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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Chemistry > Chemistry Unit 1.2 - Formulae, Equations and Moles

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Contents 1 Empirical and Molecular Formulae 1.1 Question 1.2 Question 2 Compound Formulae 3 Full and Ionic Equations 4 How to construct ionic equations 5 Reacting Masses 5.1 Question 6 Reacting gases and gas volumes 6.1 Question 7 Gaseous Reactions 7.1 Question 8 Solutions and Concentrations 9 Volumetric Calculations 9.1 Question 10 Essential Definitions 11 Comments

Empirical and Molecular Formulae

• The empirical formula of a compound tells us the simplest ratio of atoms in that compound. It is found from experimental data.

Question

A hydrocarbon was found to contain 75% carbon and 25% hydrogen by mass. Determine its empirical formula.

	C	H
Mass(g)	75	25
Divide by the relative atomic mass	75/12	25/1
Ratio of atoms	6.25	25
Divide by smallest	6.25/6.25	25/6.25
.	1	4

So, the empirical formula of this hydrocarbon is CH₄.

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• The molecular formula of a compound tells us the actual number of atoms in a compound of one mole of that compound.

Question

It was found that a hydrocarbon had an empirical formula of CH₃. It was found that its relative formula mass (r.f.m.) is 30. Determine its molecular formula.

1. Determine r.f.m. of CH₃ = (12x1) + (1x3) = 15
2. Determine the ratio of mass = 30:15 = 2:1
3. Multiply the number of atoms by the ratio to give molecular formula = 2 * CH₃ =C₂H₆

Compound Formulae

Find the Compound Formula for AB where

Step 1.

Step 2. If , they combine like this:

Step 3. If not...: cancel down this fraction into it's simplest form, like:

Step 4. Call this 'cancelled down' version of x and y as

Step 5. The Compound can be written as

Example:

1.

2. 3 is not equal to 2

3.

4.

5. Formula =

Full and Ionic Equations

A fully balanced equation or stoichiometric equation is one that shows the formulae of reactants and products and the relative number of particles reacting. e.g.

An ionic equation is one where the ions are represented separately. Only ionic compounds can be represented as ions. Covalent substances and elements cannot be represented as ions. Also, ionic solids cannot be represented as free ions because the ions are not free to move.

How to construct ionic equations

1) Write down the stoichiometric equation:

2) Split all the ionic compounds into their free ions where possible:

3) Cancel all spectator ions in the equation. These are ions that play no part in the reaction:

4) Write out what is left:

Reacting Masses

We can use chemical equations to determine the mass of reactants and products given certain information.

Question

In a reaction it was found that 6g of magnesium fully reacted with air. Determine the mass of magnesium oxide produced.

1) Write out the stoichiometric equation:

2) Convert any relevant data given into moles:

Amount(moles) = Mass(g)/r.f.m. = 6/24 = 0.25 moles

3) Relate this information to the amount in moles of the substance you are trying to work out the mass of.

Amount(moles) = 0.25 moles of MgO produced

4) Convert this amount into mass in grams:

Mass of MgO = Amount(moles) * r.f.m. = 0.25 x 40 = 10g

Reacting gases and gas volumes

One mole of any gas occupies the volume of 24dm³ or 24,000cm³ at room temperature and pressure. We need to be able to relate volumes of gases in reactants and products to answer certain questions.

Question

10g of CaCO₃ reacts fully with excess HCl. Calculate the volume of CO₂ gas produced.

1) Stoichiometric equation:

2) Work out the amount in moles of CaCO₃:

Moles of CaCO₃ = Mass/r.f.m = 10/100 = 0.1moles

3) Work out the amount in moles of CO₂:

Moles of CO₂ = Moles of CaCO₃ = 0.1moles

4) Work out the volume of CO₂:

Volume of CO₂ = Moles of CO₂ x 24 = 0.1 x 24 = 2.4dm3

Gaseous Reactions

The ratio of the volume of a gas produced in a reaction to the number of moles of this gas in the reaction is constant.

Question

Calculate the volume of water vapour produced when 10cm³ of propane reacts fully with air.

1) Write out the stoichiometric equation:

2)Work out the volume of water vapour produced:

Volume of C₃H₈ / moles of C₃H₈ = Volume of H₂O / moles of H₂O

10 / 1 = Volume of H₂O / 4

Volume of H₂O = 40cm³.

Solutions and Concentrations

The concentration of a solution is the measure of the amount of solute dissolved in 1dm³ of a solvent (usually water). It can be measured in one of two ways:

• moles/dm³ or mol dm^-3
• grams/dm^-3 or g dm^-3

In order to convert between the two different forms there is a simple equation:

Concentration( gdm^-3 ) = Concentration(Moles dm^-3) * r.f.m

Volumetric Calculations

We can determine the concentration of a solution of unknown concentration using titration. These calculations are called volumetric calculations or volumetric analyses.

Question

It was found by titration that 20cm³ of sulphuric acid exactly neutralised 25cm³ of 0.1 mol dm^-3 potassium hydroxide. Calculate the concentration of sulphuric acid in mol dm^-3.

1) Write out the stoichiometric equation:

2) Determine the amount of moles of the substance you can work out the moles of:

Amount of KOH = concentration * volume = 0.1 x 0.025 = 0.0025 moles

3) Determine the amount in moles of H₂SO₄:

Amount of H₂SO₄ = 0.5 x amount of KOH = 0.5 x 0.0025 = 0.00125 moles

4) Determine the concentration of H₂SO₄:

Concentration = moles / volume = 0.00125 / 0.02 = 0.0625 mol dm^-3

Essential Definitions

• Avogadro's constant - The number of atoms, molecules, ions or other chemical entities of a substance in one mole of that substance. L = 6.02 x 1023mol^-1.

Comments

Originally written by D Clark.