mass of one mole in grams, and it equals the formula mass written in grams. Dividing a mass by M tells you how many moles you have.

What the letters mean

n

amount of substance, in moles (mol)

m

mass, in g

M

molar mass, in g/mol

Worked examples

1. Example 1. How many moles are in 36 g of water? (M = 18 g/mol)

   1. n = m ÷ M = 36 ÷ 18

   Answer: 2 mol

2. Example 2. How many moles are in 11 g of carbon dioxide? (M = 44 g/mol)

   1. n = 11 ÷ 44

   Answer: 0.25 mol

Mole concept

Number of particles

N = n × N_A

In words: Number of particles equals moles times Avogadro's number.

One mole of any substance contains 6.022 × 10²³ particles (atoms, molecules or ions). This number is Avogadro's number. Multiply the moles by it to count the particles.

What the letters mean

N

number of particles

n

amount, in moles

N_A

Avogadro's number, 6.022 × 10^{23} per mole

Worked examples

1. Example 1. How many molecules are there in 0.5 mol of water?

   1. N = 0.5 × 6.022 × 10²³

   Answer: 3.011 × 10²³ molecules

2. Example 2. How many molecules are there in 2 mol of oxygen gas?

   1. N = 2 × 6.022 × 10²³

   Answer: 1.2044 × 10²⁴ molecules

Mole concept

Volume of a gas

V = n × 22.4 L

In words: The volume of a gas at STP equals moles times 22.4 litres.

At standard temperature and pressure (STP: 0 °C and 1 atm), one mole of any gas takes up 22.4 litres. Equal volumes of different gases at the same temperature and pressure hold the same number of molecules (Avogadro's law).

What the letters mean

V

volume of the gas at STP, in litres

n

amount of gas, in moles

Worked examples

1. Example 1. What volume do 2 mol of a gas occupy at STP?

   1. V = 2 × 22.4

   Answer: 44.8 L

2. Example 2. How many moles are in 5.6 L of a gas at STP?

   1. n = V ÷ 22.4 = 5.6 ÷ 22.4

   Answer: 0.25 mol

Watch out: Some books now define STP as 0 °C and 1 bar, where the molar volume is about 22.7 L. Use the value your book gives.

Mole concept

Percentage of an element in a compound

% of element = (mass of the element in one mole ÷ molar mass) × 100

In words: The percentage of an element is its share of the molar mass, times one hundred.

From a formula you can find how much of a compound's mass is a certain element. Add up the atomic masses of that element in the formula, divide by the molar mass of the compound and multiply by 100.

What the letters mean

mass of the element

atomic mass × number of its atoms in the formula

molar mass

mass of one mole of the compound

Worked examples

1. Example 1. Find the percentage of hydrogen in water (H₂O = 18).

   1. Hydrogen: 2 × 1 = 2
   2. 2 ÷ 18 × 100

   Answer: about 11.1%

2. Example 2. Find the percentage of calcium in calcium carbonate (CaCO₃ = 100).

   1. Calcium: 40
   2. 40 ÷ 100 × 100

   Answer: 40%

Mole concept

Reacting amounts (stoichiometry)

n_B = n_A × (coefficient of B ÷ coefficient of A)

In words: Moles of B equal moles of A times the ratio of their coefficients.

The numbers in front of the substances in a balanced equation (coefficients) give the ratio in which they react, in moles. To solve a problem: change the given mass to moles, use the ratio to find the moles of the other substance, then change to mass or volume.

What the letters mean

n_A

moles of the substance you know

n_B

moles of the substance you want

coefficient

the number written in front of a substance in the balanced equation

Worked examples

1. Example 1. CaCO₃ → CaO + CO₂. What mass of CO₂ forms when 10 g of CaCO₃ decomposes? (CaCO₃ = 100, CO₂ = 44 g/mol)

   1. Moles of CaCO₃ = 10 ÷ 100 = 0.1
   2. 1 : 1 ratio, so 0.1 mol CO₂
   3. Mass = 0.1 × 44

   Answer: 4.4 g

2. Example 2. 2H₂ + O₂ → 2H₂O. What mass of water forms from 4 g of hydrogen? (H₂ = 2, H₂O = 18 g/mol)

   1. Moles of H₂ = 4 ÷ 2 = 2
   2. 2 : 2 ratio, so 2 mol H₂O
   3. Mass = 2 × 18

   Answer: 36 g

Watch out: Always balance the equation first. The ratio comes from the coefficients, not from the masses.

