Calculations in analysis: titration and mass analysis
Week 9 · Chapter 2 (D): titration calculations with MaVa/na = MbVb/nb, the percentage of a substance in a mixture from a titration, and mass (gravimetric) analysis by volatilisation and by precipitation (water of crystallisation, mass of an ion from a precipitate) · 3 lessons
2
The percentage of a substance in a mixture
Chemistry Sec 3 Term 1, Chapter 2: pp. 43–48
GoalFind the percentage of a substance in an impure sample from a titration.
New words
impure sample · a solid that contains the substance we want together with others that do not reactpurity · the percentage of the sample that is the wanted substanceportion (aliquot) · a measured part of a solution taken out for titration
Explanation
Many samples are mixtures: NaOH mixed with salt that does not react with acid, for example. A titration with HCl counts only the NaOH. The method has four steps. (1) Find the moles of acid used: n = V × M ÷ 1000. (2) Use the equation to find the moles of the substance wanted; for NaOH this is the same number because the ratio is 1 : 1. (3) Multiply by its molar mass, 40 g/mol for NaOH, to get its mass. (4) Divide by the mass of the whole sample and multiply by 100. If only a portion of the solution was titrated, scale the moles up to the whole solution first. Always ask which whole the percentage refers to: the percentage of the sample is not the percentage of an element in a compound.
Worked examples
A direct sample. 0.50 g of impure NaOH needs 20.0 mL of 0.50 M HCl. n(HCl) = 20.0 × 0.50 ÷ 1000 = 0.0100 mol, so n(NaOH) = 0.0100 mol and its mass is 0.0100 × 40 = 0.40 g. The purity is 0.40 ÷ 0.50 × 100 = 80 %.
Step
Value
n(HCl) in 25.0 mL
0.00200 mol
n(NaOH) in 250 mL
0.0200 mol
mass of NaOH
0.80 g
percentage of sample
32 %
A portion of the solution. 2.5 g of a mixture of NaOH and NaCl is dissolved to make 250 mL. A 25.0 mL portion needs 20.0 mL of 0.100 M HCl. n in the portion = 20.0 × 0.100 ÷ 1000 = 0.00200 mol. The whole solution is ten times larger: 0.0200 mol, so m(NaOH) = 0.0200 × 40 = 0.80 g and the percentage is 0.80 ÷ 2.5 × 100 = 32 %.
Which whole? In pure NaOH (Na = 23, O = 16, H = 1), sodium is 23 ÷ 40 × 100 = 57.5 % of the compound. In the 0.50 g sample above, NaOH is 80 % of the sample. These are two different questions, with two different wholes: 40 g of compound, and 0.50 g of sample.
Class activity
Weigh a mixture, with a grown-up. Mix about two spoons of rice with one spoon of lentils in a bowl and weigh the whole mixture on a kitchen scale. Take the lentils out by hand, weigh them, and work out the percentage of lentils by mass: part ÷ whole × 100. Check by weighing the rice, too: the two masses should add up to the whole (small differences are normal). This is the kitchen cousin of a purity calculation. No tasting, and the food goes back to the kitchen.
Practice
1
1.00 g of impure NaOH needs 22.0 mL of 0.500 M HCl. Find the percentage of NaOH (Na = 23, O = 16, H = 1).
