REVISION NOTES

IGCSE Edexcel Chemistry

1.5 Chemical Formulae, Equations and Calculations

1.5.1 Write word equations and balanced chemical equations (including state symbols):

  • For reactions studied in this specification
  • For unfamiliar reactions where suitable information is provided.

State symbols used in chemical equation:

WORD EQUATION

Sodium + Hydrochloric acid → Sodium chloride + Hydrogen

BALANCED CHEMICAL EQUATION

2Na + 2HCl → 2NaCl + H2

LAW OF CONSERVATION

  • There should be equal number of atoms of the same type
  • Number should be added in the front of the compound to balance

1.5.2 Calculate relative formula masses (including relative molecular masses) (Mr) from relative atomic masses (Ar)

RELATIVE ATOMIC MASS (AR)

  • Relative mass of an atom 
  • Can be found in the periodic table

RELATIVE FORMULA MASS (MR)

  • Sum of the relative atomic masses of the atoms in the compound formula
  • Relative molecular mass: Mr of covalent compounds

EXAMPLE

What is the relative formula mass (Mr) of a water molecule (H2O)?

Ar of H = 1

Ar of O = 16

Mr of H2O = H x 2 + O x 1

                  = 1 x 2 + 16 x 1

                  = 18

1.5.3 Know that the mole (mol) is the unit for the amount of a substance

MOLE (SYMBOL:N)

  • Amount of a substance
  • It is used to count the number of particles
  • Unit: mol
  • One mole of a substance contains the same number of the stated particles, atoms, molecules or ions as one mole of any other substance

AVOGADRO’S CONSTANT

  • Avogadro’s constant = 6.02 x 1023
  • It means: there are 6.02 x 1023 particle in 1 mole
  • E.g. there are 2 x 6.02 x 1023 in 2 moles of H2O molecules

MOLAR MASS

  • The mass of one mole of a substance in grams is numerically equal to its relative formula mass
    • E.g. Ar of O is 16, it means one mole of O atoms weighs 16g
    • E.g. Mr of O2 is 32, it means one mole of O2 molecules weighs 32g
  • The mass of one mole of a substance can be called molar mass

1.5.4 Understand how to carry out calculations involving amount of substance, relative atomic mass (Ar) and relative formula mass (Mr)

Molar mass can be:

  • Relative molecular mass (if compounds are discussed)
  • Relative atomic mass (if atoms are discussed)

EXAMPLE

How many moles are there in 64g of O2?

  • Mole = mass ÷ relative formula mass
  • Mole = 64 ÷ (16 x 2) = 64 ÷ 32 = 2g

1.5.5 Calculate reacting masses using experimental data and chemical equations

Information of moles can be obtained from chemical equations

2Na + 2HCl → 2NaCl + H2

Information 1: 

Every 2 moles of Na react with 2 moles of HCl to produce 2 moles of NaCl and 1 mole of H2

Information 2: 

0.5 mole of Na reacts with 0.5 mole of HCl to produce 0.5 mole of NaCl and 0.25 mole of H2

The number of moles reacted do not have to be whole numbers

Information 3: 

mole of Na : HCl : NaCl : H2 = 2 : 2 : 2 : 1

The coefficients in chemical equations show the simplest mole ratio

REACTING MASS

Masses of reactants and products can be calculated from balanced chemical equation:

  • Given: reacting mass of one reactant
  • Calculate: mass of one product formed
  • Step 1: Find moles of the reactant (mole = mass ÷ relative formula mass)
  • Step 2: Use balanced equation to find the mole of product (mole ratio)
  • Step 3: Find mass of the product (mass = mole x relative formula mass)

EXAMPLE

If 34.5 grams of sodium metal is completely reacted with excess hydrochloric acid, how many grams of hydrogen gas will be produced according to the following reaction?

