Chemistry Form 3 Notes : THE MOLE – Introduction to the mole, molar masses and Relative atomic masses

THE MOLE

Introduction to the mole, molar masses and Relative atomic masses

1. The mole is the SI unit of the amount of substance.

2. The number of particles, e.g., atoms, ions, molecules, electrons, cows, cars, are all measured in terms of moles.

3. The number of particles in one mole is called the Avogadro’s Constant. It is denoted “L“.

The Avogadro’s Constant contains 6.023 x 10²³ particles, i.e.,

1 mole = 6.023 x 10²³ particles = 6.023 x 10²³

2 moles = 2 x 6.023 x 10²³ particles = 1.205 x 10²⁴

0.2 moles = 0.2 x 6.023 x 10²³ particles = 1.205 x 10²²

0.0065 moles = 0.0065 x 6.023 x 10²³ particles = 3.914 x 10²¹

4. The mass of one mole of a substance is called molar mass. The molar mass of:

(i) an element has mass equal to relative atomic mass / RAM (in grams) of the element, e.g.,

Molar mass of carbon (C) = relative atomic mass = 12.0 g

6.023 x 10²³ particles of carbon = 1 mole = 12.0 g

Molar mass of sodium (Na) = relative atomic mass = 23.0 g

6.023 x 10²³ particles of sodium = 1 mole = 23.0 g

Molar mass of iron (Fe) = relative atomic mass = 56.0 g

6.023 x 10²³ particles of iron = 1 mole = 56.0 g

(ii) a molecule has mass equal to relative molecular mass / RMM (in grams) of the molecule. Relative molecular mass is the sum of the relative atomic masses of the elements making the molecule.

The number of atoms making a molecule is called atomicity. Most gaseous molecules are diatomic (e.g., O₂, H₂, N₂, F₂, Cl₂, Br₂, I₂). Noble gases are monoatomic (e.g., He, Ar, Ne, Xe). Ozone gas (O₃) is triatomic, e.g.,

Molar mass of Oxygen molecule (O₂) = relative molecular mass = (16.0 x 2) g = 32.0 g

6.023 x 10²³ particles of oxygen molecule = 1 mole = 32.0 g

Molar mass of chlorine molecule (Cl₂) = relative molecular mass = (35.5 x 2) g = 71.0 g

6.023 x 10²³ particles of chlorine molecule = 1 mole = 71.0 g

Molar mass of nitrogen molecule (N₂) = relative molecular mass = (14.0 x 2) g = 28.0 g

6.023 x 10²³ particles of nitrogen molecule = 1 mole = 28.0 g

(iii) a compound has mass equal to relative formula mass / RFM (in grams) of the molecule. Relative formula mass is the sum of the relative atomic masses of the elements making the compound, e.g.,

(i) Molar mass of Water (H₂O) = relative formula mass = [(1.0 x 2) + 16.0] g = 18.0 g

6.023 x 10²³ particles of water molecule = 1 mole = 18.0 g

6.023 x 10²³ particles of water molecule has:

• 2 x 6.023 x 10²³ particles of hydrogen atoms
• 1 x 6.023 x 10²³ particles of oxygen atoms

(ii) Molar mass of sulphuric(VI) acid (H₂SO₄) = relative formula mass = [(1.0 x 2) + 32.0 + (16.0 x 4)] g = 98.0 g

6.023 x 10²³ particles of sulphuric(VI) acid = 1 mole = 98.0 g

6.023 x 10²³ particles of sulphuric(VI) acid has:

• 2 x 6.023 x 10²³ particles of hydrogen atoms
• 1 x 6.023 x 10²³ particles of sulphur atoms
• 4 x 6.023 x 10²³ particles of oxygen atoms

(iii) Molar mass of sodium carbonate(IV) (Na₂CO₃) = relative formula mass = [(23.0 x 2) + 12.0 + (16.0 x 3)] g = 106.0 g

6.023 x 10²³ particles of sodium carbonate(IV) = 1 mole = 106.0 g

6.023 x 10²³ particles of sodium carbonate(IV) has:

• 2 x 6.023 x 10²³ particles of sodium atoms
• 1 x 6.023 x 10²³ particles of carbon atoms
• 3 x 6.023 x 10²³ particles of oxygen atoms

(iv) Molar mass of Calcium carbonate(IV) (CaCO₃) = relative formula mass = [(40.0) + 12.0 + (16.0 x 3)] g = 100.0 g

