Stoichiometry of Reactions I
Revision

Stoichiometry is simply the quantitative relationship between reactants and products.

Relative atomic mass

The relative atomic mass, A of an element is the number of times the average mass of one atom of an element is heavier than one twelfth the mass of one atom of carbon-12. The relative atomic mass is measured accurately using a mass spectrometer.

Element Atomic Number Atomic Mass
Hydrogen 1 1.008
Helium 2 4.0026
Lithium 3 6.94
Beryllium 4 9.0122
Boron 5 10.81
Carbon 6 12.01
Nitrogen 7 14.01
Oxygen 8 16.00
Fluorine 9 19.00
Neon 10 20.18
Sodium 11 22.99
Magnesium 12 24.31
Aluminum 13 26.98
Silicon 14 28.09
Phosphorus 15 30.97
Sulfur 16 32.07
Chlorine 17 35.45
Argon 18 39.95
Potassium 19 39.10
Calcium 20 40.08
Scandium 21 44.96
Titanium 22 47.87
Vanadium 23 50.94
Chromium 24 51.99
Manganese 25 54.94
Iron 26 55.85
Cobalt 27 58.93
Nickel 28 58.69
Copper 29 63.55
Zinc 30 65.38
Relative Molecular Mass

The relative molecular mass, M of an element or a 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 individual relative atomic masses of the constituent elements that make up the compound.

Example 1 : Calculate the relative molecular mass of Baking soda NaHCO3(Na = 23, H = 1, C = 12, O = 16)

Solution

M = 23 + 1 + 12 + (16 × 3)
= 23 + 1 + 12 + 48
= 84g/mol



Example 2: Calculate the Relative atomic mass of copper tetraoxosulphate(VI) decahydrate CuSO4.10H20
(Cu = 63.5, S = 32, O = 16, H = 1)

Solution

M = 63.5 + 32 + (16 × 3) + 10(1 ×2 +16)
= 63.5 + 32 + 48 + 180
= 323.5g/mol

Mole

One mole of a substance is the amount containing as many elementary entities as the number of atoms in exactly 12 grams of carbon-12. It is denoted with n.

Formulas for number of mole
  1. $$\text{number of moles} = \frac{m}{M}$$
  2. $$\text{number of moles} = \frac{CV}{1000}$$
  3. $$\text{number of moles= CV}$$
  4. $$n = \frac{PV}{RT}$$
  5. $$n = \frac{\text{Number of particles}}{Avogadro's number}$$
    6. where m = mass in gram
    M = molar mass in g/mol
    P = pressure in atm
    C = Concentration in mol/dm³
    V = Volume in cm³ or dm³
    R = molar gas constant
    T = Temperature in kelvin
    Avogadro's number = 6.02 × 10²³
    Note: n = CV when V is in dm³
    n = CV/1000 when V is in cm³

Example 1: Find the number of moles present in 20g of CaCO3

Solution

$$\text{number of moles} = \frac{m}{M}$$ $$\text{M of }CaCO_3= 40 + 12 + 48$$ $$\text{M} = 100g/mol$$ $$\text{number of moles} = \frac{20}{100}$$ $$\text{number of moles} = 0.2 \text{ mol}$$

Example 2: What is the concentration of a solution containing 2g of NaOH in 100Cm³ of solution? [Na = 23, H =1, O = 16](Jamb)

Solution

$$\text{number of moles} = \frac{CV}{1000}$$ $$\text{number of moles} = \frac{m}{M}$$ $$\text{M of NaOH} = 23 + 16 + 1 $$ $$ = 40g/mol $$ $$\text{number of mole} = \frac{2}{40} $$ $$ = 0.05 mol $$ $$\text{making C subject formula }$$ $$ C = \frac{1000n}{V} $$ $$ C = \frac{1000 × 0.05}{100} $$ $$ Concentration = 0.5moldm^-³ $$

Example 3: What is the molar mass of a substance, if 0.4 mole of the Substance has a mass of 25.0g (Jamb)

Solution

$$\text{number of moles} = \frac{mass}{\text {molar mass}} $$ $$\text{mass} = 25g $$ $$\text{number of moles} = \text{0.4 mol}$$ $$\text{Molar mass} = \frac{mass}{ \text{number of moles}} $$ $$\text{molar mass} = \frac{25}{0.4} $$ $$\text{molar mass} = 62.5g/mol $$