Acids, bases and salts

The pH scale

pH = −log [H⁺] pH + pOH = 14

In words: pH is minus the logarithm of the hydrogen ion concentration; at 25 degrees Celsius pH plus pOH equals 14.

pH measures how acidic or basic a solution is. Below 7 is acidic, 7 is neutral and above 7 is basic. Each step of 1 on the scale means ten times more or fewer hydrogen ions. The sum pH + pOH = 14 holds at 25 °C.

What the letters mean

[H⁺]

concentration of hydrogen ions, in mol/L

pH

acidity: low means acidic, high means basic

pOH

the same measure for hydroxide ions

Worked examples

1. Example 1. Find the pH of a solution with [H⁺] = 1 × 10⁻³ mol/L.

   1. pH = −log (10⁻³)

   Answer: pH = 3 (acidic)

2. Example 2. A solution has pH 9. Find its pOH at 25 °C.

   1. pOH = 14 − pH = 14 − 9

   Answer: pOH = 5 (basic)

Chemical reactions

Burning of methane (combustion)

CH₄ + 2O₂ → CO₂ + 2H₂O

In words: One methane and two oxygen make one carbon dioxide and two water.

Methane, the main part of natural gas, burns in oxygen and gives out heat and light. Burning a fuel is called combustion, and it is an exothermic reaction, which means it releases heat.

The substances

CH₄

methane, the fuel

O₂

oxygen

CO₂

carbon dioxide

H₂O

water (as steam)

Worked examples

1. Example 1. How many moles of oxygen are needed to burn 2 moles of methane?

   1. The equation shows 2 O₂ for each CH₄
   2. 2 × 2

   Answer: 4 moles

Chemical reactions

Formation of water

2H₂ + O₂ → 2H₂O

In words: Two hydrogen and one oxygen make two water.

Hydrogen burns in oxygen to make only water, which is why it is seen as a clean fuel. It is a combination reaction and releases a lot of heat.

The substances

H₂

hydrogen gas

O₂

oxygen gas

H₂O

water

Worked examples

1. Example 1. What mass of water forms when 4 g of hydrogen burns completely? (H₂ = 2, H₂O = 18 g/mol)

   1. Moles of H₂ = 4 ÷ 2 = 2
   2. 2 mol H₂ give 2 mol H₂O
   3. Mass = 2 × 18

   Answer: 36 g

Chemical reactions

Electrolysis of water (decomposition)

2H₂O → 2H₂ + O₂

In words: Two water split into two hydrogen and one oxygen.

Passing electricity through water splits it into hydrogen and oxygen. A reaction in which one substance breaks into two or more simpler ones is a decomposition reaction. Twice as much hydrogen as oxygen is collected by volume.

The substances

H₂O

water, with a little acid added so that it conducts

H₂

hydrogen, collected at the negative electrode

O₂

oxygen, collected at the positive electrode

Worked examples

1. Example 1. What volume of hydrogen at STP forms from 36 g of water? (H₂O = 18 g/mol)

   1. Moles of water = 36 ÷ 18 = 2
   2. 2 mol H₂O give 2 mol H₂
   3. V = 2 × 22.4

   Answer: 44.8 L

Chemical reactions

Heating limestone (thermal decomposition)

CaCO₃ → CaO + CO₂

In words: Calcium carbonate breaks down into calcium oxide and carbon dioxide.

Strong heating breaks limestone into quicklime and carbon dioxide gas. This is how quicklime is made for cement. A decomposition that needs heat is called thermal decomposition.

The substances

CaCO₃

calcium carbonate (limestone)

CaO

calcium oxide (quicklime)

CO₂

carbon dioxide gas

Worked examples

1. Example 1. What mass of quicklime forms from 100 g of limestone? (CaCO₃ = 100, CaO = 56 g/mol)

   1. Moles of CaCO₃ = 100 ÷ 100 = 1
   2. 1 : 1 ratio, so 1 mol CaO
   3. Mass = 1 × 56

   Answer: 56 g

Chemical reactions

Slaking of lime

CaO + H₂O → Ca(OH)₂

In words: Calcium oxide and water make calcium hydroxide.