2Na + 2HCl → 2NaCl + H2

Given: mass (Na) = 34.5g

Calculate: mass (H2)

STEP 1: FIND THE MOLE OF NA

Mole (Na) = mass (Na) ÷ Mr (Na)

                 = 34.5g ÷ 23

                 = 1.5 mol

STEP 2: FIND THE MOLE OF H2 PRODUCED

Mole (Na) : Mole (H2) = 2 : 1

(this can be found from the coefficients in the balanced chemical equation)

If mole (Na) = 1.5 mol from step 1

Then mole (H2) = 1.5 ÷ 2 = 0.75 mol

STEP 3: FIND THE MASS OF H2 PRODUCED

Mass (H2) = mole (H2) x Mr (H2)

                  = 0.75 mol x (1 x 2)

                  = 0.75 x 2

                  = 1.5 grams

1.5.6 Calculate percentage yield

  • Actual yield: what is obtained after the experiment (given in question)
  • Theoretical yield: what is supposed to be obtained (from reacting mass calculation)

In practice, you never get 100% yield in a chemical process for several reasons 

  • Some reactants being left behind in the equipment
  • The reaction may be reversible
  • Product may also be lost during separation stages

1.5.7 Understand how the formulae of simple compounds can be obtained experimentally, including metal oxides, water and salts containing water of crystallisation

CASE 1: METAL OXIDES(MAGNESIUM OXIDE)

  • Experimental procedures:
    1. Weigh some pure magnesium
    2. Heat magnesium to burning in a crucible to form magnesium oxide (magnesium reacts with oxygen in the air)
    3. Weigh the mass of magnesium oxide
  • Given: mass of magnesium, mass of magnesium oxide
  • Calculate: formula of magnesium oxide
  • Calculation steps:
    1. Find the mass of magnesium
    2. Mole of magnesium = mass of magnesium ÷ molar mass of magnesium
    3. Mass of oxygen = mass of magnesium oxide – mass of magnesium
    4. Mole of oxygen = mass of oxygen ÷ molar mass of oxygen
    5. Calculate ratio of mole of magnesium : mole of oxygen (divide both moles by the smaller number)
    6. Use the ratio to form empirical formula

EXAMPLE

Some magnesium (Mg) was heated in a crucible to form magnesium oxide. Use the following information to deduce the formula of magnesium oxide.  

STEP 1: MASS OF MAGNESIUM

Mass of Mg = mass (crucible + magnesium) – mass (crucible)

                    = 21.66 – 21.26

                    = 0.40 grams

STEP 2: MASS OF OXYGEN

Mass of O = mass (crucible + magnesium oxide) – mass (crucible + magnesium)

                  = 21.92 – 21.66

                  = 0.26 grams

Step 3: mole ratio of Mg and O

So the formula is MgO.

CASE 2: WATER OF CRYSTALLISATION

  • Experimental procedures:
    1. Weigh some hydrated salt
    2. Heat hydrated salt to remove water to form anhydrous salt
    3. Weigh the mass of anhydrous salt
  • Given: mass of hydrated salt, mass of anhydrous salt 
  • Calculate: formula of hydrated salt
  • Calculation steps:
    1. Mass of water = mass of hydrated salt – mass of anhydrous salt
    2. Mole of water = mass of water / molar mass of water
    3. Mole of anhydrous salt = mass of anhydrous salt / molar mass of anhydrous salt
    4. Calculate ratio of mole of anhydrous salt : mole of water (divide both moles by the smaller number)
    5. Use the ratio to form empirical formula

EXAMPLE

11.25 g of hydrated copper sulphate, CuSO4・xH2O, is heated until it loses all of its water. Its new mass is found to be 7.19 g. What is the value of x? 

STEP 1: MASS OF WATER

Mass of water = mass of hydrated salt – mass of anhydrous salt

                         = 11.25 – 7.19

                         = 4.06 grams

Step 2: mole ratio of CuSO4 and H2O

So the formula is CuSO4・5H2O, the value of x is 5.

1.5.8 Know what is meant by the terms empirical formula and molecular formula

EXPIRICAL FORMULA

  • Simplest whole number ratio of atoms of each element in a compound
  • Remember:
    • It is a ratio
    • Simplest 
    • Whole number

MOLECULAR FORMULA

  • Actual number of atoms of each element in a compound

Example

1.5.9 Calculate empirical and molecular formulae from experimental data

CALCULATION OF EMPIRICAL FORMULA

  • Given: mass of each element in the compound
  • Calculate: empirical formula
  • Calculation steps:
    1. Mole of each element (mole = mass ÷ molar mass)
    2. Divide all numbers by the smallest number to find the simplest ratio
    3. Use the ratio to write the empirical formula

EXAMPLE

0.150 g of copper reacts with oxygen from 0.188 g of copper oxide.  Find the empirical formula of copper oxide.

