Edexcel IAL Chemistry Unit 1 - Topic 1: Formulae, Equations and Amount of Substance

Define atom

The smallest part of an element that has the properties of that element

Define element

A substance made of only one type of atom

Define ion

A species consisting of one or more atoms joined together and have a positive or negative charge.

Define molecule

A particle made of two or more atoms bonded together

Define compound

A substance containing atoms of different elements combined together

Define empirical formula

The simplest whole number ratio of atoms of each element present in a compound

Define molecular formula

The actual number of atoms of each element in a molecule

What is a mole?

The amount of substance that contains the same number of particles as the number of carbon atoms in exactly 12g of the carbon-12 isotope.

Amount of substance = n

Calculating using moles

The mole is used to count atoms, molecules, ions and electrons.

Amount of substance (mol)= mass(g) / molar mass

What is the amount of substance in 6.51g of sodium chloride?

n = 6.51 / 58.5

= 0.111 mol

Avogadro's constant L

L = 6.022 x 10²³ mol-1

The number of particles in one mole of any substance

Calculate the number of particles in a given mass of a substance

1. Amount (mol) = Mass of Substance (g) / Molar mass (g/mol)

2. Number of Particles = Amount (mol) x L

How many H2O molecules are there in 1.25g of water?

n = 1.25 / 18

= 0.0694 mol

number of molecules = 6.02 x 10²³ x 0.0694 = 4.18 x 10^22

Calculate the mass of a given number of particles of a substance

1. Divide the number of particles by the Avogadro constant

2. Multiply by molar mass

What is the mass of 100 million atoms of gold?

n = (100x10^6) / (6.02x10²³) = 1.66 x 10^-16 mol

m = 1.66 x 10^-16 x 197 = 3.27 x 10^-14

Diatomic molecules

H2, N2, O2, F2, Cl2, Br2, I2

Have No Fear Of Ice Cold Beer

Sodium hydroxide

NaOH

Nitric acid

HNO3

Iron (II) sulphate

FeSO4

Iron (III) oxide

Fe2O3

Calcium carbonate

CaCO3

Balanced full and ionic equations

1. Convert words into formulae and decide which are reactants and which are products

2. Balance the equation. Add up the numbers of all the atoms to make sure that, for each element, the totals are the same on both the left and right side of the equation

Simplifying full equations

1. Start with the full equation for the reaction

2. Replace the formulae of ionic compounds by their separate ions

3. Delete any ions that appear identically on both sides

What is the simplest ionic equation for the neutralisation of sodium hydroxide solution by nitric acid?

1. NaOH(aq) + HNO3(aq) -> NaNO3(aq) + H2O(l)

2. Na+(aq) + OH-(aq) + H+(aq) + NO3-(aq) -> Na+(aq) + NO3-(aq) + H2O(l)

3. H+(aq) + OH-(aq) -> H2O(l)

Ionic half equations

Written for reactions involving oxidation and reduction, and they usually show what happens to only one reactant.

State symbols

solid (s)

liquid (l)

gas (g)

aqueous (aq)

Define relative atomic mass (Ar)

The average mass of an atom compared to 1/12 the mass of an atom of carbon-12

Define relative molecular mass (Mr)

The average mass of a molecule compared to one twelfth of the mass of one atom of carbon-12.

What is the relative molecular mass of sulphuric acid, H2SO4?

Mr = (2x1) + 32.1 + (4x16)

Mr = 98.1

Define relative formula mass

The sum of the relative atomic masses of all the atoms shown in the formula.

What is the relative formula mass of hydrated copper (II) sulphate, CuSO4.5H2O?

Mr = 63.5 + 32.1 + (4x16) + 5((2x1)+16)

Mr = 249.6

Molar mass

The mass per mole of a substance

Amount (mol = Mass of substance (g) / Molar mass (g mol-1)

Concentration (ppm)

(Mass of Solute x 10^6) / Mass of Solvent

A mass of 23 mg of sodium chloride is dissolved in 900g of water. What is the concentration of sodium chloride in the solution in ppm?

Mass of sodium chloride = 23/1000 = 0.023 g

Concentration = (0.023x1000000)/900 = 26 ppm

Concentration (ppm)

(Volume of Gas x 1000000) / Volume of Air

5000 dm³ of air is found to contain ozone with a concentration of 87 ppm. What volume of ozone is in this sample of air?

Volume of gas = (87x5000)/1000000 = 0.435 dm³

Mass Concentration (mol/dm³)

Mass of Solute (g) / Volume of Solution (dm³)

A chemistry uses 280g of a solute to make a solution of concentration 28.4 g/dm³. What volume of solution does he make?