6.023 x 10²³ particles of calcium carbonate(IV) = 1 mole = 100.0 g

6.023 x 10²³ particles of calcium carbonate(IV) has:

• 1 x 6.023 x 10²³ particles of calcium atoms
• 1 x 6.023 x 10²³ particles of carbon atoms
• 3 x 6.023 x 10²³ particles of oxygen atoms

Practice

1. Calculate the number of moles present in:

(i) 0.23 g of sodium atoms

Molar mass of sodium atoms = 23 g

Moles = mass in grams / molar mass = 0.23 g / 23 = 0.01 moles

(ii) 0.23 g of chlorine atoms

Molar mass of chlorine atoms = 35.5 g

Moles = 0.23 g / 35.5 = 0.0065 moles (6.5 x 10^-3 moles)

(iii) 0.23 g of chlorine molecules

Molar mass of chlorine molecules = (35.5 x 2) = 71.0 g

Moles = 0.23 g / 71 = 0.0032 moles (3.2 x 10^-3 moles)

(iv) 0.23 g of dilute sulphuric(VI) acid

Molar mass of H₂SO₄ = [(2 x 1) + 32 + (4 x 16)] = 98.0 g

Moles = 0.23 g / 98 = 0.0023 moles (2.3 x 10^-3 moles)

2. Calculate the number of atoms present in: (Avogadro’s constant L = 6.0 x 10²³)

(i) 0.23 g of dilute sulphuric (VI) acid

Method I

Molar mass of H₂SO₄ = 98.0 g

Moles = 0.23 g / 98 = 0.0023 moles

1 mole has 6.0 x 10²³ atoms

0.0023 moles has (0.0023 x 6.0 x 10²³) = 1.38 x 10²¹ atoms

Method II

98.0 g = 1 mole has 6.0 x 10²³ atoms

0.23 g therefore has (0.23 g x 6.0 x 10²³) / 98 = 1.38 x 10²¹ atoms

(ii) 0.23 g of sodium carbonate(IV) decahydrate

Molar mass of Na₂CO₃.10H₂O = [(2 x 23) + 12 + (3 x 16) + (10 x 2) + (10 x 16)] = 286.0 g

Method I

Moles = 0.23 g / 286 = 0.0008 moles (8.3 x 10^-4 moles)

1 mole has 6.0 x 10²³ atoms

0.0008 moles has (0.0008 x 6.0 x 10²³) = 4.98 x 10²⁰ atoms

Method II

286.0 g = 1 mole has 6.0 x 10²³ atoms

0.23 g therefore has (0.23 g x 6.0 x 10²³) / 286 = 4.98 x 10²⁰ atoms

(iii) 0.23 g of oxygen gas

Molar mass of O₂ = (2 x 16) = 32.0 g

Method I

Moles = 0.23 g / 32 = 0.00718 moles (7.18 x 10^-3 moles)

1 mole has 2 x 6.0 x 10²³ atoms in O₂

0.00718 moles has (0.00718 x 2 x 6.0 x 10²³) = 8.616 x 10²¹ atoms

Method II

32.0 g = 1 mole has 2 x 6.0 x 10²³ atoms in O₂

0.23 g therefore has (0.23 g x 2 x 6.0 x 10²³) / 32 = 8.616 x 10²¹ atoms

(iv) 0.23 g of carbon(IV) oxide gas

Molar mass of CO₂ = [12 + (2 x 16)] = 44.0 g

Method I

Moles = 0.23 g / 44 = 0.00522 moles (5.22 x 10^-3 moles)

1 mole has 3 x 6.0 x 10²³ atoms in CO₂

0.00522 moles has (0.00522 x 3 x 6.0 x 10²³) = 9.396 x 10²¹ atoms

Method II

44.0 g = 1 mole has 3 x 6.0 x 10²³ atoms in CO₂

0.23 g therefore has (0.23 g x 3 x 6.0 x 10²³) / 44 = 9.409 x 10²¹ atoms

Empirical and molecular formula

1. The empirical formula of a compound is its simplest formula. It is the simplest whole number ratios in which atoms of elements combine to form the compound.

2. It is mathematically the lowest common multiple (LCM) of the atoms of the elements in the compound.

3. Practically, the empirical formula of a compound can be determined as in the following examples.

To determine the empirical formula of copper oxide

(a) Method 1: From copper to copper(II) oxide

Procedure:

Weigh a clean dry covered crucible (M₁). Put two spatula full of copper powder into the crucible. Weigh again (M₂). Heat the crucible on a strong Bunsen flame for five minutes. Lift the lid, and swirl the crucible carefully using a pair of tongs. Cover the crucible and continue heating for another five minutes. Remove the lid and stop heating. Allow the crucible to cool. When cool, replace the lid and weigh the contents again (M₃).

