### The Mole & Avogadro’s Constant

**The mole**

- This is the mass of a substance containing the same number of fundamental units as there are atoms in exactly 12.000 g of
^{12}C - The mole is the unit representing the amount of atoms, ions, or molecules
- One mole is the amount of a substance that contains
**6.02 x 10**particles (Atoms, Molecules or Formulae) of a substance (6.02 x 10^{23}^{23 }is known as the**Avogadro Number)**

**Examples**

- 1 mole of Sodium (Na) contains 6.02 x 10
^{23}**Atoms**of Sodium - 1 mole of Hydrogen (H
_{2}) contains 6.02 x 10^{23}**Molecules**of Hydrogen - 1 mole of Sodium Chloride (NaCl) contains 6.02 x 10
^{23 }**Formula units**of Sodium Chloride

**Linking the mole and the atomic mass**

- One mole of any element is equal to the relative atomic mass of that element in grams
- For example one mole of carbon, that is if you had 6.02 x 10
^{23}atoms of carbon in your hand, it would have a mass of 12g - So one mole of helium atoms would have a mass of 4g, lithium 7g etc
- For a compound we add up the relative atomic masses
- So one mole of water would have a mass of 2 x 1 + 16 = 18g
- Hydrogen which has an atomic mass of 1 is therefore equal to
^{1}/_{12}the mass of a^{12}C atom - So one carbon atom has the same mass as 12 hydrogen atoms

### The Mole & the Volume of Gases

**Molar volume**

- This is the volume that one mole of any gas (be it molecular such as CO
_{2}or monoatomic such as helium) will occupy - It’s value is 24dm
^{3}or 24,000 cm^{3}at room temperature and pressure (r.t.p.)

**Calculations involving gases**

**General equation:**

Amount of gas (mol) = Volume of gas (dm^{3}) ÷ 24

or

Amount of gas (mol) = Volume of gas (cm^{3}) ÷ 24000

**1. Calculating the volume of gas that a particular amount of moles occupies**

**Equation:**

Volume of gas (dm^{3}) = Amount of gas (mol) x 24

or

Volume of gas (cm^{3}) = Amount of gas (mol) x 24000

**Example:**

**2. Calculating the moles in a particular volume of gas**

**Equation:**

Amount of gas (mol) = Volume of gas (dm^{3}) ÷ 24

or

Amount of gas (mol) = Volume of gas (cm^{3}) ÷ 24000

**Example:**

### Calculating Reacting Masses, Solutions & Concentrations of Solutions in g/dm3 & mol/dm3

**Calculating percentage composition, moles, mass and relative formula mass**

*Formula triangle for moles, mass and formula mass*

* *

**1. Calculating Moles**

**Equation:**

Amount in Moles = Mass of Substance in grams ÷ M_{r}(or A_{r})

**Example:**

**2. Calculating Mass**

**Equation:**

Mass of substance (grams) = Moles x M_{r}(or A_{r})

**Example:**

**3. Calculating Relative Formula Mass**

**Equation:**

M_{r}(or A_{r}) = Mass of Substance in Grams ÷ Moles

**Example:**

10 moles of Carbon Dioxide has a Mass of 440 g. What is the Relative Formula Mass of Carbon Dioxide?

Relative Formula Mass = Mass ÷ Number of Moles

Relative Formula Mass = 440 ÷ 10 = 44

**Relative Formula Mass of Carbon Dioxide =** 44

** **

**4. Calculating Percentage Composition**

- The percentage composition is found by calculating the percentage by mass of each particular element in a compound

** ****Example:**

Calculate the percentage of oxygen in CO_{2}

**Step 1 – **Calculate the molar mass of the compound

Molar mass CO_{2} = (2 x 16) + 12 = 44

**Step 2 – **Add the atomic masses of the element required as in the question (oxygen)

16 + 16 = 32

**Step 3 – **Calculate the percentage

% of oxygen in CO_{2} = 32/44 x 100 = 72.7%

** **

**Calculations of solutions: moles, concentration and volume**

**General Equation:**

Concentration (mol / dm^{3}) = Amount of substance (mol) ÷ Volume of solution (dm^{3})

This general equation is rearranged for the term as is asked in the question.

**1. Calculating Moles**

**Equation:**

Amount of Substance (mol) = Concentration x Volume of Solution (dm^{3})

**Example:**

Calculate the Moles of Solute Dissolved in 2 dm^{3} of a 0.1 mol / dm^{3} Solution

**Concentration of Solution : **0.1 mol / dm^{3}

**Volume of Solution : **2 dm^{3}

Moles of Solute = 0.1 x 2 = 0.1 mol (the dm^{3} above and below the line cancel out)

**Amount of Solute =** 0.2 mol

**2. Calculating Concentration**

**Equation:**

Concentration (mol / dm^{3}) = Amount of substance (mol) ÷ Volume of solution (dm^{3})

**Example:**

25.0 cm^{3} of 0.050 mol / dm^{3} sodium carbonate was completely neutralised by 20.00 cm^{3} of dilute hydrochloric acid. Calculate the concentration in mol / dm^{3} of the hydrochloric acid.

