molar mass

1.
A certain sample of calcium carbonate(CaCO3) contains 4.86 mol.
a) What is the mass in grams of this sample?
b) What is the mass of the CO3²⁻ ions present?

a)
Molar mass of CaCO3 = 40 + 12 + 16x3 = 100 g/mol
Mass of CaCO3­ = (4.86 mol) x (100 g/mol) = 486 g

b)
No. of moles of CO3²⁻ ions = No. of moles of CaCO3 = 4.86 mol
Molar mass of CO3²⁻ ions = 12 + 16x3 = 60 g/mol
Molar mass of CO3²⁻ ions = (4.86 mol) x (60 g/mol) = 291.6 g


= = = = =
2.
calculate the mass of magnesium oxide (MgO) formed when 2.43g of maganesium (Mg) are burnt with
a) excess oxygen
b)1.28g of oxygen

a)
Molar mass of Mg = 24.3 g/mol
Molar mass of O2 = 16x2 = 32 g/mol
Molar mass of MgO = 24.3 + 16 = 40.3 g/mol

2Mg + O2 → 2MgO
Molar Mass Mg : MgO = 2 : 2 = 1 : 1

No. of moles of Mg = (2.43 g) / (24.3 g/mol) = 0.1 mol
No. of moles of MgO formed = 0.1 mol
Mass of MgO formed = (0.1 mol) x (40.3 g/mol) = 4.03 g

b)
2Mg + O2 → 2MgO
Molar Mass Mg : O2 : MgO = 2 : 1 : 2

No. of moles of Mg present = 0.1 mol
No. of moles of O2 present = (1.28 g)/ (32 g/mol) = 0.04 mol

To complete react with 0.04 mol of O2,
no. of moles of Mg needed = (0.04 mol) x 2 = 0.08 mol
Hence, Mg is in excess, and O2 completely reacts (limiting reagent).

No. of moles of O2 reacted = 0.04 mol
No. of moles of MgO formed =- (0.04 mol) x 2 = 0.08 mol
Mass of MgO formed = (0.08 mol) x (40.3 g/mol) = 3.224 g


= = = = =
3.
A mixture containing 2.8 g of iron and 2 g sulphur is heated together. What is the mass of iron(II) sulphide, FeS, is produced?
Fe(s) + S(s) → FeS(s)

Molar mass of Fe = 55.8 g/mol
Molar mass of S = 32.1 g/mol
Molar mass of FeS = 55.8 + 32.1 = 87.9 g/mol

Mole ratio Fe : S : FeS = 1 : 1 : 1

No. of moles of Fe present = 2.8/55.8 = 0.0502 mol
No. of moles of S present = 2/32.1 = 0.0623 mol
Hence, Fe is the limiting reagent (completely reacted).

No. of moles of Fe reacted = 0.0502 mol
No. of moles of FeS formed = 0.0502 mol
Mass of FeS formed = 0.0502 x 87.9 = 4.41 g