Mass = Moles x M r Moles = Concentration x Volume To convert volume from cm 3 → dm 3, divide by 1000.

Mass = Moles x Mr

Moles = Concentration x Volume

To convert volume fromcm3 → dm3, divide by 1000

Example 1: 3.00 g of lawn sand containing an iron(II) salt was shaken with dilute sulphuric acid

The resulting solution needed 25.00 cm3 of 0.0200 mol dm-3 KMnO4 to oxidise the Fe2+ ions in the solution to Fe3+ ions

Use this information to calculate the percentage by mass of Fe2+ ions in this sample of lawn sand

1) Equation:

MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+

2) Moles of MnO4- = Conc x Vol

= 0.200 x (25 ÷ 1000) = 5 x 10-3 mol

3) Moles of Fe2+ = Moles of MnO4- x 5 (Ratio 1:5)

= 5 x 10-3 x 5 = 0.025 mol

4) Mass of Fe2+ = Moles x Mr

= 0.025 x 55.8 = 1.40 g

5) % by mass of Fe2+ = (1.40 ÷ 3.00) x 100 = 46.7 %

The most important part is to write the equation for reaction as it allows you to work out the ratio (step 3)

Example 2: In a titration, 0.321 g of a moss killer reacted with 23.60 cm3 of acidified 0.0218 mol dm-3 K2Cr2O7 solution

Calculate the percentage by mass of iron in the moss killer

Assume that all of the iron in the moss killer is in the form of iron(II)

1) Equation:

Cr2O72- + 14H+ + 6Fe2+ → 2Cr3+ + 7H2O + 6Fe3+

2) Moles of Cr2O72- = Conc x Vol

= 0.0218 x (23.60 ÷ 1000) = 5.145 x 10-3 mol

3) Moles of Fe2+ = Moles of Cr2O72- x 6 (Ratio 1:6)

= 5.145 x 10-3 x 6 = 3.087 x 10-3 mol

4) Mass of Fe2+ = Moles x Mr

= 3.087 x 10-3 x 55.8 = 0.172 g

5) % by mass of Fe2+ = (0.172 ÷ 0.321) x 100 = 53.6 %

The most important part is to write the equation for reaction as it allows you to work out the ratio (step 3)

Example 3: A 1.27g sample of impure iron was reacted with an excess of dilute sulphuric acid

The solution formed was made up to 250 cm3

A 25.0 cm3 sample of this solution reacted completely with exactly 19.6 cm3 of a 0.0220 mol dm-3 solution of potassium manganate(VII)

Calculate the percentage by mass of iron in the sample

1) Equation:

MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+

2) Moles of MnO4- = Conc x Vol

= 0.0220 x (19.6 ÷ 1000) = 4.31 x 10-4 mol

3) Moles of Fe2+ = Moles of MnO4- x 5 (Ratio 1:5)

= 4.31 x 10-4 x 5 = 2.16 x 10-3 mol in 25 cm3

4) Moles of Fe2+ in 250 cm3 = 2.16 x 10-3 x 10 = 2.16 x 10-2 mol

5) Mass of Fe2+ = Moles x Mr

= 2.16 x 10-2 x 55.8 = 1.21 g

6) % by mass of Fe2+ = (1.21 ÷ 1.27) x 100 = 95.3 %

Example 4: A solution of iron(II) sulfate was prepared by dissolving 10.00 g of FeSO4.7H2O (Mr = 277.9) in water and making up to 250 cm3 of solution

The solution was left to stand, exposed to air, and some of the iron(II) ions became oxidised to iron(III) ions

A 25.0 cm3 sample of the partially oxidised solution required 23.70 cm3 of 0.0100 mol dm-3 potassium dichromate(VI) solution for complete reaction in the presence of an excess of dilute sulfuric acid

Calculate the percentage of iron(II) ions that had been oxidised by the air

1) Equation:

Cr2O72- + 14H+ + 6Fe2+ → 2Cr3+ + 7H2O + 6Fe3+

2) Moles of Cr2O72- = Conc x Vol

= 0.0100 x (23.70 ÷ 1000) = 2.37 x 10-4 mol

3) Moles of Fe2+ = Moles of Cr2O72- x 6 (Ratio 1:6)

= 2.37 x 10-4 x 6 = 1.42 x 10-3 mol in 25 cm3

4) Moles of Fe2+ in 250 cm3 = 1.42 x 10-3 x 10 = 1.42 x 10-2 mol

5) Original moles of Fe2+ = Mass ÷ Mr

= 10.00 ÷ 277.9 = 0.036 mol

6) Moles of Fe2+ oxidised = Original - Fe2+ remaining = 0.036 - 1.42 x 10-2

= 0.0218 mol

7) % by mass of Fe2+ oxidised = (0.0218 ÷ 0.036) x 100 = 60.6 %