Sample results

Sample questions

1. Calculate the mass of copper powder used.

Mass of crucible + copper before heating (M₂) = 18.4 g

Less Mass of crucible (M₁) = -15.6 g

Mass of copper = 2.8 g

2. Calculate the mass of oxygen used to react with copper.

Method I

Mass of crucible + copper after heating (M₃) = 19.1 g

Mass of crucible + copper before heating (M₂) = -18.4 g

Mass of oxygen = 0.7 g

Method II

Mass of crucible + copper after heating (M₃) = 19.1 g

Mass of crucible = -15.6 g

Mass of copper(II) oxide = 3.5 g

Mass of copper = -2.8 g

Mass of oxygen = 0.7 g

3. Calculate the number of moles of:

(i) copper used (Cu = 63.5)

Number of moles of copper = mass used / molar mass = 2.8 / 63.5 = 0.0441 moles

(ii) oxygen used (O = 16.0)

Number of moles of oxygen = mass used / molar mass = 0.7 / 16.0 = 0.0441 moles

4. Determine the mole ratio of the reactants

Moles of copper = 0.0441 moles = 1

Moles of oxygen = 0.0441 moles = 1

Mole ratio Cu : O = 1 : 1

5. What is the empirical formula of copper oxide formed?

CuO (copper(II) oxide)

6. State and explain the observations made during the experiment.

Observation: Colour change from brown to black

Explanation: Copper powder is brown. On heating, it reacts with oxygen from the air to form black copper(II) oxide.

7. Explain why magnesium ribbon/shavings would be unsuitable in a similar experiment as the one above.

Hot magnesium generates enough heat energy to react with both oxygen and nitrogen in the air, forming a white solid mixture of magnesium oxide and magnesium nitride. This causes experimental mass errors.

(b) Method 2: From copper(II) oxide to copper

Procedure:

Weigh a clean dry porcelain boat (M₁). Put two spatula full of copper(II) oxide powder into the boat. Reweigh the porcelain boat (M₂). Put the porcelain boat in a glass tube and set up the apparatus as below;

Pass slowly (to prevent copper(II) oxide from being blown away) a stream of either dry hydrogen, ammonia, laboratory gas, or carbon(II) oxide gas for about two minutes from a suitable generator.

When all the air in the apparatus is driven out, heat the copper(II) oxide strongly for about five minutes until there is no further change. Stop heating.

Continue passing the gases until the glass tube is cool.

Turn off the gas generator.

Carefully remove the porcelain boat from the combustion tube.

Reweigh (M₃).

Sample results

Sample questions

1. Calculate the mass of copper(II) oxide used.

Mass of boat before heating (M₂) = 19.1 g

Mass of empty boat (M₁) = -15.6 g

Mass of copper(II) oxide = 3.5 g

2. Calculate the mass of:

(i) Oxygen

Mass of boat before heating (M₂) = 19.1 g

Mass of boat after heating (M₃) = -18.4 g

Mass of oxygen = 0.7 g

(ii) Copper

Mass of copper(II) oxide = 3.5 g

Mass of oxygen = 0.7 g

Mass of copper = 2.8 g

3. Calculate the number of moles of:

(i) Copper used (Cu = 63.5)

Number of moles of copper = mass used / molar mass = 2.8 / 63.5 = 0.0441 moles

(ii) Oxygen used (O = 16.0)

Number of moles of oxygen = mass used / molar mass = 0.7 / 16.0 = 0.0441 moles

4. Determine the mole ratio of the reactants

Moles of copper = 0.0441 moles = 1

Moles of oxygen = 0.0441 moles = 1

Mole ratio Cu : O = 1 : 1

5. What is the empirical formula of copper oxide formed?

CuO (copper(II) oxide)

6. State and explain the observations made during the experiment.

Observation: Colour change from black to brown

Explanation: Copper(II) oxide powder is black. On heating, it is reduced by a suitable reducing agent to brown copper metal.

7. Explain why magnesium oxide would be unsuitable in a similar experiment as the one above.

Magnesium is high in the reactivity series. None of the above reducing agents is strong enough to reduce the oxide to the metal.

