All practice pages

Stoichiometry

Reading mole ratios off a balanced equation, grams to grams, the limiting reactant and what is left over, theoretical and percent yield, titration, and gravimetric analysis. Checked box by box for value and significant figures.

Chapter 4, part 3

Stoichiometry, limiting reactants, yields, titrations and quantitative analysis, from slides 30 to 41 of the Chapter 4 partial notes and the third lecture video.

Homework: Reaction Stoichiometry and Yields and Quantitative Chemical Analysis, in the Chapter 4 folder.

A balanced equation is a recipe. Its coefficients give the ratio in which moles react and form, so knowing the amount of one substance tells you the amount of any other. It has to be balanced first.

Grams to grams goes through moles

Grams of A 1. ÷ molar mass of A Moles of A 2. × mole ratio, B over A, from the coefficients Moles of B 3. × molar mass of B Grams of B There is no step from grams of A straight to grams of B: the equation counts moles, not grams.

  1. What mass of ammonia is produced from 1.00 kg of hydrogen with excess nitrogen? Balanced equation:

    3H₂(g) + N₂(g) → 2NH₃(g)

  2. Grams of H₂ to moles: 1.00 kg is 1000 g, and H₂ is 2(1.008) = 2.016 g/mol.

    1000 g ÷ 2.016 g/mol = 496.0 mol H₂

  3. The coefficients say 2 mol NH₃ for every 3 mol H₂.

    496.0 mol H₂ × 2 mol NH₃3 mol H₂ = 330.7 mol NH₃

  4. Moles of NH₃ to grams: 14.01 + 3(1.008) = 17.03 g/mol. Three significant figures, from 1.00 kg.

    330.7 mol × 17.03 g/mol = 5632 g = 5.63 × 10³ g NH₃

As one set-up: 1.00 kg H₂ × 1000 g / 1 kg × 1 mol H₂ / 2.016 g H₂ × 2 mol NH₃ / 3 mol H₂ × 17.03 g NH₃ / 1 mol NH₃. The units cancel down to grams of ammonia.

The limiting reactant

Limiting

The reactant used up first. It limits how much product can form.

In excess

The other reactant: some of it is left over when the reaction stops.

How to tell

You need the amount of every reactant. Work out how much product each one could make; the one that makes less is limiting.

  1. What mass of ammonia forms from 1.00 kg of H₂ and 1.00 kg of N₂? Moles of each:

    1000 g ÷ 2.016 g/mol = 496.0 mol H₂, 1000 g ÷ 28.02 g/mol = 35.69 mol N₂

  2. The ammonia each could make:

    496.0 mol H₂ × 23 = 330.7 mol NH₃, 35.69 mol N₂ × 21 = 71.38 mol NH₃

  3. Nitrogen makes less, so N₂ is limiting and hydrogen is in excess.

    71.38 mol × 17.03 g/mol = 1216 g = 1.22 × 10³ g NH₃

  4. How much hydrogen is left over? The nitrogen uses 3 × 35.69 = 107.1 mol H₂.

    496.0 − 107.1 = 388.9 mol H₂, 388.9 mol × 2.016 g/mol = 784 g H₂

Theoretical and percent yield

Theoretical yield

The most product the limiting reactant can make: the amount you calculate.

Actual yield

The amount you really get, measured. It is less than the theoretical yield.

percent yield = actual yieldtheoretical yield × 100%

  1. If 7.81 × 10² g of NH₃ is obtained from the reaction above, the percent yield is

    781 g ÷ 1216 g × 100% = 64.2%

Titrations

Titration

Adding a solution from a buret until the reaction with the sample is just complete, to find an unknown amount or concentration.

Standard solution

A solution whose concentration is known exactly.

Equivalence point

The point where the reactants are present in exactly the ratio of the balanced equation.

Indicator

A substance that changes colour near the equivalence point.

End point

Where the indicator changes colour. It should be very close to the equivalence point.

Not M₁V₁ = M₂V₂. That is for dilution, where the moles stay the same. In a titration two different substances react, so go through the mole ratio of the balanced equation.

  1. What is the molarity of a NaOH solution if 17.36 mL of it neutralizes 10.00 mL of 1.00 M H₂SO₄?

    H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)

  2. Moles of the acid, from its volume in liters and its molarity.

    0.01000 L × 1.00 mol/L = 0.0100 mol H₂SO₄

  3. Each H₂SO₄ uses two NaOH.

    0.0100 mol × 2 mol NaOH1 mol H₂SO₄ = 0.0200 mol NaOH

  4. Moles over liters, to the three significant figures of 1.00 M.

    0.0200 mol ÷ 0.01736 L = 1.15 M NaOH

Quantitative analysis

Gravimetric analysis

The substance being measured is turned into a precipitate, which is filtered, dried and weighed. Its mass and formula give the amount in the sample. Lab 6 works this way.

Combustion analysis

A compound is burned in oxygen and the CO₂ and H₂O are weighed, which gives the carbon and hydrogen. The empirical formula tab practises it.

  1. A 0.5000 g sample containing chloride is dissolved, and excess silver nitrate precipitates 0.9620 g of AgCl. What percent of the sample is chloride? AgCl is 107.87 + 35.45 = 143.32 g/mol.

    0.9620 g ÷ 143.32 g/mol = 0.006712 mol AgCl

  2. One Cl in each AgCl, so the same moles of chloride.

    0.006712 mol × 35.45 g/mol = 0.2379 g Cl, 0.2379 g ÷ 0.5000 g × 100% = 47.59% Cl