**Step 1 – **Calculate the amount, in moles, of sodium carbonate reacted by rearranging the equation for amount of substance (mol) and dividing by 1000 to convert cm^{3} to dm^{3}

Amount of Na_{2}CO_{3} = (25.0 x 0.050) ÷ 1000 = 0.00125 mol

**Step 2 – **Calculate the amount, in moles, of hydrochloric acid reacted

1 mol of Na_{2}CO_{3} reacts with 2 mol of HCl, so the Molar Ratio is 1 : 2

Therefore 0.00125 moles of Na_{2}CO_{3} react with 0.00250 moles of HCl

**Step 3 – **Calculate the concentration, in mol / dm^{3} of the Hydrochloric Acid

1 dm^{3} = 1000 cm^{3}

^{ }Volume of HCl = 20 ÷ 1000 = 0.0200 dm^{3}

Concentration HCl (mol / dm^{3}) = 0.00250 ÷ 0.0200 = 0.125

**Concentration of Hydrochloric Acid =** 0.125 mol / dm^{3}

**3. Calculating Volume**

**Equation:**

Volume (dm^{3}) = Amount of substance (mol) ÷ Concentration (mol / dm^{3})

**Example:**

Calculate the volume of hydrochloric acid of concentration 1.0 mol / dm^{3} that is required to react completely with 2.5g of calcium carbonate.

**Step 1 – **Calculate the amount, in moles, of calcium carbonate that reacts

M_{r} of CaCO_{3} is 100

Amount of CaCO_{3} = (2.5 ÷ 100) = 0.025 mol

**Step 2 – **Calculate the moles of hydrochloric acid required

CaCO_{3} + 2HCl → CaCl_{2} + H_{2}O + CO_{2}

1 mol of CaCO_{3} requires 2 mol of HCl

So 0.025 mol of CaCO_{3} Requires 0.05 mol of HCl

**Step 3 – **Calculate the volume of HCl Required

Volume = (Amount of Substance(mol) ÷ Concentration (mol / dm^{3})

= 0.05 ÷ 1.0

= 0.05 dm^{3 }(the moles cancel out above and below the line)

**Volume of Hydrochloric Acid =** 0.05 dm^{3}

**The limiting reactant and reacting masses**

**Limiting reactant**

- The limiting reactant is the reactant which is
**not present in excess**in a reaction - It is always the first reactant to be used up which then causes the reaction to stop
- In order to determine which reactant is the limiting reagent in a reaction, we have to consider the ratios of each reactant in the balanced equation

**Example:**

9.2g of sodium is reacted with 8.0g of sulfur to produce sodium sulfide, NaS. Which reactant is in excess and which is the limiting reactant?

**Step 1 – **Calculate the moles of each reactant

Moles = Mass ÷ A_{r }

Moles Na = 9.2/23 = 0.40

Moles S = 8.0/32 = 0.25

**Step 2 – **Write the balanced equation and determine the molar ratio

2Na + S → Na_{2}S so the molar ratios is 2 : 1

**Step 3 – **Compare the moles. 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.

**Calculating reacting masses**

- Chemical equations can be used to calculate the moles or masses of reactants and products
- Use information from the question to find the amount in moles of the substances being considered
- Identify the ratio between the substances using the balanced chemical equation
- Apply mole calculations to find answer

**Example 1:**

**Calculate the Mass of Magnesium Oxide that can be made by completely burning 6 g of Magnesium in Oxygen**

Magnesium (s) + Oxygen (g)→Magnesium Oxide (s)

**Symbol Equation: **

2Mg + O_{2}→ 2MgO

**Relative Formula Mass: **Magnesium : 24 Magnesium Oxide : 40

**Step 1 – **Calculate the moles of Magnesium Used in reaction

Moles = Mass ÷ M_{r }Moles = 6 ÷ 24 = 0.25

**Step 2 – **Find the Ratio of Magnesium to Magnesium Oxide using the balanced Chemical Equation

**Step 3 – **Find the Mass of Magnesium Oxide

Moles of Magnesium Oxide = 0.25

Mass = Moles x M_{r }Mass = 0.25 x 40 = 10 g

**Mass of Magnesium Oxide Produced =** 10 g

**Example 2:**

**Calculate the Mass, in Tonnes, of Aluminium that can be Produced from 51 Tonnes of Aluminium Oxide**

Aluminium Oxide (s)→Aluminium (s) + Oxygen (g)