8. Write the equation for the reaction that would take place when the reducing agent is:

(i) Hydrogen

CuO(s) + H₂(g) -> Cu(s) + H₂O(l)

(Black) (brown) (colourless liquid formed on cooler parts)

(ii) Carbon(II) oxide

CuO(s) + CO(g) -> Cu(s) + CO₂(g)

(Black) (brown) (colourless gas, forms white ppt with lime water)

(iii) Ammonia

3CuO(s) + 2NH₃(g) -> 3Cu(s) + N₂(g) + 3H₂O(l)

(Black) (brown) (colourless liquid formed on cooler parts)

9. Explain why the following is necessary during the above experiment:

(i) A stream of dry hydrogen gas should be passed before heating copper(II) oxide.

Air combines with hydrogen in presence of heat causing an explosion.

(ii) A stream of dry hydrogen gas should be passed after heating copper(II) oxide has been stopped.

Hot metallic copper can be re-oxidized back to copper(II) oxide.

(iii) A stream of excess carbon(II) oxide gas should be ignited to burn.

Carbon(II) oxide is highly poisonous/toxic. On ignition, it burns to form less toxic carbon(IV) oxide gas.

10. State two sources of error in this experiment.

(i) All copper(II) oxide may not be reduced to copper.

(ii) Some copper(II) oxide may be blown out of the boat by the reducing agent.

Theoretical determination of empirical formula

(a) An oxide of copper contains 80% by mass of copper. Determine its empirical formula. (Cu = 63.5, O = 16.0)

% of oxygen = 100% – % of copper = 100 – 80 = 20% oxygen

Empirical formula is CuO

(b) 1.60 g of an oxide of magnesium contains 0.84 g by mass of magnesium. Determine its empirical formula (Mg = 24.0, O = 16.0)

Mass of oxygen = 1.60 – 0.84 = 0.56 g

Empirical formula is MgO

(c) An oxide of silicon contains 47% by mass of silicon. What is its empirical formula? (Si = 28.0, O = 16.0)

Mass of oxygen = 100 – 47 = 53%

Empirical formula is SiO₂

(d) A compound contains 70% by mass of iron and 30% oxygen. What is its empirical formula? (Fe = 56.0, O = 16.0)

Mass of oxygen = 30%

Multiply both by 2 to get whole numbers: Fe₂O₃

Empirical formula is Fe₂O₃

Hydrated copper (II) sulphate(VI) crystals

During heating of hydrated copper (II) sulphate(VI) crystals, the following readings were obtained:

• Mass of evaporating dish = 300.0 g
• Mass of evaporating dish + hydrated salt = 305.0 g
• Mass of evaporating dish + anhydrous salt = 303.2 g

Calculate the number of water of crystallization molecules in hydrated copper (II) sulphate(VI) (Cu = 64.5, S = 32.0, O = 16.0, H = 1.0)

Working:

Mass of hydrated salt = 305.0 g – 300.0 g = 5.0 g

Mass of anhydrous salt = 303.2 g – 300.0 g = 3.2 g

Mass of water in hydrated salt = 5.0 g – 3.2 g = 1.8 g

Molar mass of water (H₂O) = 18.0 g

Molar mass of anhydrous copper (II) sulphate(VI) (CuSO₄) = 160.5 g

The empirical formula of hydrated salt = CuSO₄·5H₂O

Hydrated salt has five (5) molecules of water of crystallization.

Molecular formula

The molecular formula is the actual number of each kind of atom present in a molecule of a compound.

The empirical formula of an ionic compound is the same as the chemical formula, but for simple molecular structured compounds, the empirical formula may not be the same as the chemical formula.

The molecular formula is a multiple of the empirical formula. It is determined from the relationship:

(i) n = Relative formula mass / Relative empirical formula mass

where n is a whole number.

(ii) Relative empirical formula mass x n = Relative formula mass where n is a whole number.

Practice sample examples

1. A hydrocarbon was found to contain 92.3% carbon and the remaining hydrogen.

If the molecular mass of the compound is 78, determine the molecular formula (C = 12.0, H = 1.0)

Mass of hydrogen = 100 – 92.3 = 7.7%

Empirical formula is CH

The molecular formula is thus determined:

n = Relative formula mass / Relative empirical formula mass = 78 / 13 = 6

The molecular formula is (CH) x 6 = C₆H₆

2. A compound of carbon, hydrogen and oxygen contains 54.55% carbon, 9.09% hydrogen and remaining 36.36% oxygen.

If its relative molecular mass is 88, determine its molecular formula (C = 12.0, H = 1.0, O = 16.0)

Empirical formula is C₂H₄O

The molecular formula is thus determined:

n = Relative formula mass / Relative empirical formula mass = 88 / 44 = 2

The molecular formula is (C₂H₄O) x 2 = C₄H₈O₂

Molar gas volume

The volume occupied by one mole of all gases at the same temperature and pressure is a constant. It is:

(i) 24 dm³ / 24 litres / 24000 cm³ at room temperature (25°C / 298 K) and pressure (r.t.p).

i.e., 1 mole of all gases = 24 dm³ / 24 litres / 24000 cm³ at r.t.p.