V = Mass of solute / Mass concentration

280 / 28.4 = 9.86 dm³

Calculations using molar concentration

Amount = Mass / Molar mass

Molar concentration = Amount / Volume

A chemist makes 500 cm³ of a solution of nitric acid of concentration 0.800 mol/dm³. What mass of HNO3 does she need?

n = c x v

0.800 x 0.500 = 0.400 mol

m = n x M

0.400 x 63.0 = 25.2 g

Concentration (mol/dm³)

Number of Moles (mol) / Volume (dm³)

Concentration (g dm-3)

Mass of solute (g) / volume of solution (dm³)

Calculations from equations using concentration and mass

Use an equation to calculate the mass of a reactant or product if given the volume and molar concentration of another substance.

A mass of 47.8g of magnesium carbonate reacts with 2.50 mol/dm³ hydrochloric acid. What volume of acid is needed?

MgCO3 + 2HCl -> MgCl2 + H2O + CO2

n = 47.8 / 84.3 = 0.567 mol

MgCO3 : HCl 1:2

n(HCl) = 2 x 0.567 = 1.134 mol

V(HCl) = 1.134 / 2.50 = 0.454 dm³

Determining the formula of an oxide of copper using experimental data

1. Place a known mass of the oxide of copper in the tube

2. Heat the oxide in a stream of hydrogen gas.

3. The gas reacts with the oxygen in the copper oxide and forms steam.

4. The colour of the solid gradually changes to orange-brown, which is the colour of copper

5. The excess gas is burned off at the end of the tube for safety reasons.

6. After cooling, remove and weigh the solid copper.

7. It is good practice to heat the solid again in the stream of the gas to check whether its mass changes. Heating to a constant mass suggests that the conversion to copper is complete.

Calculating empirical formula

1. Divide the mass of each element by its relative atomic mass

2. Divide the answers by the smallest of the numbers

Calculate the empirical formula for these results:

Mass of copper oxide = 4.28g

Mass of copper = 3.43g

Mass of oxygen removed = 4.28-3.43 = 0.85g

Mass of element (g) =

Cu - 3.43, O - 0.85

Relative atomic mass =

Cu - 63.5, O - 16.0

Divide by Ar =

Cu - 0.0540, O - 0.0531

Ratio =

Cu - 1, O - 1

CuO

The compound has the percentage composition by mass C=38.4%, H=4.8%, Cl=56.8%

Percentage of element =

C - 38.4, H - 4.8, Cl - 56.8

Ar =

C - 12.0, H - 1.0, Cl - 35.5

Divide by Ar =

C - 3.2, H - 4.8, Cl - 1.6

Ratio =

C - 2, H - 3, Cl - 1

C2H3Cl

A 1.87g sample of an organic compound was completely burned, forming 2.65g of carbon dioxide and 1.63g of water.

Mass of C = (2.65x12)/44 = 0.723g

Mass of H = (1.63x2)/18 = 0.181g

Mass of O = 18.7-(0.723+0.181) = 0.966g

Mass of element =

C - 0.723, H - 0.181, O - 0.966

Ar =

C - 12.0, H - 1.0, O - 16.0

Divide by Ar =

C - 0.0603, H - 0.181, O - 0.0604

Ratio =

C - 1, H - 3, O - 1

CH3O

A compound contains the percentage composition by mass Na=34.3%, C=17.9%, O=47.8% and has a molar mass of 134g/mol

Percentage of element=

Na - 34.3, C- 17.9, O- 47.8

Ar =

Na - 23.0, C- 12.0, O- 16.0

Divide by Ar =

Na - 1.49, C- 1.49, O- 2.99

Ratio =

Na - 1, C- 1, O- 2

Empirical formula = NaCO2

Formula mass = 67

134/67 = 2

Molecular formula = Na2C2O4

Ideal gas equation

PV=nRT

SI units

Pressure (Pa)

Volume (m³)

Temperature (K)

Amount of substance (mol)

Gas constant (8.31 J/mol/K)

A compound has the percentage composition by mass C=52.2%, H=13.0%, O=34.8%. A sample containing 0.173g of the compound had a volume of 95.0 cm³ when measured at 105kPA and 45°C. What is the molecular formula of this compound?

