Yield calculations and an introduction to aqueous reactions: electrolytes, precipitation

Last week we saw the “ideal” side of stoichiometry: as much product formed as you calculated. In reality it does not. The first half of this week is the measure of that, the yield. The second half is the medium where most of chemistry happens: water. You will learn how substances that dissolve in water behave, which ions form a solid when they meet, and how to write that.

Golden rules:

1. The theoretical yield is computed from the limiting reactant. The actual yield is the value weighed in the experiment.
2. When using percent yield backward, divide. A low yield needs more reactant.
3. An ionic compound splits into its ions in water, and the subscripts give the number of ions.
4. Net ionic equation: only the species that change remain.

Theoretical, actual and percent yield

• Theoretical yield: the amount of product that would form if all of the limiting reactant turned into product. It is found by calculation.
• Actual yield: the amount really obtained in the experiment. It is usually less because of side reactions, losses and incomplete reactions.
• Percent yield:

$\%\ \text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$

Example 1. $\mathrm{Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2}$. From 100.0 g $\mathrm{Fe_2O_3}$ (CO in excess), 61.2 g of Fe is obtained.

$100.0\ \text{g} \times \frac{1\ \text{mol Fe}_2\text{O}_3}{159.69\ \text{g}} \times \frac{2\ \text{mol Fe}}{1\ \text{mol Fe}_2\text{O}_3} \times \frac{55.845\ \text{g}}{1\ \text{mol Fe}} = 69.94\ \text{g Fe (theoretical)}$

$\%\ \text{yield} = \frac{61.2}{69.94}\times100 = 87.5\%$

Example 2 (backward). With the same yield, how many grams of $\mathrm{Fe_2O_3}$ are needed to get 50.0 g of Fe?

Fe that must be theoretical = 50.0 / 0.875 = 57.14 g
mol Fe = 57.14 / 55.845 = 1.023 mol  → mol Fe2O3 = 1.023 / 2 = 0.5116 mol
m(Fe2O3) = 0.5116 × 159.69 = 81.7 g

PITFALL · multiplying by the percentage. In the backward calculation, if you do 50.0 × 0.875 you get less reactant. A sanity check: if the yield is below 100%, you must use more reactant than needed for 100% yield.

If the percent yield exceeds 100%, the product is not dry or contains impurities. Physically, more product than the theoretical yield cannot form.

In multi-step syntheses, the overall yield is the product of the step yields: 90% and 90% → 81%.

Aqueous solutions and electrolytes

A substance that conducts electricity when dissolved in water is called an electrolyte. The conductivity comes from the ions moving freely in the solution.

A one-way arrow is written for a strong electrolyte and a two-way arrow for a weak one:

$\mathrm{MgCl_2(s) \rightarrow Mg^{2+}(aq) + 2Cl^-(aq)} \qquad \mathrm{CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)}$

In the laboratory you see this with a light bulb and two electrodes: in a strong electrolyte solution the bulb glows brightly, in a weak electrolyte dimly, and in a nonelectrolyte solution not at all. Pure water also hardly conducts; the dissolved ions provide the conductivity.

PITFALL · confusing solubility with electrolyte strength. “Strong electrolyte” says that the part that dissolves ionizes completely, not that it dissolves a lot. AgCl dissolves very little, but because the tiny part that dissolves ionizes completely, it counts as a strong electrolyte. Acetic acid mixes with water in any proportion but is a weak electrolyte.

Ion concentrations

In a strong electrolyte, the concentration of each ion is the molarity of the compound times the number of that ion in the formula. For 0.0500 M $\mathrm{Al_2(SO_4)_3}$:

$[\mathrm{Al^{3+}}] = 2 \times 0.0500 = 0.100\ \text{M} \qquad [\mathrm{SO_4^{2-}}] = 3 \times 0.0500 = 0.150\ \text{M}$

If two solutions are mixed (with no reaction), add the moles of each ion separately and divide by the total volume. Adding or averaging the molarities directly is wrong.

Precipitation reactions

When two soluble salts are mixed, one cation and the anion of the other may form a compound that is insoluble in water. This solid settles to the bottom as a precipitate. This reaction is called a precipitation reaction.