Examples:

• 1 mole of O₂ = 32 g = 6.0 x 10²³ particles = 24 dm³ / 24 litres / 24000 cm³ at r.t.p.
• 1 mole of H₂ = 2 g = 6.0 x 10²³ particles = 24 dm³ / 24 litres / 24000 cm³ at r.t.p.
• 1 mole of CO₂ = 44 g = 6.0 x 10²³ particles = 24 dm³ / 24 litres / 24000 cm³ at r.t.p.
• 1 mole of NH₃ = 17 g = 6.0 x 10²³ particles = 24 dm³ / 24 litres / 24000 cm³ at r.t.p.
• 1 mole of CH₄ = 16 g = 6.0 x 10²³ particles = 24 dm³ / 24 litres / 24000 cm³ at r.t.p.

(ii) 22.4 dm³ / 22.4 litres / 22400 cm³ at standard temperature (0°C / 273 K) and pressure (s.t.p).

i.e., 1 mole of all gases = 22.4 dm³ / 22.4 litres / 22400 cm³ at s.t.p.

Examples:

• 1 mole of O₂ = 32 g = 6.0 x 10²³ particles = 22.4 dm³ / 22.4 litres / 22400 cm³ at s.t.p.
• 1 mole of H₂ = 2 g = 6.0 x 10²³ particles = 22.4 dm³ / 22.4 litres / 22400 cm³ at s.t.p.
• 1 mole of CO₂ = 44 g = 6.0 x 10²³ particles = 22.4 dm³ / 22.4 litres / 22400 cm³ at s.t.p.
• 1 mole of NH₃ = 17 g = 6.0 x 10²³ particles = 22.4 dm³ / 22.4 litres / 22400 cm³ at s.t.p.
• 1 mole of CH₄ = 16 g = 6.0 x 10²³ particles = 22.4 dm³ / 22.4 litres / 22400 cm³ at s.t.p.

The volume occupied by one mole of a gas at r.t.p or s.t.p is commonly called the molar gas volume. Whether the molar gas volume is at r.t.p or s.t.p must always be specified.

From the above, therefore, a lesser or greater volume can be determined as in the examples below.

Practice examples

1. Calculate the number of particles present in:

(Avogadro’s constant = 6.0 x 10²³ mole^-1)

(i) 2.24 dm³ of oxygen.

22.4 dm³ -> 6.0 x 10²³

2.24 dm³ -> (2.24 x 6.0 x 10²³) / 22.4 = 6.0 x 10²² molecules = 2 x 6.0 x 10²² = 1.2 x 10²³ atoms

(ii) 2.24 dm³ of carbon(IV) oxide.

22.4 dm³ -> 6.0 x 10²³

2.24 dm³ -> (2.24 x 6.0 x 10²³) / 22.4 = 6.0 x 10²² molecules = (3 x 6.0 x 10²²) = 1.8 x 10²³ atoms

2. 0.135 g of a gaseous hydrocarbon X on complete combustion produces 0.41 g of carbon(IV) oxide and 0.209 g of water. 0.29 g of X occupy 120 cm³ at room temperature and 1 atmosphere pressure. Name X and draw its molecular structure. (C = 12.0, O = 16.0, H = 1.0, 1 mole of gas occupies 24 dm³ at r.t.p)

Molar mass CO₂ = 44 g mol^-1

Molar mass H₂O = 18 g mol^-1

Molar mass X = (0.29 x 24 x 1000) / 120 = 58 g mol^-1

Mass of carbon in CO₂ = (12 x 0.41) / 44 = 0.1118 g

Mass of hydrogen in H₂O = (2 x 0.209) / 18 = 0.0232 g

Empirical formula is C₁H_2.5, multiply by 2 to get whole numbers: C₂H₅

Molecular formula is C₂H₅

Molecule name: Butane

Molecular structure:

H H H H H H

H C C C C C C H

H H H H