Reacting Masses & Volumes

Mole Calculations

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

$number of moles = \frac{mass of a substance (g)}{molar mass (g {mol}^{- 1})}$

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 CaF₂ 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:

$percentage yield = \frac{actual yield}{predicted or theoretical yield} \times 100$

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

Calculate the mass of magnesium oxide that can be produced by completely burning 6 g of magnesium in oxygen.

magnesium + oxygen → magnesium oxide

Answer

Step 1: The symbol equation is:
- 2Mg (s) + O₂(g) → 2MgO (s)

Step 2: The relative atomic and formula masses are:
- Magnesium: 24.3
- Oxygen: 32
- Magnesium oxide: 40.3

Step 3: Calculate the moles of magnesium used in the reaction:
- Number of moles = $\frac{6.0 g}{24.3 g {mol}^{- 1}}$ = 0.2469 moles

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

	Magnesium	Magnesium oxide
Moles	2	2
Ratio	1	1
Change in moles	- 0.2469	+ 0.2469

- Therefore, 0.2469 mol of MgO is formed
Step 5: Find the mass of magnesium oxide
- mass = mol x M_r
- mass = 0.2469 mol x 40.3 g mol^-1
- mass = 9.95 g
Therefore, the mass of magnesium oxide produced is 9.95 g

Worked example

Calculate % yield using moles

In an experiment to displace copper from copper(II) sulfate, 6.54 g of zinc was added to an excess of copper(II) sulfate solution.

The copper was filtered off, washed and dried.

The mass of copper obtained was 4.80 g.

Calculate the percentage yield of copper.

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
- Number of moles = $\frac{6.54 g}{65.4 g {mol}^{- 1}}$ = 0.10 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 M_r
- Mass = 0.10 mol x 63.5 g mol^-1
- Mass = 6.35 g

Step 5: Calculate the percentage yield of copper
- Percentage yield = $\frac{4.80 g}{6.35 g}$ x 100 = 75.6 %

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

9.2 g of sodium is reacted with 8.0 g of sulfur to produce sodium sulfide, Na₂S.

Which reactant is in excess and which is the limiting reagent?

Answer

Step 1: Calculate the moles of each reactant
- Number of moles (Na) = $\frac{9.2 g}{23.0 g {mol}^{- 1}}$ = 0.40 mol
- Number of moles (S) = $\frac{8.0 g}{32.1 g {mol}^{- 1}}$ = 0.25 mol
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
- To completely react 0.40 moles of Na requires 0.20 moles of S
- Since there are 0.25 moles of S, then S is in excess
- Therefore, Na is the limiting reactant

Volumes of gases

Avogadro suggested that ‘equal volumes of gases contain the same number of molecules’
- This is also called Avogadro’s hypothesis
At room temperature and pressure one mole of any gas has a volume of 24.0 dm³
- Room temperature is 20 ^oC
- Room pressure is 1 atmosphere
This molar gas volume of 24.0 dm³can 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 24.0

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

$amount of gas (mol) = \frac{volume of gas ({dm}^{3})}{24.0}$

Worked example

Calculating the volume of gas using excess & limiting reagents

Calculate the volume that the following gases occupy:

Hydrogen (3 mol)
Carbon dioxide (0.25 mol)
Oxygen (5.4 mol)
Ammonia (0.02 mol)

Calculate the moles in the following volumes of gases:

Methane (225.6 dm³)
Carbon monoxide (7.2 dm³)
Sulfur dioxide (960 dm³)

Answer

Gas	Amount of gas (mol)	Volume of gas (dm³)
Hydrogen	3.0	3.0 x 24.0 = 72.0
Carbon dioxide	0.25	0.25 x 24.0 = 6.0
Oxygen	5.4	5.4 x 24.0 = 129.6
Ammonia	0.02	0.02 x 24.0 = 0.48
Methane	$\frac{225.6}{24.0}$ = 9.4	225.6
Carbon monoxide	$\frac{7.2}{24.0}$ = 0.30	7.2
Sulfur dioxide	$\frac{960}{24.0}$ = 40	960

Volumes & concentrations of solutions

The concentration of a solution is the amount of solute dissolved in a solvent to make 1 dm³of solution
- The solute is the substance that dissolves in a solvent to form a solution
- The solvent is often water

$concentration (mol {dm}^{- 3}) = \frac{number of moles of solute (mol)}{volume of solution ({dm}^{3})}$

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^-3the following points need to be considered:
- Change mass in grams to moles
- Change cm³to 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

Calculate the volume of 1.0 mol dm^-3 hydrochloric acid required to completely react with 2.5 g of calcium carbonate.