Percentage of element=

C - 52.2, H - 13.0, O - 34.8

Ar =

C - 12.0, H - 1.0, O - 16.0

Divide by Ar =

C - 4.35, H - 13.0, O - 2.175

Ratio =

C - 2, H - 6, O - 1

Empirical formula = C2H6O

n = pV / RT

(105x10^3 x 95x10^-6)/ 8.31x318 = 0.00377 mol

M = m / n

0.173/0.00377 = 45.9 g/mol

Formula mass = 46

Molecular formula = C2H6O

Calculating reacting masses from equations

1. Calculate the molar masses of all substances told about and asked amount

2. Calculate the amount of substance

3. Use the reaction ratio in the equation to work out the amount of substance needed

4. Calculate the mass of product

What mass of sulphur trioxide is needed to form 75.0 g of sulphuric acid?

SO3 + H2O -> H2SO4

Molar mass (SO3) = 80.1 g/mol

Molar mass (H2SO4) = 98.1 gm/mol

n = 75 / 98.1 = 0.765 mol

1:1 n(SO3) = 0.765 mol

m = 0.765 x 80.1 = 61.2 g

Working out formulae and equations from reacting masses

1. Calculate the molar masses of the relevant substances

2. Calculate the amounts of these substances

3. Use these amounts to calculate the simplest whole number ratio for these substances

4. Use this ratio to work out the equation for the reaction

A 16.7g sample of hydrate of sodium carbonate (Na2CO3 . 10H2O) is heated at a constant temperature for a specified time until the reaction is complete. A mass of 3.15 g of water is obtained. What is the equation for the reaction occurring?

M(Na2CO3 . 10H2O) = 286.1 g/mol

M (H2O) = 18.0 g/mol

Na2CO3 . 10H2O n = 16.7 / 286.1 = 0.0584 mol

H2O n = 3.15 / 18.0 = 0.175 mol

1:3

Na2CO3 . 10H2O -> Na2CO3 . 7H2O + 3H2O

Volume of molar volume

24 dm³/mol

At room temperature and standard pressure

Calculating molar volume from a single gas

Volume of molar volume = Volume (dm³) / Amount (mol)

What is the amount in moles of CO2 in 500 cm³ of carbon dioxide?

Vm = 500 / 24000 = 0.021

Calculating molar volume from gases, solids and liquids

1. Calculate the amount in moles from either the mass or the volume, depending on which one is given

2. Use the relevant reaction ratio in the equation to calculate the amount of the other substance

3. Convert this amount to a mass or a volume, depending on what the question asks

Theoretical yield

Always assume that the reaction goes to completion with no losses.

1. Calculate the amount of starting material

2. Use the reacting ratio to calculate the amount of desired product.

3. Calculate the mass of desired product

Actual yield

The actual mass obtained by weighing the product obtained.

Percentage yield

(Actual Yield/ Theoretical Yield) x 100

The closer the value is to 100%, the better.

A higher percentage means that less of the starting materials are lost or end up as unwanted products.

A manufacturer uses this reaction to obtain methanol from carbon monoxide and hydrogen. The manufacturer obtains 4.07 tonnes of methanol starting from 4.32 tonnes of carbon monoxide. What is the percentage yield?

CO + 2H2 -> CH3OH

n(CO) = (4.32x10^6) / 280 = 1.54x10^5 mol

n(CH3OH) - 1.54x10^5 mol

1:1

m = 1.54 x 10^5x32 = 4.94x10^6 mol

% yield = (4.07x10^6/4.94x10^6) x 100 = 82.4%

Assessing the suitability of an industrial process

Availability or scarcity of non-renewable raw materials

Cost of raw materials

Quantity of energy needed

Percentage yield

Percentage atom economy

(Molar mass of the desired product / Sum of the molar masses of all products) x 100

Reaction types and their atom economy

Addition reactions have 100% atom economy

Elimination and substitution reactions have lower atom economies

Multistep reactions may have even lower atom economies.

Acids with metals

Metal + Acid -> Salt + Hydrogen

Metal must be sufficiently reactive.

Bubbles of hydrogen gas are formed

If salt formed is soluble then a solution forms as well.

These may appear to be examples of neutralisation reactions but H+ ions gain electrons from the metal and are converted to H2(g) so H+ ions are reduced.

Acid with metals example

Mg + 2HCl -> MgCl2 + H2

Mg(s) + 2H+(aq) -> Mg2+(aq) + H2(g)

Acids with metal oxides

Metal Oxide + Acid -> Salt + Water

Formation of solution

Neutralisation reactions because H+ ions react with O2- ions.

Acids with metal oxides example

CuO + H2SO4 -> CuSO4 + H2O

CuO(s) + 2H+(aq) -> Cu2+(aq) + H2O(l)

Acids with insoluble metal hydroxides

Metal Hydroxide + Acid -> Salt + Water

Formation of solution

Neutralisation reactions because H+ ions react with OH- ions.

