Listed below is the molar mass of the following substances:

(a) Molar mass of Ethyne C2H2= 2 x Mass of C+2 x Mass of H = (2×12)+(2×1)=24+2=26g

(b) Molar mass of Sulphur molecule S8 = 8 x Mass of S = 8  x 32 = 256g

(c) Molar mass of  Phosphorus molecule, P4 = 4 x Mass of P = 4 x 31 = 124g

(d) Molar mass of Hydrochloric acid, HCl = Mass of H+ Mass of Cl = 1+35.5 = 36.5g

(e) Molar mass of Nitric acid, HNO3 =Mass of H+ Mass of Nitrogen + 3 x Mass of O = 1 + 14+

3×16 = 63g

The following are the names of the elements present in the following compounds:

(a) Quick lime – Calcium and oxygen (CaO)

(b) Hydrogen bromide – Hydrogen and bromine (HBr)

(c) Baking powder – Sodium, Carbon, Hydrogen, Oxygen (NaHCO3)

(d) Potassium sulphate – Sulphur, Oxygen, Potassium (K2SO4)

The following are the chemical formula of the above-mentioned list:

(a) Magnesium chloride – MgCl2

(b) Calcium oxide – CaO

(c) Copper nitrate – Cu(NO3)2

(d) Aluminium chloride – AlCl3

(e) Calcium carbonate – CaCO3

Polyatomic ions are ions that contain more than one atom but they behave as a single unit

Example: CO32-, H2PO4–

11.00g of carbon dioxide is formed when 3.00g carbon is burnt in 8.00g of oxygen.

Carbon and oxygen are combined in the ratio 3:8 to give carbon dioxide using up all the carbon and

oxygen

Hence, for 3g of carbon and 50g of oxygen, 8g of oxygen is used and 11g of carbon is formed, the

left oxygen is unused i.e., 50-8=42g of oxygen is unused.

This depicts the law of definite proportions – The combining elements in compounds are present in

definite proportions by mass

Mass of the sample compound = 0.24g, mass of boron = 0.096g, mass of oxygen = 0.144g

To calculate percentage composition of the compound:

Percentage of boron = mass of boron / mass of the compound x 100

= 0.096g / 0.24g x 100  = 40%

Percentage of oxygen = 100 – percentage of boron

= 100 – 40 = 60%

Atomic mass of Na=23u, Atomic mass of Fe= 56u

To calculate the number of atoms in 100g of sodium:

23g of Na contains = 6.022 x 1023 atoms

1g of Na contains = 6.022 x 1023 atoms / 23

100g of Na contains = 6.022 x 1023 atoms x 100 / 23

= 2.6182 x 1024 atoms

To calculate the number of atoms in 100g of sodium:

56g of Fe contains = 6.022 x 1023 atoms

1g of Fe contains = 6.022 x 1023 atoms / 56

100g of Fe contains =  6.022 x 1023 atoms x 100 / 56

= 1.075 x 1024 atoms

Hence, through comparison, it is evident that 100g of Na has more atoms.

1 mole of carbon weighs 12g

1 mole of carbon atoms = 6.022 x 1023

Molecular mass of carbon atoms = 12g = an atom of carbon mass

Hence, mass of 1 carbon atom = 12 / 6.022 x 1023  = 1.99 x 10-23g

Atomic mass of Zn = 65u

Atomic mass of Na = 23u

Atomic mass of K = 39u

Atomic mass of C = 12u

Atomic mass of O = 16u

The formula unit mass of ZnO= Atomic mass of Zn + Atomic mass of O = 65u + 16u = 81u

The formula unit mass of Na2O = 2 x Atomic mass of Na + Atomic mass of O = (2 x 23)u + 16u = 46u + 16u = 62u

The formula unit mass of K2CO3 = 2 x Atomic mass of K + Atomic mass of C + 3 x Atomic mass of O = (2 x 39)u + 12u + (3 x 16)u = 78u + 12u + 48u = 138u

The following are the molecular masses:

The molecular mass of H2 – 2 x atoms atomic mass of H = 2 x 1u = 2u

The molecular mass of O2 – 2 x atoms atomic mass of O = 2 x 16u = 32u

The molecular mass of Cl2 – 2 x atoms atomic mass of Cl = 2 x 35.5u = 71u

The molecular mass of CO2 – atomic mass of C + 2 x atomic mass of O = 12 + ( 2×16)u = 44u

The molecular mass of CH4 – atomic mass of C + 4 x atomic mass of H = 12 + ( 4 x 1)u = 16u

The molecular mass of C2H6– 2 x atomic mass of C + 6 x atomic mass of H = (2 x 12) +

(6 x 1)u=24+6=30u

The molecular mass of C2H4– 2 x atomic mass of C + 4 x atomic mass of H = (2x 12) +

(4 x 1)u=24+4=28u

The molecular mass of NH3– atomic mass of N + 3 x atomic mass of H = (14 +3 x 1)u= 17u

The molecular mass of CH3OH – atomic mass of C + 3x atomic mass of H + atomic mass of O + atomic mass of H = (12 + 3×1+16+1)u=(12+3+17)u = 32u