Quicklime reacts violently with water, giving out a lot of heat, and turns into slaked lime. The reaction is exothermic. Slaked lime is used for whitewashing walls.

The substances

CaO

calcium oxide (quicklime)

H₂O

water

Ca(OH)₂

calcium hydroxide (slaked lime)

Worked examples

1. Example 1. What mass of slaked lime forms from 56 g of quicklime? (CaO = 56, Ca(OH)₂ = 74 g/mol)

   1. Moles of CaO = 56 ÷ 56 = 1
   2. 1 : 1 ratio, so 1 mol Ca(OH)₂
   3. Mass = 1 × 74

   Answer: 74 g

Chemical reactions

Iron displaces copper (displacement)

Fe + CuSO₄ → FeSO₄ + Cu

In words: Iron and copper sulphate make iron sulphate and copper.

A more reactive metal pushes a less reactive metal out of its salt solution. An iron nail dipped in blue copper sulphate solution turns brown as copper settles on it, and the blue colour fades. This is a displacement reaction.

The substances

Fe

iron, the more reactive metal

CuSO₄

copper sulphate, a blue solution

FeSO₄

iron sulphate, a pale green solution

Cu

copper, the brown metal that settles

Worked examples

1. Example 1. What mass of copper is displaced by 5.6 g of iron? (Fe = 56, Cu = 63.5 g/mol)

   1. Moles of Fe = 5.6 ÷ 56 = 0.1
   2. 1 : 1 ratio, so 0.1 mol Cu
   3. Mass = 0.1 × 63.5

   Answer: 6.35 g

Chemical reactions

Zinc and hydrochloric acid

Zn + 2HCl → ZnCl₂ + H₂

In words: Zinc and two hydrochloric acid make zinc chloride and hydrogen.

Metals above hydrogen in the reactivity series react with dilute acids and give off hydrogen gas. The test for hydrogen is that it burns with a squeaky pop when a flame is brought near.

The substances

Zn

zinc

HCl

hydrochloric acid

ZnCl₂

zinc chloride, a salt

H₂

hydrogen gas

Worked examples

1. Example 1. What volume of hydrogen at STP forms when 6.5 g of zinc reacts with excess acid? (Zn = 65)

   1. Moles of Zn = 6.5 ÷ 65 = 0.1
   2. 1 : 1 ratio, so 0.1 mol H₂
   3. V = 0.1 × 22.4

   Answer: 2.24 L

Chemical reactions

Barium sulphate precipitate (double displacement)

Na₂SO₄ + BaCl₂ → BaSO₄ + 2NaCl

In words: Sodium sulphate and barium chloride make barium sulphate and two sodium chloride.

When two solutions are mixed, the ions swap partners. Here a white solid, barium sulphate, comes out of the solution as a precipitate. Reactions in which the ions of two compounds exchange places are double displacement reactions.

The substances

Na₂SO₄

sodium sulphate solution

BaCl₂

barium chloride solution

BaSO₄

barium sulphate, the white precipitate

NaCl

sodium chloride, which stays dissolved

Worked examples

1. Example 1. What mass of precipitate forms from 0.1 mole of BaCl₂ with excess sodium sulphate? (BaSO₄ = 233 g/mol)

   1. 1 : 1 ratio, so 0.1 mol BaSO₄
   2. Mass = 0.1 × 233

   Answer: 23.3 g

Chemical reactions

Silver chloride precipitate

AgNO₃ + NaCl → AgCl + NaNO₃

In words: Silver nitrate and sodium chloride make silver chloride and sodium nitrate.

A white, curdy precipitate of silver chloride shows that chloride ions are present. It turns grey in sunlight, which is the basis of old photography. It is another double displacement reaction.

The substances

AgNO₃

silver nitrate solution

NaCl

sodium chloride solution

AgCl

silver chloride, the white precipitate

NaNO₃

sodium nitrate, which stays dissolved

Worked examples

1. Example 1. What mass of silver chloride forms from 0.01 mole of NaCl with excess silver nitrate? (AgCl = 143.5 g/mol)

   1. 1 : 1 ratio, so 0.01 mol AgCl
   2. Mass = 0.01 × 143.5

   Answer: 1.435 g

Chemical reactions

Baking soda and an acid

NaHCO₃ + HCl → NaCl + H₂O + CO₂

In words: Sodium hydrogencarbonate and hydrochloric acid make salt, water and carbon dioxide.