Acids with insoluble metal hydroxides example

Zn(OH)2 + H2SO4 -> ZnSO4 + 2H2O

Zn(OH)2(s) + 2H+(aq) -> Zn2+(aq) + 2H2O(l)

Acids with alkalis

Alkali + Acid -> Salt + Water

No visible changes except temperature rise.

Neutralisation reactions because H+ ions react with OH- ions.

Acids with alkalis example

NaOH + H3PO4 -> NaH2PO4 + H2O

H+(aq) + OH-(aq) -> H2O(l)

Acids with carbonates

Metal Carbonate + Acid -> Salt + Water + Carbon Dioxide

Bubbles of carbon dioxide gas

If salt formed is soluble, then a solution is formed.

Neutralisation reactions because the H+ ions react with CO32- ions.

Acids with carbonates example

Li2CO3 + 2HCl -> 2LiCl + H2O + CO2

CO32-(aq) + 2H2+(aq) -> H2O(l) + CO2(g)

Acids with hydrogencarbonates

Metal Hydrogencarbonate + Acid -> Salt + Water + Carbon Dioxide

Bubbles of carbon dioxide gas

If salt formed is soluble, then a solution is formed.

Neutralisation reactions because the H+ ions react with CO32- ions.

Hydrogencarbonate ion (HCO3-)

Acids with hydrogencarbonates example

Sodium hydrogen carbonate + Citric acid -> Sodium citrate + Water + Carbon dioxide

Test for presence of carbonate or hydrogen carbonate ions in solid or solution

Add aqueous acid

Test gas produced with limewater

Displacement reaction involving metals

1. Mg(s) + CuSO4(aq) -> Cu(s) + MgSO4(aq)

2. 2Al(s) + Fe2O3(s) -> 2Fe(s) + Al2O3(s)

Both reactions involve one metal reacting with compound of different metal

Both produce metal and different metal compound

Both are redox reactions

Metal element on reactant side has taken place of the metal in the metal compound on the reactants side.

Reaction 1 takes place in aqueous solution, Reaction involves only solids.

Reaction 1 occurs without the need for energy, Reaction 2 requires very high temperature to start.

Reaction 1 is done in laboratory, Reaction 2 is done for a specific purpose in industry.

Metal displacement reactions in aqueous solution

A more reactive metal displaces a less reactive metal from one of its salts.

Mg(s) + Cu2+(aq) + SO42-(aq) -> Cu(s) + Mg2+(aq) + SO42-(aq)

Mg(s) + Cu2+(aq) -> Cu(s) + Mg2+(aq)

When magnesium metal is added to copper (II) sulphate solution, the blue colour of the solution becomes paler.

When excess of magnesium is added, solution becomes colourless and magnesium sulphate forms.

Magnesium changes from silvery to brown as copper forms on it.

Redox reaction as electrons are transferred from magnesium atoms to copper (II) ions, magnesium atoms are oxidised and copper (II) ions are reduced.

Metal displacement reactions in solid state

2Al(l) + 2Fe3+(l) + 3O2-(l) -> 2Fe(l) + 3Al3+(l) + 3O2-(l)

2Al(l) + 2Fe3+(l) -> 2Fe(l) + 3Al3+(l)

Redox reaction because electrons are transferred from aluminium atoms to iron (III) ions, aluminium atoms are oxidised and iron (III) ions are reduced.

Displacement reactions involving halogens

More reactive halogens can displace less reactive halogens from their compounds.

Cl2(aq) + 2KBr(aq) -> Br2(aq) + 2KCl(aq)

Cl2(aq) + 2K+(aq) + 2Br-(aq) -> Br2(aq) + 2K+(aq) + 2Cl-(aq)

Cl2(aq) + 2Br-(aq) -> Br2(aq) + 2Cl-(aq)

Redox reaction because electrons are transferred from bromide ions to chlorine, so bromide ions are oxidised and chlorine is reduced.

Test for presence of sulphate ions in solution

Add barium chloride/nitrate

White precipitate formed is barium sulphate

Na2SO4(aq) + BaCl2(aq) -> BaSO4(s) + 2NaCl(aq)

SO42-(aq) + Ba2+(aq) -> BaSO4(s)

Test for presence of halide ions in solution

Add silver nitrate solution

Precipitate formed is silver halides

NaCl(aq) + AgNO3(aq) -> AgCl(s) + NaNO3(aq)

Cl-(aq) + Ag+(aq) -> AgCl(s)

Test for carbon dioxide

Carbon dioxide is bubbled through calcium hydroxide solution (limewater)

White precipitate of calcium carbonate forms (limewater goes milky)

Ca(OH)2(aq) + CO2(g) -> CaCO3(s) + H2O(l)