You predict which compound precipitates with the solubility rules:

Three kinds of equations

You can write the mixing of $\mathrm{AgNO_3}$ and NaCl in three ways:

Molecular equation, all substances as formulas:

$\mathrm{AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq)}$

Complete ionic equation: strong electrolytes (aq) are split into their ions; solids, gases, water and weak electrolytes stay as molecules:

$\mathrm{Ag^+(aq) + NO_3^-(aq) + Na^+(aq) + Cl^-(aq) \rightarrow AgCl(s) + Na^+(aq) + NO_3^-(aq)}$

Net ionic equation: the spectator ions that stand unchanged on both sides are dropped:

$\mathrm{Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)}$

The net ionic equation describes every case in which $\mathrm{Ag^+}$ meets $\mathrm{Cl^-}$, whichever salt it comes from. The net ionic equation must also be balanced in both atoms and charge.

PITFALL · splitting the solid into ions. A product marked (s) is a precipitate; in the complete ionic equation you cannot write it as $\mathrm{Ag^+ + Cl^-}$. Only strong electrolytes dissolved in water are split.

Precipitation stoichiometry and gravimetric analysis

Calculating the mass of the precipitate is the same as the solution stoichiometry of week 2: $M \times V$ → moles → ratio from the net ionic equation → grams. If two solutions are mixed, find the limiting ion.

Example 3. 25.0 mL of 0.120 M $\mathrm{BaCl_2}$ is mixed with excess $\mathrm{Na_2SO_4}$. Net ionic: $\mathrm{Ba^{2+} + SO_4^{2-} \rightarrow BaSO_4(s)}$.

mol Ba2+  = 0.0250 L × 0.120 M = 3.00×10^-3 mol
mol BaSO4 = 3.00×10^-3 mol (1:1)
m(BaSO4)  = 3.00×10^-3 × 233.38 = 0.700 g

Gravimetric analysis is the reverse: the ion in an unknown sample is completely precipitated, the precipitate is dried and weighed, and working back from the mass the percentage in the sample is found. You use the mass fraction of the element in the precipitate; for example, $35.45/143.32 = 0.2473$ of AgCl by mass is Cl.

Example 4 (two solutions, limiting ion and the remaining concentrations). 30.0 mL of 0.150 M $\mathrm{AgNO_3}$ is mixed with 20.0 mL of 0.100 M $\mathrm{Na_3PO_4}$.

First find which compound precipitates. The possible new pairs are $\mathrm{Ag_3PO_4}$ (phosphate is insoluble with Ag, it precipitates) and $\mathrm{NaNO_3}$ (soluble). Net ionic:

$\mathrm{3Ag^+(aq) + PO_4^{3-}(aq) \rightarrow Ag_3PO_4(s)}$

mol Ag+   = 0.0300 × 0.150 = 4.50×10^-3     ÷ 3 = 1.50×10^-3   ← smaller: Ag+ is limiting
mol PO4   = 0.0200 × 0.100 = 2.00×10^-3     ÷ 1 = 2.00×10^-3
Ag3PO4    = 4.50×10^-3 / 3 = 1.50×10^-3 mol × 418.58 g/mol = 0.628 g
left PO4  = 2.00×10^-3 − 1.50×10^-3 = 5.0×10^-4 mol
total V   = 0.0500 L
[PO4 3−] = 0.0100 M    [Na+] = 6.00×10^-3 / 0.0500 = 0.120 M    [NO3−] = 0.0900 M    [Ag+] ≈ 0

The moles of the spectator ions do not change, but their concentrations drop because the volume grows. This is the step most often forgotten in mixture questions.

Which compound precipitates? A shortcut for the exam

1. Split the two mixed salts into their ions.
2. Pair the cations “crosswise” (the cation of A with the anion of B and vice versa).
3. Check each new pair in the solubility table.
4. If none precipitates, write “no reaction”; the ions have only mixed.

For example, when KCl and $\mathrm{NaNO_3}$ are mixed, the possible pairs are $\mathrm{KNO_3}$ and NaCl; both are soluble, so there is no reaction.

Common mistakes