Baking soda neutralises acids and releases carbon dioxide, which is why it works as an antacid and why it makes cakes rise. The same gas is used in soda-acid fire extinguishers.

The substances

NaHCO₃

sodium hydrogencarbonate (baking soda)

HCl

hydrochloric acid

NaCl

sodium chloride

H₂O

water

CO₂

carbon dioxide gas

Worked examples

1. Example 1. What volume of CO₂ at STP forms from 8.4 g of baking soda with excess acid? (NaHCO₃ = 84 g/mol)

   1. Moles of NaHCO₃ = 8.4 ÷ 84 = 0.1
   2. 1 : 1 ratio, so 0.1 mol CO₂
   3. V = 0.1 × 22.4

   Answer: 2.24 L

Chemical reactions

Breakdown of hydrogen peroxide

2H₂O₂ → 2H₂O + O₂

In words: Two hydrogen peroxide break into two water and one oxygen.

Hydrogen peroxide slowly breaks down into water and oxygen. A catalyst such as manganese dioxide makes it fizz quickly. A catalyst speeds up a reaction without being used up.

The substances

H₂O₂

hydrogen peroxide

H₂O

water

O₂

oxygen gas

Worked examples

1. Example 1. What volume of oxygen at STP forms from 68 g of hydrogen peroxide? (H₂O₂ = 34 g/mol)

   1. Moles of H₂O₂ = 68 ÷ 34 = 2
   2. 2 mol H₂O₂ give 1 mol O₂
   3. V = 1 × 22.4

   Answer: 22.4 L

Chemical reactions

Heating potassium chlorate

2KClO₃ → 2KCl + 3O₂

In words: Two potassium chlorate give two potassium chloride and three oxygen.

On heating, with manganese dioxide as a catalyst, potassium chlorate gives off oxygen. It is a common way to prepare oxygen in the laboratory.

The substances

KClO₃

potassium chlorate

KCl

potassium chloride

O₂

oxygen gas

Worked examples

1. Example 1. How many moles of oxygen form from 2 moles of potassium chlorate?

   1. The equation shows 3 O₂ for every 2 KClO₃

   Answer: 3 moles

Chemical reactions

Burning of ethanol

C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O

In words: One ethanol and three oxygen make two carbon dioxide and three water.

Like other carbon compounds, ethanol burns in plenty of air to give carbon dioxide and water, and it releases heat. This is why ethanol is blended with petrol as a fuel.

The substances

C₂H₅OH

ethanol

O₂

oxygen

CO₂

carbon dioxide

H₂O

water

Worked examples

1. Example 1. What mass of carbon dioxide forms when 1 mole of ethanol burns? (CO₂ = 44 g/mol)

   1. 1 mol ethanol gives 2 mol CO₂
   2. Mass = 2 × 44

   Answer: 88 g

Chemical reactions

Esterification

CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O

In words: Acetic acid and ethanol make ethyl ethanoate and water.

A carboxylic acid and an alcohol, warmed with a little concentrated sulphuric acid as a catalyst, make an ester and water. Esters have sweet, fruity smells and are used in perfumes and flavours. The double arrow shows the reaction can go both ways.

The substances

CH₃COOH

acetic acid (ethanoic acid)

C₂H₅OH

ethanol

CH₃COOC₂H₅

ethyl ethanoate, a sweet-smelling ester

H₂O

water

Worked examples

1. Example 1. If 1 mole of acid and 1 mole of ethanol react completely, what mass of ester forms? (ethyl ethanoate = 88 g/mol)

   1. 1 : 1 : 1 ratio, so 1 mol ester
   2. Mass = 1 × 88

   Answer: 88 g

Chemical reactions

Addition of hydrogen to ethene

C₂H₄ + H₂ → C₂H₆

In words: Ethene and hydrogen make ethane.

Unsaturated hydrocarbons, which have a double bond, can add hydrogen when heated with a nickel catalyst. This is an addition reaction. The same idea is used to turn vegetable oils into solid fats.