**Symbol Equation:**

2Al_{2}O_{3 }→ 4Al + 3O_{2}

**A _{r} and M_{r}: **Aluminium : 27 Oxygen : 16 Aluminium Oxide : 102

**1 Tonne = 10 ^{6} g**

**Step 1 – **Calculate the moles of aluminium oxide used

Mass of Aluminium Oxide in Grams = 51 x 10^{6} = 51,000,000 g

Moles = Mass ÷ A_{r} Moles = 51,000,000 ÷ 102 = 500,000

**Step 2 – **Find the ratio of aluminium oxide to aluminium using the balanced chemical equation

**Step 3 – **Find the mass of aluminium

Moles of aluminium = 1,000,000

Mass in grams = Moles x A_{r }Mass = 1,000,000 x 27 = 27,000,000

Mass in Tonnes = 27,000,000 ÷ 10^{6} = 27 Tonnes

**Mass of Aluminium Produced =** 27 Tonnes

### Using the Mole to Determine Empirical & Molecular Formulae

**Empirical formula: **gives the **simplest whole number ratio** of atoms of each element in the compound

- Calculated from knowledge of the ratio of masses of each element in the compound

**Example:**

A compound that contains 10 g of Hydrogen and 80 g of Oxygen has an Empirical Formula of H_{2}O. This can be shown by the following calculations:

Amount of Hydrogen Atoms = Mass in grams ÷ A_{r} of Hydrogen = (10 ÷ 1) = **10 moles**

Amount of Oxygen Atoms = Mass in grams ÷ A_{r} of Oxygen = (80 ÷ 16) = **5 moles**

**The ratio of moles of hydrogen atoms to moles of oxygen atoms:**

Since equal numbers of moles of atoms contain the same number of atoms, the ratio of hydrogen atoms to oxygen atoms is 2:1

Hence the **empirical formula** is H_{2}O

**Molecular formula: **gives the exact numbers of atoms of each element present in the formula of the compound

- Divide 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

**Relationship between Empirical and Molecular Formula:**

**Example:**

The Empirical Formula of X is C_{4}H_{10}S_{1} and the Relative Formula Mass of X is 180. What is the Molecular Formula of X?

**Relative Formula Mass: **Carbon : 12 Hydrogen : 1 Sulfur : 32

**Step 1 – **Calculate Relative Formula Mass of Empirical Formula

(C x 4) + (H x 10) + (S x 1) = (12 x 4) + (1 x 10) + (32 x 1) = 90

**Step 2 – **Divide Relative Formula Mass of X by Relative Formula Mass of Empirical

Formula

180 / 90 = 2

**Step 3 – **Multiply Each Number of Elements by 2

(C_{4 x 2}) + (H_{10 x 2}) + (S_{1 x 2}) = (C_{8}) + (H_{20}) + (S_{2})

**Molecular Formula of X = **C_{8}H_{20}S_{2}

### Calculating Percentage Yield & Percentage Purity of the Product

**Percentage yield**

- This is the calculation of the percentage yield obtained from the theoretical yield
- In practice, you never get 100% yield in a chemical process for several reasons
- These include some reactants being left behind in the equipment, the reaction may be reversible or product may also be lost during separation stages

**Equation:**

Percentage Yield = (Yield Obtained / Theoretical Yield) x 100

**Example:**

In an experiment to displace copper from copper sulfate, 6.5 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 4.8 g. Calculate the percentage yield of copper.

**Equation Of Reaction: **

Zn (s) + CuSO_{4}(aq)→ ZnSO_{4}(aq) + Cu (s)

**Step 1: **Calculate the Amount, in Moles of Zinc Reacted

Moles of Zinc = 6.5 ÷ 65 = 0.10 moles

**Step 2: **Calculate the Maximum Amount of Copper that could be formed from the

Molar ratio

Maximum Moles of Copper = 0.10 moles (Molar ratio is 1:1)

**Step 3: **Calculate the Maximum Mass of Copper that could be Formed

Maximum Mass of Copper = ( 0.10 x 64 ) = 6.4 g

**Step 4: **Calculate the Percentage of Yield of Copper

Percentage Yield = ( 4.8 ÷ 6.4 ) x 100 = 75%

**Percentage Yield of Copper = **75%

**Percentage purity**

- Often the product you are trying to fabricate may become contaminated with unwanted substances such as unreacted reactants, catalysts etc.

**Equation:**

Percentage Purity = (Mass of pure substance / Mass of impure substance) x 100

**Example:**

In an experiment 7.0g of impure calcium carbonate were heated to a very high temperature and 2.5g of carbon dioxide were formed. Calculate the percentage purity of the calcium carbonate.

**Equation Of Reaction:**

CaCO_{3}(s) → CaO(s) + CO_{2}(g)

**Step 1: **Calculate the relative formula masses

1 mole CaCO_{3} → 1 mole CO_{2}

40+12+(3×16) 12+(2×16)

100 → 44

**Step 2: **Calculate the theoretical mass of calcium carbonate used if pure

From 2.5g CO_{2} we would expect 2.5/44 x 100 = 5.68g

**Step 3: **Calculate the percentage purity

(Mass of pure substance / mass of impure substance) x 100

= 5.68/7.0 x 100

= 81.1%