Calculate the molecular masses of the following compounds:

Calculate the molecular masses of the following compounds:

(Given Atomic Masses:
H = 1, O = 16, C = 12, Cl = 35.5, N = 14, Ca = 40, K = 39,
S = 32, Na = 23, Zn = 65, Mn = 55, Ag = 107.87, Cu = 63.5, Fe = 56)


(i) Hâ‚‚Oâ‚‚ (Hydrogen Peroxide)

H atoms = 2, O atoms = 2
Molecular mass = 2 × H + 2 × O = 2 × 1 + 2 × 16 = 2 + 32 = 34 g/mol


(ii) COâ‚‚ (Carbon Dioxide)

C = 1, O = 2
= 1 × 12 + 2 × 16 = 12 + 32 = 44 g/mol


(iii) CO (Carbon Monoxide)

C = 1, O = 1
= 12 + 16 = 28 g/mol


(iv) HCl (Hydrogen Chloride)

H = 1, Cl = 1
= 1 + 35.5 = 36.5 g/mol


(v) NH₃ (Ammonia)

N = 1, H = 3
= 14 + 3 = 17 g/mol


(vi) CaO (Calcium Oxide)

Ca = 1, O = 1
= 40 + 16 = 56 g/mol


(vii) CaCO₃ (Calcium Carbonate)

Ca = 1, C = 1, O = 3
= 40 + 12 + 3 × 16 = 40 + 12 + 48 = 100 g/mol


(viii) KCl (Potassium Chloride)

K = 1, Cl = 1
= 39 + 35.5 = 74.5 g/mol


(ix) KClO₃ (Potassium Chlorate)

K = 1, Cl = 1, O = 3
= 39 + 35.5 + 3 × 16 = 39 + 35.5 + 48 = 122.5 g/mol


(x) Hâ‚‚SOâ‚„ (Sulfuric Acid)

H = 2, S = 1, O = 4
= 2 × 1 + 32 + 4 × 16 = 2 + 32 + 64 = 98 g/mol


(xi) HNO₃ (Nitric Acid)

H = 1, N = 1, O = 3
= 1 + 14 + 3 × 16 = 1 + 14 + 48 = 63 g/mol


(xii) NaOH (Sodium Hydroxide)

Na = 1, O = 1, H = 1
= 23 + 16 + 1 = 40 g/mol


(xiii) ZnCO₃ (Zinc Carbonate)

Zn = 1, C = 1, O = 3
= 65 + 12 + 48 = 125 g/mol


(xiv) ZnO (Zinc Oxide)

Zn = 1, O = 1
= 65 + 16 = 81 g/mol


(xv) MnOâ‚‚ (Manganese Dioxide)

Mn = 1, O = 2
= 55 + 32 = 87 g/mol


(xvi) MnClâ‚‚ (Manganese(II) Chloride)

Mn = 1, Cl = 2
= 55 + 2 × 35.5 = 55 + 71 = 126 g/mol


(xvii) ZnSOâ‚„ (Zinc Sulphate)

Zn = 1, S = 1, O = 4
= 65 + 32 + 4 × 16 = 65 + 32 + 64 = 161 g/mol


(xviii) NaCl (Sodium Chloride)

Na = 1, Cl = 1
= 23 + 35.5 = 58.5 g/mol


(xix) NaNO₃ (Sodium Nitrate)

Na = 1, N = 1, O = 3
= 23 + 14 + 48 = 85 g/mol


(xx) AgCl (Silver Chloride)

Ag = 1, Cl = 1
= 107.87 + 35.5 = 143.37 g/mol


(xxi) AgNO₃ (Silver Nitrate)

Ag = 1, N = 1, O = 3
= 107.87 + 14 + 48 = 169.87 g/mol


(xxii) CuO (Copper(II) Oxide)

Cu = 1, O = 1
= 63.5 + 16 = 79.5 g/mol


(xxiii) Fe₂O₃ (Iron(III) Oxide)

Fe = 2, O = 3
= 2 × 56 + 3 × 16 = 112 + 48 = 160 g/mol

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top