The substances

C₂H₄

ethene, an unsaturated hydrocarbon

H₂

hydrogen

C₂H₆

ethane, a saturated hydrocarbon

Worked examples

1. Example 1. What mass of hydrogen is needed to add to 28 g of ethene? (C₂H₄ = 28, H₂ = 2 g/mol)

   1. Moles of ethene = 28 ÷ 28 = 1
   2. 1 : 1 ratio, so 1 mol H₂
   3. Mass = 1 × 2

   Answer: 2 g

Chemical reactions

Chlorination of methane (substitution)

CH₄ + Cl₂ → CH₃Cl + HCl

In words: Methane and chlorine make chloromethane and hydrogen chloride.

In sunlight, a hydrogen atom of a saturated hydrocarbon is replaced by a chlorine atom. This is a substitution reaction. Saturated hydrocarbons mostly burn or take part in substitution reactions.

The substances

CH₄

methane

Cl₂

chlorine

CH₃Cl

chloromethane

HCl

hydrogen chloride

Worked examples

1. Example 1. What mass of chloromethane forms from 1 mole of methane? (CH₃Cl = 50.5 g/mol)

   1. 1 : 1 ratio, so 1 mol CH₃Cl
   2. Mass = 1 × 50.5

   Answer: 50.5 g

Gases and solutions

Ideal gas equation

PV = nRT

In words: Pressure times volume equals moles times the gas constant times temperature.

It links the pressure, volume, amount and temperature of a gas. Temperature must be in kelvin (add 273 to the Celsius value). The value of R depends on the units: 0.0821 L atm/(mol K), or 8.314 J/(mol K).

What the letters mean

P

pressure, in atm

V

volume, in litres

n

amount of gas, in moles

R

gas constant, 0.0821 L atm mol^{-1} K^{-1}

T

temperature, in kelvin (K)

Worked examples

1. Example 1. Find the volume of 1 mol of gas at 273 K and 1 atm. (R = 0.0821)

   1. V = nRT ÷ P = 1 × 0.0821 × 273 ÷ 1

   Answer: about 22.4 L

2. Example 2. 2 mol of gas are in a 10 L vessel at 300 K. Find the pressure. (R = 0.0821)

   1. P = nRT ÷ V = 2 × 0.0821 × 300 ÷ 10

   Answer: about 4.93 atm

Gases and solutions

Boyle's law

P₁V₁ = P₂V₂

In words: At constant temperature, pressure times volume stays the same.

If you squeeze a gas into a smaller space without changing its temperature, its pressure goes up in the same proportion. Halving the volume doubles the pressure.

What the letters mean

P₁, V₁

pressure and volume at the start

P₂, V₂

pressure and volume at the end

Worked examples

1. Example 1. A gas of volume 4 L at 1 atm is compressed to 2 L at the same temperature. Find the new pressure.

   1. P₂ = P₁V₁ ÷ V₂ = 1 × 4 ÷ 2

   Answer: 2 atm

Gases and solutions

Charles's law

V₁ ÷ T₁ = V₂ ÷ T₂

In words: At constant pressure, volume divided by temperature stays the same.

A gas expands when heated at constant pressure. The volume is proportional to the temperature measured in kelvin, not in degrees Celsius.

What the letters mean

V₁, T₁

volume and temperature (in kelvin) at the start

V₂, T₂

volume and temperature (in kelvin) at the end

Worked examples

1. Example 1. 300 mL of gas at 300 K is heated to 360 K at constant pressure. Find the new volume.

   1. V₂ = V₁ × T₂ ÷ T₁ = 300 × 360 ÷ 300

   Answer: 360 mL

Watch out: Always change °C to K by adding 273 before using the gas laws.

Gases and solutions

Molarity

M = n ÷ V

In words: Molarity equals moles of solute divided by litres of solution.

Molarity is the concentration of a solution: moles of dissolved substance in each litre of solution. A 2 M solution has 2 moles in every litre.

What the letters mean

M

molarity, in mol/L

n

moles of the dissolved substance (the solute)

V

volume of the solution, in litres

Worked examples

1. Example 1. 0.5 mol of a substance is dissolved to make 250 mL of solution. Find the molarity.

   1. V = 250 mL = 0.25 L
   2. M = 0.5 ÷ 0.25

   Answer: 2 M

2. Example 2. 4 g of NaOH (M = 40 g/mol) is dissolved to make 500 mL of solution. Find the molarity.

   1. n = 4 ÷ 40 = 0.1 mol
   2. M = 0.1 ÷ 0.5

   Answer: 0.2 M

Gases and solutions

Dilution

M₁V₁ = M₂V₂

In words: Molarity times volume before dilution equals molarity times volume after.

