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Calculating Formulae (Edexcel International AS Chemistry): Revision Note

Richard Boole

Last updated

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.

Answer:

  • 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.

Answer:

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?

(Ar 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 Mr of X and the Mr of the empirical formula = 180.2 / 90.1

    • Ratio between Mr of X and the Mr of 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:

percent sign space mass space of space an space element space equals space fraction numerator A subscript straight r cross times space number space of space atoms space of space the space element over denominator M subscript straight r space of space the space compound end fraction space cross times space 100

Worked Example

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

Answer:

Calculate % Mass of an Element in a Compound WE, downloadable IGCSE & GCSE Chemistry revision notes

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. (Mr 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 Mr

fraction numerator 5.59 over denominator 159.6 end fraction = 0.035

fraction numerator 4.41 over denominator 18.0 end fraction = 0.245

Divide by the lowest figure to obtain the ratio

fraction numerator 0.035 over denominator 0.035 end fraction= 1

fraction numerator 0.245 over denominator 0.035 end fraction= 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 Tips and Tricks

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

The ideal gas equation has three:

  1. Pressure is often quoted in kPa but the calculation needs pressure in Pa

    • kPa to Pa = multiply by 1000 or 103 

  2. 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 

  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: begin mathsize 16px style fraction numerator 66.7 over denominator 12.0 end fraction space equals space end style5.558 moles

 Hydrogen: begin mathsize 16px style fraction numerator 11.1 over denominator 1.00 end fraction space equals end style11.1 moles

 Oxygen: begin mathsize 16px style fraction numerator left parenthesis 100 space minus space 66.7 space minus space 11.1 right parenthesis over denominator 16.0 end fraction space equals end style1.3875 moles

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

  Carbon: fraction numerator 5.558 over denominator 1.3875 end fraction space equals4

  Hydrogen: fraction numerator 11.1 over denominator 1.3875 end fraction space equals8

  Oxygen: fraction numerator 1.3875 over denominator 1.3875 end fraction space equals1

Step 3: State the empirical formula

  • The empirical formula is C4H8O

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

  • n = equals fraction numerator P V over denominator R T end fraction equals fraction numerator left parenthesis 101 cross times 10 cubed right parenthesis space cross times left parenthesis 56.0 cross times 10 to the power of negative 6 end exponent right parenthesis over denominator 8.31 space cross times space left parenthesis 90 space plus space 273 right parenthesis end fraction equals1.875 x 10-3 moles

Step 5: Calculate the molar mass

  • Molar mass equals fraction numerator 0.135 over denominator 1.875 space cross times 10 to the power of negative 3 end exponent end fraction equals72

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

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    Richard Boole

    Author: Richard Boole

    Expertise: Chemistry

    Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.