So the empirical formula of copper oxide is CuO.

CALCULATION OF MOLECULAR FORMULA

  • Given: molar mass of molecular formula
  • Calculate: molecular formula
  • Calculation steps:
    1. Calculate the Mr of molecular formula
    2. Calculate the Mr of empirical formula
    3. Divide step 1 by step 2 to find the multiple
    4. Multiply the empirical formula to find the molecular formula

EXAMPLE

The empirical formula of a compound of boron and hydrogen is  BH3. Its molar mass is  28g/mol . Determine the molecular formula of the compound.

So the molecular formula of the compound is B2H6.

1.5.10C Understand how to carry out calculations involving amount of substance, volume and concentration (in mol/dm³) of solution

CONCENTRATION OF SOLUTION

Solution consists of:

  • Solvent
  • Solute
  • Amount of solute (n): in mol
  • Concentration (c): in mol/dm3
  • Volume of solution (V): in dm3
    • 1 dm3 = 1000 cm3

EXAMPLE

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

Volume in dm3 = 300 cm3 ÷ 1000 = 0.3 dm3

Concentration ​=​ mole of solute ​÷​ volume of solution

                         = 0.03 mol ÷ 0.3 dm3

                         = 0.1 mol/dm3

1.5.11C Understand how to carry out calculations involving gas volumes and the molar volume of a gas (24 dm3 and 24 000 cm3 at room temperature and pressure (rtp))

VOLUME AND MOLE OF A GAS

  • Regardless of the type of gas, same amount of gas molecules occupies the same volume under same condition
  • RTP: room temperature and pressure
    • 20oC
    • 1 atmospheric pressure
  • Volume of 1 mole of any gas at rtp is 24 dm3 or 24,000 cm3

EXAMPLE

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

Volume (in cm3) = mole x 24,000

                            = 3 x 24,000

                            = 72,000 cm3

1.5.12 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)

FORMULA OF METAL OXIDE BY COMPOSITION

  • Similar to 1.31
  • Calculate the mole of each element in the compound
  • Divide all moles by the smallest number to find the simplest whole number ratio

FORMULA OF METAL OXIDE BY REDUCTION

  • Step 1: find the mass of metal
  • Step 2: calculate the mole of metal
  • Step 3: calculate the mass of oxygen
  • Step 4: calculate the mole of oxygen
  • Step 5: divide all moles by the smallest number to find the simplest ratio

EXAMPLE

An oxide of copper was reduced to copper metal by reaction with hydrogen.   

 

copper oxide + hydrogen → copper + water   

 

Use the data in the table to find the empirical formula of this oxide of copper.

STEP 1: MASS OF COPPER

Mass of Mg = mass (crucible + copper) – mass (crucible)

                     = 26.82 – 25.61

                     = 1.21 grams

STEP 2: MASS OF OXYGEN

Mass of O = mass (crucible + copper oxide) – mass (crucible + copper)

                  = 26.97 – 26.82

                  = 0.15 grams

Step 3: mole ratio of Cu and O

So the formula is Cu2O.

Back
Next

1.1 State of Matter

1.2 Elements, Compounds and Mixtures

1.3 Atomic Structure

1.4 The Periodic Table

1.5 Chemical Formulae, Equations and Calculations

1.6 Ionic Bonding

1.7 Covalent Bonding

1.8 Metallic Bonding

1.9 Electrolysis

2.1 Group 1 (Alkali Metals) – Lithium, Sodium and Potassium

2.2 Group 7 (Halogens) – Chlorine, Bromine and Iodine

2.3 Gases in the Atmosphere

2.4 Reactivity Series

2.5 Extraction and Uses of Metals

2.6 Acids, Alkalis and Titrations

2.7 Acids, Bases and Salt Preparations

2.8 Chemical Tests

3.1 Energetics

3.2 Rates of Reaction

3.3 Reversible Reactions and Equilibria

4.1 Introduction

4.2 Crude Oil

4.3 Alkanes

4.4 Alkenes

4.5 Alcohols

4.6 Carboxylic Acids

4.7 Esters

4.8 Synthetic Polymers

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