• The number of moles of a substance can be found by using the following equation:

• It is important to be clear about the type of particle you are referring to when dealing with moles
  • Eg. 1 mole of CaF₂?contains one mole of CaF_2?formula units, but one mole of Ca²⁺?and two moles of F^-?ions

Reacting masses

• The?masses?of reactants are useful to determine how much of the reactants?exactly?react with each other to prevent waste
• To calculate the reacting masses, the chemical equation is required
• This equation shows the ratio of moles of all the reactants and products, also called the?stoichiometry,?of the equation
• To find the mass of products formed in a reaction the following pieces of information is needed:
  • The mass of the reactants
  • The molar mass of the reactants
  • The balanced equation

Percentage yield

• In a lot of reactions, not all reactants react to form products which can be due to several factors:
  • Other reactions take place simultaneously
  • The reaction does not go to?completion
  • Reactants or products are?lost?to the atmosphere
• The?percentage yield?shows how much of a particular product you get from the reactants compared to the maximum theoretical amount that you can get:

• Where?actual yield?is the number of moles or mass of product obtained?experimentally
• The?predicted yield?is the number of moles or mass obtained by calculation

Worked example: Mass calculation using moles

Answer

• Step 1:?The symbol equation is:

2Mg(s)????+ ????O₂(g)??????→??????2MgO(s)

• Step 2:?The relative formula masses are:

Magnesium : 24 ??????? Oxygen : 32?????????Magnesium Oxide : 40

• Step 3:?Calculate the moles of magnesium used in reaction

• Step 4:?Find the ratio of magnesium to magnesium oxide using the balanced chemical equation

Therefore, 0.25 mol of MgO is formed

• Step 5:?Find the mass of magnesium oxide

mass =?mol?x?Mr

mass?= 0.25?mol?x 40?g mol^-1mass?= 10 g

Therefore,?mass of magnesium oxide produced is 10 g

Worked Example: Calculate % yield using moles

Answer

• Step 1:?The symbol equation is:

Zn (s) ???+ ???CuSO₄?(aq) ????→ ????ZnSO₄?(aq) ???+ Cu (s)

• ?Step 2:?Calculate the amount of zinc reacted in moles

• Step 3:?Calculate the maximum amount of copper that could be formed from the molar ratio:

Since the ratio of Zn(s) to Cu(s) is 1:1 a maximum of 0.10 moles can be produced

• Step 4:?Calculate the maximum mass of copper that could be formed (theoretical yield)

mass =?? mol?? x?? Mr

mass =?? 0.10 mol x 64 g mol^-1

mass =?? 6.4 g

• Step 5:?Calculate the percentage yield of copper

Excess & limiting reagents

• Sometimes, there is an?excess?of one or more of the reactants (excess reagent)
• The reactant which is not in excess is called the?limiting reagent
• To determine which reactant is limiting:
  • The number of moles of the reactants should be calculated
  • The ratio of the reactants shown in the equation should be taken into account eg:

C? + 2H₂????? →???? CH₄

There are 10 mol of Carbon reacting with 3 mol of Hydrogen

• Hydrogen is the?limiting reagent?and since the ratio of C : H₂?is 1:2 only 1.5 mol of C will react with 3 mol of H2

Worked example: Excess & limiting reagent

Answer

• Step 1:?Calculate the moles of each reactant

• Step 2:?Write the balanced equation and determine the molar ratio

2Na + S → Na2S

The molar ratio of Na: Na₂S is 2:1

• Step 3:?Compare the moles and determine the limiting reagent

So to react completely 0.40 moles of Na require 0.20 moles of S and since there are 0.25 moles of S, then S is in excess. Na is therefore the limiting reactant.

Volumes of gases

• Avogadro?suggested that ‘equal volumes of gases contain the same number of molecules’ (also called?Avogadro’s hypothesis)
• At room temperature (20 degrees Celsius) and pressure (1 atm)?one mole?of any gas has a volume of 24.0 dm3
• This?molar gas volume?of 24.0 dm3?can be used to find:
  • The volume of a given mass or number of moles of gas:

volume of gas (dm3) = amount of gas (mol) x 24

• The mass or number of moles of a given volume of gas:

Worked example: Calculation volume of gas using excess & limiting reagents

Answer

Volumes & concentrations of solutions

• The?concentration?of a solution is the amount of?solute?dissolved in a?solvent?to make 1 dm3?of??solution
  • The solute is the substance that dissolves in a solvent to form a solution
  • The solvent is often water

• A?concentrated?solution is a solution that has a?high?concentration of solute
• A?dilute?solution is a solution with a?low?concentration of solute
• When carrying out calculations involve concentrations in mol dm^-3?the following points need to be considered:
  • Change mass in grams to?moles
  • Change cm³?to dm³
• To calculate the?mass?of a substance present in solution of known?concentration and volume:
  • Rearrange the concentration equation

number of moles (mol) = concentration (mol dm^-3) x volume (dm³)

• • Multiply the moles of solute by its molar mass

mass of solute (g) = number of moles (mol) x molar mass (g mol^-1)

Worked example: Calculating volume from concentration

Answer

• Step 1:?Write the balanced symbol equation

CaCO₃??+ ?2HCl ?→ ?CaCl₂??+ ?H₂O ?+ ?CO₂

• Step 2:?Calculate the amount, in moles, of calcium carbonate that reacts

• Step 3:?Calculate the moles of hydrochloric acid required using the reaction’s stoichiometry

1 mol of CaCO₃?requires 2 mol of HCl

So 0.025 mol of CaCO₃?requires 0.05 mol of HCl

• Step 4:?Calculate the volume of HCl required

Volume of hydrochloric acid = 0.05 dm³

Worked example: Neutralisation calculation

Answer

• Step 1:?Write the balanced symbol equation

Na₂CO₃??+ ?2HCl ?→ ?Na₂Cl₂??+ ?H₂O ?+ ?CO₂

• Step 2:?Calculate the amount, in moles, of sodium carbonate reacted by rearranging the equation for amount of substance (mol) and dividing by the volume by 1000 to convert cm³?to dm³

amount (Na₂CO₃) = 0.025 dm³?x 0.050 mol dm^-3?= 0.00125 mol

• Step 3:?Calculate the moles of hydrochloric acid required using the reaction’s stoichiometry

1 mol of Na₂CO₃?reacts with 2 mol of HCl, so the molar ratio is 1 : 2

Therefore 0.00125 moles of Na₂CO₃?react with 0.00250 moles of HCl

• Step 4:?Calculate the concentration, in mol dm-3?of hydrochloric acid

concentration (HCl) (mol dm^-3) = 0.125 mol dm^-3

Stoichiometric relationships

• The stoichiometry of a reaction can be found if the?exact?amounts?of reactants and products formed are known
• The?amounts?can be found by using the following equation:

• The gas volumes can be used to deduce the?stoichiometry?of a reaction
  • Eg. in the?combustion?of 50 cm3?of propane reacting with 250 cm³?of oxygen, 150 cm³??of carbon dioxide is formed suggesting that the ratio of propane:oxygen:carbon dioxide is 1:5:3

C₃H₈?(g) + 5O2?(g) → 3CO2?(g) + H2O (l)