Adding water to a solution does not change the amount of solute, only the volume. So moles before (M₁V₁) equal moles after (M₂V₂).

What the letters mean

M₁, V₁

molarity and volume of the concentrated solution

M₂, V₂

molarity and volume of the diluted solution

Worked examples

1. Example 1. 50 mL of a 2 M solution is diluted to 0.5 M. What is the final volume?

   1. V₂ = M₁V₁ ÷ M₂ = 2 × 50 ÷ 0.5

   Answer: 200 mL

Energy and equilibrium

Gibbs energy and spontaneity

ΔG = ΔH − TΔS

In words: The change in Gibbs energy equals the change in enthalpy minus temperature times the change in entropy.

A reaction is spontaneous (it can happen by itself) when ΔG is negative. ΔH is the heat change and ΔS the change in disorder. Temperature is in kelvin, and ΔH and TΔS must use the same energy unit.

What the letters mean

ΔG

change in Gibbs energy, in kJ

ΔH

change in enthalpy (heat), in kJ

T

temperature, in K

ΔS

change in entropy, in kJ/K

Worked examples

1. Example 1. For a reaction ΔH = −90 kJ and ΔS = −0.2 kJ/K. Find ΔG at 300 K and say whether it is spontaneous.

   1. ΔG = −90 − 300 × (−0.2)
   2. ΔG = −90 + 60

   Answer: −30 kJ, so it is spontaneous

Energy and equilibrium

Equilibrium constant

For aA + bB ⇌ cC + dD: K_c = [C]^c[D]^d ÷ [A]^a[B]^b

In words: The equilibrium constant is the product of the concentrations of the products over the product of the reactants, each raised to its coefficient.

At equilibrium the forward and backward reactions go at the same rate and the concentrations stop changing. The ratio Kc is fixed at a given temperature. A large Kc means the products are favoured; a small Kc means the reactants are.

What the letters mean

[A], [B], [C], [D]

concentrations at equilibrium, in mol/L

a, b, c, d

coefficients in the balanced equation

K_c

the equilibrium constant

Worked examples

1. Example 1. For N₂ + 3H₂ ⇌ 2NH₃ the equilibrium concentrations are [N₂] = 1, [H₂] = 2 and [NH₃] = 4 mol/L. Find Kc.

   1. Kc = [NH₃]² ÷ ([N₂][H₂]³)
   2. Kc = 16 ÷ (1 × 8)

   Answer: 2

Watch out: Pure solids and liquids are left out of the expression.

Rates of reaction

Rate law and half-life

rate = k[A]^m[B]ⁿ t_½ = 0.693 ÷ k (first order)

In words: The rate equals the rate constant times the concentrations raised to their orders; for a first-order reaction the half-life is 0.693 divided by k.

The rate law says how the speed of a reaction depends on concentration. The orders m and n are found by experiment, not from the equation. For a first-order reaction, the half-life (time to halve the amount) does not depend on the starting amount.

What the letters mean

rate

speed of the reaction, in mol L^{-1} s^{-1}

k

rate constant

[A], [B]

concentrations, in mol/L

m, n

orders of reaction with respect to A and B

t_½

half-life, in s

Worked examples

1. Example 1. For rate = k[A][B], k = 0.5, [A] = 0.2 and [B] = 0.1. Find the rate.

   1. rate = 0.5 × 0.2 × 0.1

   Answer: 0.01 mol/(L s)

2. Example 2. A first-order reaction has k = 0.0693 per second. Find its half-life.

   1. t½ = 0.693 ÷ 0.0693

   Answer: 10 s

Electrochemistry

Faraday's law of electrolysis

m = M I t ÷ (z F)

In words: The mass deposited equals molar mass times current times time, divided by the charge per ion times the Faraday constant.

The mass of a substance produced at an electrode is proportional to the electric charge passed (I × t). The Faraday constant F is the charge of one mole of electrons, about 96500 C. The number z is how many electrons each ion needs.

