Mass/Volume Relationship: mole and Molar quantities

CONTENT

• Mole and Molar Quantities
• Relative Atomic Mass and Relative Molecular Mass.
• Calculations involving Mass and Volume.

MOLE AND MOLAR QUANTITIES

THE MOLE

A mole is a number of particles of a substance which may be atoms, ions, molecules or electrons. This number of particles is approximately 6.02 x 10²³ in magnitude and is known as Avogadro’s number of particles.

The mole is defined as the amount of a substance which contains as many elementary units as there are atoms in 12g of Carbon-12.

RELATIVE ATOMIC MASS

The relative atomic mass of an element is the number of time the average mass of one atom of that element is heavier than one twelfth the mass of one atom of Carbon-12. It indicates the mass of an atom of an element. For e.g, the relative atomic mass of hydrogen, oxygen, carbon, sodium and calcium are 1, 16, 12, 23, and 40 respectively.

The atomic mass of an element contains the same number of atoms which is 6.02 x 10²³atoms; 1 mole of hydrogen having atomic mass of 2.0g contains 6.02 x 10²³ atoms.

RELATIVE MOLECULAR MASS

The relative molecular mass of an element or compound is the number of times the average mass of one molecule of it is heavier than one-twelfth the mass of one atom of Carbon-12

It is the sum of the relative atomic masses of all atoms in one molecule of that substance. It is also called the formula mass. The formula mass refers not only to the relative mass of a molecule but also that of an ion or radical.

CALCULATION

Calculate the relative molecular mass of:

1. Magnesium chloride
2. Sodium hydroxide
3. Calcium trioxocarbonate

       [Mg=24, Cl=35.5, Na=23, O=16, H=1, Ca=40,C=12]

Solution:

1. MgCl ₂ = 24 + 35.5×2 = 24 + 71 = 95gmol^-1
2. NaOH = 23 + 16 + 1 = 40gmol^-1
3. CaCO₃ = 40 + 12 +16×3 = 100gmol^-1

EVALUATION

1. What is relative molecular mass of a compound?
2. Calculate the relative molecular mass of (a) NaNO₃ (b) CuSO₄.5H₂O

MOLAR VOLUME OF GASES

The volume occupied by 1 mole of a gas at standard conditions of temperature and pressure (s.t.p) is 22.4 dm³. Thus 1 mole of oxygen gas of molar mass 32.0gmol^-1 occupies a volume of 22.4dm³ at s.t.p and 1 mole of helium gas of molar mass 4.0gmol^-1 occupies a volume of 22.4 dm³ at s.t.p.

Note: When the conditions of temperature and pressure are altered, the molar volume will also change. Also, standard temperature = 273K and standard pressure = 760mmHg.

RELATIONSHIP BETWEEN QUANTITIES

EVALUATION

1. Define the molar volume of a gas
2. How many molecules are contained in 1.12dm³ of hydrogen gas at s.t.p?

STOICHIOMETRY OF REACTION

The calculation of the amounts (generally measured in moles or grams) of reactants and products involved in a chemical reaction is known as stoichiometry of reaction. In other words, the mole ratio in which reactants combine and products are formed gives the stoichiometry of the reactions.

From the stoichiometry of a given balanced chemical equation, the mass or volume of the reactant needed for the reaction or products formed can be calculated.

CALCULATION OF MASSES OF REACTANTS AND PRODUCTS

EVALUATION

1. What does the term ‘Stoichiometry of reaction’ mean?

Ethane [C₂H₆] burns completely in oxygen. What amount in moles of CO₂will be produced when 6.0g of ethane are completely burnt in oxygen?

CALCULATION OF VOLUME OF REACTING GASES

1. In an experiment, 10cm³ of ethene [C₂H₄] was burnt in 50cm³ of oxygen.
2. Which gas was supplied in excess? Calculate the volume of the excess gas remaining at the end of the reaction.
3. Calculate the volume of CO₂ gas produced

Solution:

The equation for the reaction is:

C₂H_4(g) + 3O_2(g) → 2CO_2(g) + 2H₂O_(g)

1. From the equation,

1 mole of ethene reacts with 3mole of oxygen

1 volume of ethene reacts with 3 volumes of oxygen

10cm³ of ethene will react with 30cm³ of oxygen

Since 50cm³ of oxygen was supplied, oxygen was in excess

Hence volume of the excess gas = initial volume – volume used up = 50-30 = 20cm³

• 1 volume of ethene produces 2 volumes of CO₂

10 cm3 of ethene will produce 20cm3 of CO₂

Therefore, 20cm³ of CO₂ was produced

• 20cm³ of CO was mixed and sparked with 200cm³ of air containing 21% of O₂. If all the volumes are measured at s.t.p, calculate the total volume of the resulting gases.

Solution:

In 200cm³ of air,

Volume of O₂ = 21   x 200cm³ = 42cm³

                        100

Volume of N₂ and rare gases = 200-42 = 158cm³

The equation for the reaction is:

                        2CO_(g) + O_2(g) → 2CO_2(g)

Volume ratio                2       :   1     :     2

Before sparking        20cm³   42cm³

Reacting volume        20cm³   10cm³

After sparking                    32cm³   20cm³

Volume of resulting gases = 32 + 20 + 158 = 210cm³

GENERAL EVALUATION/REVISION

1. Find the volume of oxygen produced by 1 mole of KClO₃ at s.t.p in the following reaction: 2KClO_3(s)  → 2KCl_(s) + 30_2(g)
2. Define the term ‘Relative atomic mass’
3. Balance the following redox equations I^–  +  MnO₄^–        →             IO₃^–  +  MnO₂ in basic medium
4. Write the symbols of the following elements: mercury, silver, gold, lead, tin, antimony.
5. Define valency.

READING ASSIGNMENT

New School Chemistry for Senior Secondary School byO. Y. Ababio, Pg 156-164

WEEKEND ASSIGNMENT

SECTION A: Write the correct option ONLY

1. Amount of a substance is expressed in a. mole b. grams c. kilograms d. mass
2. Determine the mass of CO₂ produced by burning 104g of ethyne [C₂H₂]a. 256g b.352g c. 416g d. 512g
3. The mole ratio in which reactants combine and products are formed is known as a. rate of reaction b. stoichiometry of reaction C. equation of reaction d. chemical reaction
4. The unit for relative molecular mass is a. mole b. gmol^-1 c. grams d. mass
5. What mass of Pb(NO₃)₂ would be required to 9g of PbCl₂ on the addition of excess NaCl solution? [Pb=207, Na=23, O=16, N=14] a. 10.7g b. 1.2g c. 6.4g d. 5.2g

SECTION B

1. Calculate the number of molecules of CO₂ produced when 10g of CaCO₃ is treated with 100cm³ of 0.20moldm^-3HCl
2. Calculate the volume of nitrogen that will be produced at s.t.p from the decomposition of 9.60g ammonium dioxonitrate (III), NH₄NO₂.