Theoretical and Percent Yield理论产率与百分产率

Big Idea · Scale, Proportion, and Quantity (SPQ)AP Science Practice 5 · Mathematical RoutinesAP link · Unit 4~35 min

Learning goals

  • Distinguish theoretical yield from actual yield, and find the theoretical yield from the limiting reactant
  • Calculate percent yield, and use a percent yield to predict the actual yield
  • Explain common reasons for yields below 100% and why a yield above 100% signals a problem

Before you start

  • Limiting and excess reactants (2.6.6)
  • Error and percent error (2.1.3)

The recipe says 24 cookies

A recipe promises 24 cookies, but when the tray comes out you count only 20: some dough stuck to the bowl and one cookie burned. Chemical reactions are the same. The last few lessons calculated the most product you could get. In a real lab or factory you almost always get less, and percent yield measures how much less.

Theoretical yield, actual yield, and percent yield

  • Theoretical yield理论产量: the maximum amount of product, calculated from the limiting reactant and the balanced equation. This is what you found in 2.6.5 and 2.6.6.
  • Actual yield实际产量: the amount of product actually collected and weighed in the experiment.
  • Percent yield产率: the actual yield as a percentage of the theoretical yield.
% yield=actual yieldtheoretical yield×100%\%\ \text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%

Use the same units for both yields, either grams or moles. Percent yield looks like percent error (2.1.3), but it answers a different question: not “how far off was my measurement?” but “how much of the possible product did I collect?”

Why not 100%?

  • The reaction is incomplete. It wasn’t heated long enough, or the reactants didn’t mix well. Reversible reactions, such as making ammonia, never go all the way to completion.
  • Side reactions happen. Some reactant turns into a different product. When magnesium burns in air, for example, a little of it reacts with nitrogen to form magnesium nitride.
  • Product is lost in handling. Some is lost during filtering, transferring, or evaporating.

A percent yield can never really exceed 100%. If yours does, the product is impure: a precipitate still holds water because it wasn’t dried, or something else got mixed in. A yield over 100% is a reason to look for a mistake, not a reason to celebrate.

Chinese problems sometimes also give a 转化率 (conversion), the percentage of a reactant that reacted. That is a different idea: percent yield looks at the product, conversion looks at a reactant.

This lesson uses the atomic masses on the AP periodic table, such as H 1.008, C 12.01, N 14.01, O 16.00, Na 22.99, Mg 24.31, Cl 35.45, Fe 55.85, Cu 63.55, and Ag 107.87.

Choose an experiment and find the theoretical yield. Then press “Run the experiment” to see an actual yield and its percent yield. You can also type your own actual yield, including one larger than the theoretical yield:

Try it: percent yield lab

Example 1: percent yield for burning magnesium

2.40 g of magnesium ribbon burns completely in air, and 3.65 g of magnesium oxide is collected. What is the percent yield?

2 Mg(s)+OX2(g)→2 MgO(s)\ce{2Mg(s) + O2(g) -> 2MgO(s)}
Step-by-step solution

  1. Limiting reactant. Oxygen from the air is in excess, so Mg is limiting and sets the theoretical yield.

  2. Theoretical yield. Mg and MgO have equal coefficients, so equal moles:

    n(Mg)=2.40/24.31=0.09872 moln(MgO)=0.09872 molm(MgO)=0.09872×40.31=3.98 g\begin{aligned} n(\ce{Mg}) &= 2.40 / 24.31 \\ &= 0.09872\ \text{mol} \\ n(\ce{MgO}) &= 0.09872\ \text{mol} \\ m(\ce{MgO}) &= 0.09872 \times 40.31 \\ &= 3.98\ \text{g} \end{aligned}

  3. Percent yield. Actual ÷ theoretical × 100%:

    3.653.98×100%=91.7%\frac{3.65}{3.98} \times 100\% = \boxed{91.7\%}

  4. Likely reasons. Some MgO escaped as white smoke, and a little Mg formed magnesium nitride instead.

  5. Check. A percent yield can’t exceed 100%; if it does, the product is usually wet or impure.

Example 2: from percent yield to actual yield

A blast furnace reduces 1.00 metric ton (1.00 × 10⁶ g) of iron(III) oxide with carbon monoxide. The percent yield of iron is 90.0%. How many metric tons of iron are produced?

