Mole ConceptChapter 5

LEARNING OUTCOMES

Define the mole Derive empirical and molecular formulae State Avogadro’s Law Apply the mole concept to ionic and molecular

equations

A mole is the number of atoms or molecules in 1 g of hydrogen or 12 g of carbon.

The number 6 x 1023 is called one mole or Avogadro’s constant in honour of Amedeo Avogadro.

1 mole of atoms of any element will have a mass equal to its relative atomic mass, expressed in grams.

Introducing the Mole

Chapter 5Mole Concept

Chapter 5

Formulas

1. Mass of 1 mole of atoms = Ar in grams

2. Number of moles of atoms =

3. Mass of 1 mole of molecules = Mr in grams

4. Number of moles of molecules =

5. Mass of substance containing 1 mole of particles = Molar mass

6. Percentage yield =

Mass of the element in grams Relative atomic mass, Ar

Mass of the substance in grams Relative molecular mass, Mr

Actual mass of product obtained Theoretical mass of product obtainable

Mole Concept

The empirical formula is the simplest formula. It shows the simplest ratio of the elements present in a

compound. Examples of empirical formulae are:

H2O, CO2, H2SO4, CH2, CH3

The following are not empirical formulae: (a) C2H4, (b) C2H6, (c) C2H4O2

because they can be reduced to:(a) CH2, (b) CH3, (c) CH2O

Chapter 5

Empirical Formula

Mole Concept

The molecular formula is the true formula. It shows all the atoms present in the molecule. Examples of molecular formulae are:

H2O, H2O2, CO2, H2SO4, Cu(NO3)2

Note that H2O is water, and the molecular formula is the same as the empirical formula.

H2O2 is hydrogen peroxide. Its empirical formula is HO.

Chapter 5

Molecular Formula

Mole Concept

Let the molecular formula of propene be (CH2)n. Since the Mr is 42, (12 + 1x2)n = 42

14n = 42 n = 42 = 3 14

Hence the molecular formula is 3x(CH2) = C3H6

Chapter 5

Finding the Molecular Formula

Propene has the empirical formula CH2. The relative molecular mass of propene is 42. Find the molecular formula of propene.

Worked example 1

Solution

Mole Concept

C : H (a) 85.7 : 14.3 85.7 : 14.3

12 1 7.14 : 14.3 7.14 7.14 1 : 2 The empirical formula is CH2. (b) Let the molecular formula be (CH2)n. Since the Mr is 56, (12 + 1x2)n = 56

14n = 56 n = 56 = 4 14

Hence the molecular formula is 4 x (CH2) = C4H8

Chapter 5

Finding the Molecular FormulaWorked example 1

Solution

A hydrocarbon consists of 85.7% of carbon and 14.3% of hydrogen by mass. (a) Find the empirical formula of the compound. (b) If the molecular mass is 56, find the molecular formula.

Mole Concept

Formula of a compound A pure compound has a fixed chemical composition;

hence it can be represented by a chemical formula. For example, a molecule of water is made up of 2

atoms of hydrogen and 1 atom of oxygen, and its molecular formula is H2O.

We can find the formula of a compound from its percentage composition.

Chapter 5

Mole Concept

Step 4 : Write down the formula: Na2SO4

A compound of sodium contains the following percentage composition by mass: 32.4% of sodium, 22.6% of sulphur and 45.0% of oxygen. Find the formula of the compound.

Worked example 1

Solution Na : S : O

Step 1: Write down the percentage: 32.4 : 22.6 : 45.0Step 2: Divide each by the Ar: 32.4 : 22.6 : 45.0 (to get number of moles) 23 32 16 1.41 : 0.706 : 2.81Step 3: Divide by the smallest number: 1.41 : 0.706 : 2.81 0.706 0.706 0.706 2 : 1 : 4

Chapter 5

Finding the formula of a compound

Mole Concept

Step 4 : Write down the formula: CO2H2 (or HCOOH)

A compound contains 1.2 g of carbon, 3.2 g of oxygen and 0.2g ofhydrogen. Find the formula of the compound.

C : O : HStep 1: Write down the mass ratio: 1.2 : 3.2 : 0.2Step 2: Divide each mass by the Ar: 1.2 : 3.2 : 0.2 12 16 1 0.1 : 0.2 : 0.2Step 3: Divide by the smallest number: 0.1 : 0.2 : 0.2 0.1 0.1 0.1 1 : 2 : 2

Chapter 5

Finding the formula of a compoundWorked example 2

Solution

Mole Concept

Step 1: Find the % of hydrogen: 100 – 48.6 – 43.2 = 8.2 %

Step 2: Write down the % ratio: C : O : H 48.6 : 43.2 : 8.2Step 3: Divide each mass by the Ar: 48.6 : 43.2 : 8.2 12 16 1 4.05 : 2.7 : 8.2Step 4: Divide by the smallest number: 4.05 : 2.7 : 8.2 2.7 2.7 2.7 1.5 : 1 : 3Step 5: Multiply each number by 2: 3 : 2 : 6Step 6 : Write down the formula: C3O2H6 (or C2H5COOH)

A compound contains 48.6% of carbon, 43.2% of oxygen, with the remainder being hydrogen. Find the formula of the compound.

Chapter 5

Finding the formula of a compoundWorked example 3

Solution

Mole Concept

Chapter 5

Molar Volume of GasesAvogadro’s Law states that equal volume of gases under the same temperature and pressure contain the same number of molecules.

Volume of 1 mole of gas = 24 dm3

Volume of gas = Number of moles x 24 dm3

Number of moles = Volume of gas in dm3

24 dm3

Mole Concept

Limiting Reactants 2H2(g) + O2(g) 2H2O(g)

2 moles of hydrogen gas react with one or more moles of oxygen to form 2 moles of steam or water vapour.

Therefore we say that oxygen is in excess and hydrogen is called the limiting reactant because the reaction stops when hydrogen is used up.

Chapter 5

Mole Concept

Find the formula of each of the following:1. A compound containing 75% carbon and 25% hydrogen by mass.2. A compound containing 46.7% silicon and 53.3% oxygen by mass.3. A compound consisting of 43.4% sodium, 11.3% carbon and 45.3%

oxygen by mass.4. A compound consisting of 2.8 g of iron combined with 1.2 g of oxygen.5. (a) A compound containing 18.9% lithium,

64.9% oxygen and the rest carbon by mass.(b) Give the name of this compound.

Solution

Chapter 5

Quick check 1

Mole Concept

1. A hydrocarbon consists of 80% carbon and 20% hydrogen by mass. (a) Find the empirical formula of the compound.(b) If the molecular mass is 30, find the molecular formula.

2. An acid contains 40% carbon, 6.67% hydrogen and 53.33% oxygen by mass. (a) Find the empirical formula of the acid. (b) If the molecular mass of the acid is 60, what is its molecular formula?

3. A compound called borazine has the following percentage composition by mass: 40.74% boron, 51.85% nitrogen and the rest hydrogen. Find the molecular formula of borazine, given it has a relative molecular mass of 81.

Solution

Chapter 5

Quick check 2

Mole Concept

1. CH4

2. SiO2

3. Na2CO3

4. Fe2O3

5. (a) Li2CO3

(b) lithium carbonate

Return

Chapter 5

Solution to Quick check 1

Mole Concept

1. (a) Empirical formula: CH3 (b) Molecular formula: C2H6

2. (a) Empirical formula: CH2O (b) Molecular formula: C2H4O2 (CH3COOH)

3. Molecular formula: B3N3H6

Chapter 5

Solution to Quick check 2

Return

Mole Concept