Calculating Formulae | Edexcel International AS Chemistry Revision Notes 2017

Empirical & Molecular

Empirical formula

Empirical formula is the simplest whole number ratio of the elements present in one molecule or formula unit of the compound
It is calculated from knowledge of the ratio of masses of each element in the compound
The empirical formula can be found by determining the mass of each element present in a sample of the compound
It can also be deduced from data that gives the percentage compositions by mass of the elements in a compound

Worked example

Empirical formula from mass

Determine the empirical formula of a compound that contains 10 g of hydrogen and 80 g of oxygen.

The above example shows how to calculate empirical formula from the mass of each element present in the compound
The example below shows how to calculate the empirical formula from percentage composition

Worked example

Empirical formula from %

Determine the empirical formula of a compound that contains 85.7% carbon and 14.3% hydrogen.

Molecular formula

The molecular formula gives the exact numbers of atoms of each element present in the formula of the compound
The molecular formula can be found by dividing the relative formula mass of the molecular formula by the relative formula mass of the empirical formula
Multiply the number of each element present in the empirical formula by this number to find the molecular formula

Worked example

Calculating molecular formula

The empirical formula of X is C₄H₁₀S and the relative molecular mass of X is 180.2

What is the molecular formula of X?

(A_r data: C = 12.0, H = 1.0, S = 32.1 )

Answer

Step 1: Calculate relative mass of the empirical formula

- Relative empirical mass = (C x 4) + (H x 10) + (S x 1)
- Relative empirical mass = (12.0 x 4) + (1.0 x 10) + (32.1 x 1)
- Relative empirical mass = 90.1

Step 2: Divide relative formula mass of X by relative empirical mass

- Ratio between M_r of X and the M_rof the empirical formula = 180.2 / 90.1
- Ratio between M_r of X and the M_rof the empirical formula = 2

Step 3: Multiply each number of elements by 2

- (C₄ x 2) + (H₁₀ x 2) + (S x 2) = (C₈) + (H₂₀) + (S₂)
- Molecular Formula of X is C₈H₂₀S₂

Percentage Composition by Mass

The percentage by mass of an element in a compound can be calculated using the following equation:

$% mass of an element = \frac{A_{r} \times number of atoms of the element}{M_{r} of the compound} \times 100$

Worked example

Calculate the percentage by mass of calcium in calcium carbonate, CaCO₃

Answer

Water of Crystallisation

Water of crystallisation is when some compounds can form crystals which have water as part of their structure
A compound that contains water of crystallisation is called a hydrated compound
The water of crystallisation is separated from the main formula by a dot when writing the chemical formula of hydrated compounds
- E.g. hydrated copper(II) sulfate is CuSO₄∙5H₂O

A compound which doesn’t contain water of crystallisation is called an anhydrous compound
- E.g. anhydrous copper(II) sulfate is CuSO₄
A compound can be hydrated to different degrees
- E.g. cobalt(II) chloride can be hydrated by six or two water molecules
- CoCl₂ ∙6H₂O or CoCl₂ ∙2H₂O

The conversion of anhydrous compounds to hydrated compounds is reversible by heating the hydrated salt:

Hydrated: CuSO₄•5H₂O ⇌ CuSO₄ + 5H₂O :Anhydrous

The degree of hydration can be calculated from experimental results:
- The mass of the hydrated salt must be measured before heating
- The salt is then heated until it reaches a constant mass
- The two mass values can be used to calculate the number of moles of water in the hydrated salt - known as the water of crystallisation

Worked example

Calculating water of crystallisation

10.0 g of hydrated copper sulfate are heated to a constant mass of 5.59 g. Calculate the formula of the original hydrated copper sulfate.

(M_r data: CuSO₄ = 159.6, H₂O = 18.0)

Answer

List the components	CuSO₄	H₂O
Note the mass of each component	5.59 g	10 - 5.59 = 4.41 g
Divide the component mass by the components M_r	$\frac{5.59}{159.6}$ = 0.035	$\frac{4.41}{18.0}$ = 0.245
Divide by the lowest figure to obtain the ratio	$\frac{0.035}{0.035}$ = 1	$\frac{0.245}{0.035}$ = 7
Hydrated salt formula	CuSO₄•7H₂O

Ideal Gas Equation

The ideal gas equation is:

PV = nRT

- P = pressure in pascals (Pa)
- V = volume in cubic metres (m³)
- n = the amount of substance in moles (mol)
- R = the gas constant, which is given in the Data Booklet as 8.31 J mol^-1 K^-1
- T = temperature in Kelvin (K)

Examiner Tip

There are several calculations in Chemistry where you need to convert the units to or from SI units

The ideal gas equation has three:

Pressure is often quoted in kPa but the calculation needs pressure in Pa
- kPa to Pa = multiply by 1000 or 10³
Volume is usually quoted in cm³ or dm³ but the calculation needs volume in m³
- cm³ to m³ = divide by 1000000 or multiply by 10^-6
- dm³ to m³ = divide by 1000 or multiply by 10^-³
Temperature can be quoted in Kelvin or Celsius
- Celsius to Kelvin = + 273