FeX2OX3(s)+3 CO(g)→Δ2 Fe(l)+3 COX2(g)\ce{Fe2O3(s) + 3CO(g) ->[\Delta] 2Fe(l) + 3CO2(g)}
Step-by-step solution

  1. Moles of Fe₂O₃. M(Fe₂O₃) = 159.70 g/mol:

    n(FeX2OX3)=1.00×106159.70=6.262×103 mol\begin{aligned} n(\ce{Fe2O3}) &= \frac{1.00 \times 10^{6}}{159.70} \\ &= 6.262 \times 10^{3}\ \text{mol} \end{aligned}

  2. Theoretical yield. 1 mol Fe₂O₃ gives 2 mol Fe:

    n(Fe)=2×6.262×103=1.252×104 molm(Fe)=1.252×104×55.85=6.99×105 g=0.699 t\begin{aligned} n(\ce{Fe}) &= 2 \times 6.262 \times 10^{3} \\ &= 1.252 \times 10^{4}\ \text{mol} \\ m(\ce{Fe}) &= 1.252 \times 10^{4} \times 55.85 \\ &= 6.99 \times 10^{5}\ \text{g} \\ &= 0.699\ \text{t} \end{aligned}

  3. Actual yield. Actual = theoretical × percent yield:

    0.699×0.900=0.629 t0.699 \times 0.900 = \boxed{0.629\ \text{t}}

  4. Another route. With large masses, a mass ratio is quick: 159.70 t of Fe₂O₃ gives 2 × 55.85 = 111.70 t of Fe, so 1.00 t gives 0.699 t, the same theoretical yield.

Practice: which value is closest to the percent yield?
  1. Theoretical yield 10.0 g, actual yield 8.0 g
  2. Theoretical yield 4.00 g, actual yield 3.00 g
  3. Theoretical yield 20.0 g, actual yield 19.0 g
  4. Theoretical yield 5.0 g, product weighed 5.4 g
  5. Theoretical yield 12.0 g, actual yield 6.0 g
  6. Theoretical yield 2.50 g, actual yield 2.25 g
  7. 0.100 mol Mg burns completely; 3.22 g MgO is collected
  8. 6.50 g Zn reacts with excess HCl; 0.150 g H₂ is collected

Common mistakes

Mistake 1: “Percent yield = mass of product ÷ mass of reactant.” In Example 1 that gives 3.65 g ÷ 2.40 g = 152%, which is meaningless, because the product also contains mass from oxygen. The denominator must be the theoretical yield calculated from the equation.

Mistake 2: “A yield over 100% means a great experiment.” Conservation of mass means you can’t make more product than the theoretical yield. A yield over 100% means what you weighed was not pure product, most often because it wasn’t completely dry.

Mistake 3: “Any reactant works for the theoretical yield.” The theoretical yield must come from the limiting reactant (2.6.6). Using an excess reactant makes the theoretical yield too large and the percent yield too small.

Quick check

A reaction has a theoretical yield of 25.0 g, and 20.0 g of product is collected. What is the percent yield?

A reaction’s percent yield is 75.0% and its theoretical yield is 40.0 g. What is the actual yield?

Summary

  • Theoretical yield comes from the limiting reactant and the equation, actual yield is what you collect, and percent yield = actual ÷ theoretical × 100%.
  • Incomplete reactions, side reactions, and handling losses keep yields below 100%. A yield above 100% means the product is impure.
  • Given a percent yield, actual yield = theoretical yield × percent yield. To make a set amount of product, plan for extra starting material.
On to AP (for later)

AP Chemistry Unit 4 often combines percent yield with limiting reactants in one problem: find the limiting reactant, then the theoretical yield, then the percent yield. AP lab questions also ask whether a mistake, such as not drying a precipitate or losing some during filtering, makes the measured yield too high or too low. The AP lessons cover that kind of error analysis in detail.

Practice this chapter: 2.6 · The Mole and Stoichiometry · 10 questions

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