What the letters mean

m

mass deposited, in g

M

molar mass of the substance, in g/mol

I

current, in A

t

time, in s

z

electrons needed per ion (for Cu^{2+}, z = 2)

F

Faraday constant, 96500 C/mol

Worked examples

1. Example 1. A current of 2 A flows for 965 s through copper sulphate solution. Find the mass of copper deposited. (Cu = 63.5)

   1. m = 63.5 × 2 × 965 ÷ (2 × 96500)

   Answer: 0.635 g

Electrochemistry

Cell potential

E°_cell = E°_cathode − E°_anode

In words: The standard cell potential equals the cathode potential minus the anode potential.

A galvanic cell turns chemical energy into electrical energy. The electrode with the higher (more positive) reduction potential is the cathode, where reduction happens. The cell voltage is the difference between the two potentials.

What the letters mean

E°_cell

standard voltage of the cell, in volts

E°_cathode

standard reduction potential at the cathode

E°_anode

standard reduction potential at the anode

Worked examples

1. Example 1. In a zinc–copper cell, E° for Cu²⁺/Cu is +0.34 V and for Zn²⁺/Zn is −0.76 V. Find the cell voltage.

   1. Copper has the higher potential, so it is the cathode
   2. E° = 0.34 − (−0.76)

   Answer: 1.10 V

Chemical reactions

Haber process (reversible reaction)

N₂ + 3H₂ ⇌ 2NH₃

In words: One nitrogen and three hydrogen are in balance with two ammonia.

Ammonia is made industrially from nitrogen and hydrogen at high pressure with an iron catalyst. The reaction is reversible, and it is exothermic. By Le Chatelier's principle, high pressure favours ammonia because there are fewer gas molecules on the right.

The substances

N₂

nitrogen, from the air

H₂

hydrogen, from natural gas

NH₃

ammonia

Worked examples

1. Example 1. How many moles of ammonia can form from 6 moles of hydrogen?

   1. The equation shows 2 NH₃ for every 3 H₂
   2. 6 × 2 ÷ 3

   Answer: 4 moles

Chemical reactions

Burning of propane

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

In words: One propane and five oxygen make three carbon dioxide and four water.

Propane, a main part of LPG cooking gas, burns completely in enough oxygen. In a shortage of air the flame turns yellow and poisonous carbon monoxide forms instead.

The substances

C₃H₈

propane

O₂

oxygen

CO₂

carbon dioxide

H₂O

water

Worked examples

1. Example 1. What mass of CO₂ forms when 11 g of propane burns? (C₃H₈ = 44, CO₂ = 44 g/mol)

   1. Moles of propane = 11 ÷ 44 = 0.25
   2. 1 mol propane gives 3 mol CO₂, so 0.75 mol
   3. Mass = 0.75 × 44

   Answer: 33 g

Chemical reactions

Fermentation of glucose

C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

In words: One glucose makes two ethanol and two carbon dioxide.

Enzymes in yeast turn sugar into ethanol and carbon dioxide when there is no oxygen. This is the process behind brewing, and the gas makes bread dough rise.

The substances

C₆H₁₂O₆

glucose

C₂H₅OH

ethanol

CO₂

carbon dioxide

Worked examples

1. Example 1. What mass of ethanol forms from 180 g of glucose? (glucose = 180, ethanol = 46 g/mol)

   1. Moles of glucose = 180 ÷ 180 = 1
   2. 1 mol glucose gives 2 mol ethanol
   3. Mass = 2 × 46

   Answer: 92 g

Chemical reactions

Sodium burning in chlorine (redox)

2Na + Cl₂ → 2NaCl

In words: Two sodium and one chlorine make two sodium chloride.

Sodium gives away an electron (it is oxidised) and chlorine takes it (it is reduced). A reaction in which electrons move from one substance to another is a redox reaction. The result is the ionic compound in table salt.

The substances

Na

sodium metal

Cl₂

chlorine gas

NaCl

sodium chloride

Worked examples

1. Example 1. What mass of sodium chloride forms from 46 g of sodium? (Na = 23, NaCl = 58.5 g/mol)

   1. Moles of Na = 46 ÷ 23 = 2
   2. 2 mol Na give 2 mol NaCl
   3. Mass = 2 × 58.5

   Answer: 117 g