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Revision:Chemistry Unit 1.2 - Formulae, Equations and Moles

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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.



  • 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 \mathsf{A^{x+} , B^{y-}}

Step 1. \mathsf{{A^{x+} + B^{y-}}}

Step 2. If \mathsf{ x = y} , they combine like this: \mathsf{ AB}

Step 3. If not...: \mathsf{ \frac{x}{y}} cancel down this fraction into it's simplest form, like: \mathsf{\frac{2}{4}} = \frac{1}{2}

Step 4. Call this 'cancelled down' version of x and y as \mathsf{ \frac{m}{n}}

Step 5. The Compound can be written as \mathsf{A_n(B)_m}

Example:

1.\mathsf{Al^{3+} , S_2O_3^{2-}}

2. 3 is not equal to 2

3. \mathsf{\frac{3}{2}}

4. \mathsf{\frac{3}{2} = \frac{m}{n}}

5. \mathsf{ \therefore} Formula = \mathsf{Al_2(S_2O_3)_3}


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.

\mathsf{AgNO_{3(aq)} + KI_{(aq)} \longrightarrow AgI_{(s)} + KNO_{3(aq)}}

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:

\mathsf{AgNO_{3(aq)} + KI_{(aq)} \longrightarrow AgI_{(s)} + KNO_{3(aq)}}


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

\mathsf{Ag^+_{(aq)} + NO^-_{3(aq)} + K^+_{(aq)} + IH^-SO_{4(aq)} \longrightarrow AgI_{(s)} + K^+_{(aq)} + NO^-_{3(aq)}}


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

\mathsf{Ag^+_{(aq)} + NO^-_{3(aq)} + K^+_{(aq)} + I^-_{(aq)} \longrightarrow AgI_{(s)} + K^+_{(aq)} + NO^-_{3(aq)}}


4) Write out what is left:

\mathsf{Ag^+_{(aq)} + I^-_{(aq)} \longrightarrow AgI_{(s)}}

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:

\mathsf{2Mg_{(s)} + O_{2(g)} \longrightarrow 2MgO_{(s)}}


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:

\mathsf{CaCO_{3(s)} + 2HCl_{(aq)} \longrightarrow CaCl_{2(aq)} + CO_{2(g)} + H_2O_{(l)}}


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:

\mathsf{C_3H_8_{(g)} + 5O_{2(g)} \longrightarrow 3CO_{2(g)} + 4H_2O_{(g)}}


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:

\mathsf{H_2SO_{4(aq)} + 2KOH_{(aq)} \longrightarrow K_2SO_{4(aq)} + 2H_2O_{(aq)}}


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.

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