This is why you should always show your working! Examiners can't take all of your marks for one error, if you show your working then they should check through for errors and award marks accordingly

The ideal gas equation can be used to find the amount of moles in a gaseous substance
- It can also be used for volatile liquids at temperatures above their boiling point
If the mass of the substance is known, then the molar mass can be calculated
- This can then be used with empirical formula data to determine the molecular formula of a compound

Worked example

An unknown compound was analysed and found to contain 66.7% of carbon, 11.1% hydrogen and the remainder was oxygen

0.135 g of the unknown compound had a volume of 56.0 cm³ at a temperature of 90 ^oC and a pressure of 101 kPa

Determine the molecular formula of the unknown compound

Answer

Step 1: Calculate the number of moles of carbon, hydrogen and oxygen

Carbon: $\frac{66.7}{12.0} =$ 5.558 moles

Hydrogen: $\frac{11.1}{1.00} =$ 11.1 moles

Oxygen: $\frac{(100 - 66.7 - 11.1)}{16.0} =$ 1.3875 moles

Step 2: Divide by the smallest answer to get the ratio

Carbon: $\frac{5.558}{1.3875} =$ 4

Hydrogen: $\frac{11.1}{1.3875} =$ 8

Oxygen: $\frac{1.3875}{1.3875} =$ 1

Step 3: State the empirical formula

- The empirical formula is C₄H₈O

Step 4: Calculate the amount in moles, using PV = nRT

- n = $= \frac{P V}{R T} = \frac{(101 \times 10^{3}) \times (56.0 \times 10^{- 6})}{8.31 \times (90 + 273)} =$ 1.875 x 10^-3 moles

Step 5: Calculate the molar mass

- Molar mass $= \frac{0.135}{1.875 \times 10^{- 3}} =$ 72

Step 6: Deduce the molecular formula

- The empirical formula, C₄H₈O, has a mass of (4 x 12.0) + (8 x 1.0) + 16.0 = 72.0
- Therefore, the molecular formula is also C₄H₈O

Calculating Formulae (Edexcel International AS Chemistry)

Revision Note

Empirical formula

Molecular formula

Percentage Composition by Mass

Water of Crystallisation

Ideal Gas Equation

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1. Structure, Bonding & Introduction to Organic Chemistry

1.1 Formulae & Equations

1.1.1 Formulae & Mass

1.1.2 Avogadro & the Mole

1.1.3 Full & Ionic Equations

1.1.4 Calculating Formulae

1.2 Amount of Substance

1.2.1 Concentration Calculations

1.2.2 Reacting Mass Calculations

1.2.3 Reacting Volume Calculations

1.2.4 Calculations of Product

1.3 Atomic Structure

1.3.1 Sub-Atomic Particles

1.3.2 Isotopes & Mass Spectra

1.4 Electrons & Ions

1.4.1 Endothermic Ionisation Energy

1.4.2 Electronic Structures

1.4.3 Shapes of Orbitals

1.4.4 Electronic Configurations & Chemical Properties

1.5 The Periodic Table

1.5.1 Periodicity - Ionisation Energy

1.5.2 Periodicity - Thermal Trends

1.6 Ionic & Metallic Bonding & Structure

1.6.1 Evidence of Ions

1.6.2 Ionic Bonding & Structures

1.6.3 Ionic Bond Strength

1.6.4 Polarisation

1.6.5 Metallic Bonding & Lattices

1.7 Covalent Bonding & Structure

1.7.1 Evidence of Covalent Bonding

1.7.2 Covalent Bonding

1.7.3 Carbon Allotropes

1.7.6 Predicting Shapes & Angles

1.8 Introductory Organic Chemistry

1.8.2 Functional Groups & Homologous Series

1.8.3 Nomenclature & Classification

1.8.4 Isomers - Structural

1.9 Alkanes

1.9.1 Alkanes - Introduction

1.9.2 Alkanes as Fuels

1.9.3 Combustion of Alkanes

1.9.4 Alternative Fuels

1.9.5 Free Radicals & Fission

1.9.6 The Free Radical Substitution Mechanism

1.10 Alkenes

1.10.1 Alkenes - Introduction

1.10.2 Isomers - Geometric

1.10.3 Reactions of Alkenes

1.10.4 Saturation Test

1.10.5 Electrophilic Addition - Mechanism

1.10.6 Addition Polymerisation

1.10.7 Polymer Disposal

2. Energetics, Group Chemistry, Halogenoalkanes & Alcohols

2.1 Energetics

2.1.1 Enthalpy Level Diagrams

2.1.2 Enthalpy Change - Definitions

2.1.3 Using Calorimetry

2.1.4 Using Hess Cycles

2.1.5 Bond Enthalpies

2.2 Intermolecular Forces

2.2.1 Intermolecular Forces - Introduction

2.2.2 Hydrogen Bonding

2.2.3 Physical Properties

2.2.4 Solvent Choice

2.3 Redox Chemistry & Acid-Base Titrations

2.3.1 Oxidation Numbers - Introduction

2.3.2 Oxidation & Reduction

2.3.3 Disproportionation Reactions

2.3.4 Ionic Equations

2.3.5 Acid-Base Titrations with Indicators