Answer

Step 1: Write the balanced symbol equation
- CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
Step 2: Calculate the amount, in moles, of calcium carbonate:
- Number of moles (CaCO₃) = $\frac{2.5 g}{100 g {mol}^{- 1}}$ = 0.025 mol

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 (HCl) = $\frac{amount (mol)}{concentration (mol {dm}^{- 3})}$ = $\frac{0.05 mol}{1.0 mol {dm}^{- 3}}$ = 0.05 dm³
- So, the volume of hydrochloric acid required is 0.05 dm³

Worked example

Neutralisation calculation

25.0 cm³ of 0.050 mol dm^-3 sodium carbonate solution was completely neutralised by 20.0 cm³ of dilute hydrochloric acid.

Calculate the concentration, in mol dm^-3, of the hydrochloric acid.

Answer

Step 1: Write the balanced symbol equation:
- Na₂CO₃ + 2HCl → Na₂Cl₂ + H₂O + CO₂
Step 2: Calculate the amount, in moles, of sodium carbonate reacted
- Rearrange the equation for the amount of substance (mol) and divide the volume by 1000 to convert cm³ to dm³
- Number of moles (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) = $\frac{amount (mol)}{volume ({dm}^{3})}$ = $\frac{0.00250}{0.0200}$ = 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:

$number of moles = \frac{mass of a substance (g)}{molar mass (g {mol}^{- 1})}$

The gas volumes can be used to deduce the stoichiometry of a reaction
- Eg. in the combustion of 50 cm³of propane reacting with 250 cm³of oxygen, 150 cm³of 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)

Reacting Masses & Volumes (CIE A Level Chemistry)

Revision Note

Reacting masses

Percentage yield

Excess & limiting reagents

Volumes of gases

Volumes & concentrations of solutions

Stoichiometric relationships

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1. Atomic Structure

1.1 Particles in the Atom & Atomic Radius

Atomic Structure & Subatomic Particles

Determining Subatomic Structure

Variations in Atomic & Ionic Radius

1.2 Isotopes

Isotopes

1.3 Electrons, Energy Levels & Atomic Orbitals

Electronic Structure

Sub-shells & Orbitals

Electronic Configuration

Determining Electronic Configuration

1.4 Ionisation Energy

Defining Ionisation Energy

Ionisation Energy Trends

Ionisation Energy & Electronic Configuration

2. Atoms, Molecules & Stoichiometry

2.1 Relative Masses of Atoms & Molecules

Relative Masses

2.2 The Mole & the Avogadro Constant

The Mole & the Avogadro Constant

2.3 Formulas

Formulae

Balancing Equations

Empirical & Molecular Formulae

Water of Crystallisation

2.4 Reacting Masses & Volumes (of Solutions & Gases)

3. Chemical Bonding

3.1 Electronegativity & Bonding

Electronegativity

Trends in Electronegativity

Electronegativity & Bonding

3.2 Ionic Bonding

Ionic Bonding

Ionic Bonding Examples

3.3 Metallic Bonding

Metallic Bonding

3.4 Covalent Bonding & Coordinate (Dative Covalent) Bonding

Covalent Bonding

Coordinate Bonding

Hybridisation

Bond Energy & Length

3.5 Shapes of Molecules

Shapes of Molecules

3.6 Intermolecular Forces, Electronegativity & Bond Properties

Hydrogen Bonding

Bond Polarity & Dipole Moments

Van der Waals' Forces

Inter & Intramolecular Forces

3.7 Dot-&-Cross Diagrams

Dot-&-Cross Diagrams

4. States of Matter

4.1 The Gaseous State: Ideal & Real Gases & pV = nRT

Gas Pressure

Ideal Gases

4.2 Bonding & Structure

Lattice Structures

Bonding & Structure

5. Chemical Energetics

5.1 Enthalpy Change, ΔH

Enthalpy Change, ΔH

Reaction Pathway Diagrams

Standard Enthalpy Change

Enthalpy & Bond Energies

5.2 Hess’s Law

Hess’s Law

6. Electrochemistry

6.1 Redox Processes: Electron Transfer & Changes in Oxidation Number (Oxidation State)