Stoichiometry (Mole Ratios)计量计算（摩尔比）

Big Idea · Scale, Proportion, and Quantity (SPQ)AP Science Practice 5 · Mathematical RoutinesAP link · Unit 4~40 min

Learning goals

  • Explain that the coefficients in a balanced equation give the ratio of particles and of moles, not of masses
  • Solve stoichiometry problems in three steps, mass → moles → moles → mass
  • Compare the mole method with the mass-ratio method from Stage 1 and see that both give the same answer

Before you start

  • Calculations from chemical equations (1.6.5)
  • Converting between mass, moles, and particles (2.6.2)

A more direct route

In 1.6.5 you solved equation problems with mass ratios: under each substance you wrote coefficient × formula mass, then set up a proportion. Moles make this more direct, because the coefficients in an equation already are a ratio of moles. Later, when the given quantity is a gas volume (Chapter 2.8) or a solution concentration (Chapter 2.9), you will convert it to moles and use the same ratio. Calculating amounts from a balanced equation is called stoichiometry化学计量计算.

Coefficients are mole ratios

Take the burning of hydrogen:

2 HX2(g)+OX2(g)→2 HX2O(l)\ce{2H2(g) + O2(g) -> 2H2O(l)}
H₂O₂H₂O
Particles2 molecules1 molecule2 molecules
Moles2 mol1 mol2 mol
Mass4.032 g32.00 g36.03 g

The coefficients give the ratio of particles, and particles scale up to moles, so the coefficients give the mole ratio物质的量之比 directly: 2 mol H₂ : 1 mol O₂ : 2 mol H₂O. The masses, 4.032 : 32.00 : 36.03, are not in a 2 : 1 : 2 ratio, because each kind of molecule has a different molar mass.

Three steps across the mole bridge

Given the mass of substance A, find the mass of substance B:

  1. Mass → moles: divide by the molar mass of A.
  2. Moles A → moles B: multiply by the mole ratio (coefficient of B / coefficient of A). This is the only step that uses the equation.
  3. Moles → mass: multiply by the molar mass of B.

As a single dimensional-analysis chain (2.1.2), with a and b the coefficients of A and B:

g A×1 mol AMA g A×b mol Ba mol A×MB g B1 mol B=g B\begin{aligned} \text{g A} &\times \frac{1\ \text{mol A}}{M_\text{A}\ \text{g A}} \times \frac{b\ \text{mol B}}{a\ \text{mol A}} \\ &\times \frac{M_\text{B}\ \text{g B}}{1\ \text{mol B}} = \text{g B} \end{aligned}

If the given or wanted quantity is a number of particles, step 1 or step 3 uses NA instead of the molar mass (2.6.2).

How Chinese textbooks set it up

Chinese textbooks write the moles under each substance in the equation and set up a proportion, much like the mass-ratio layout of 1.6.5. They also allow mixed units as long as each substance keeps the same unit top and bottom, for example 65 g for Zn and 1 mol for H₂. US textbooks write the mole ratio as a conversion factor, (1 mol H₂ / 1 mol Zn), and chain it with molar masses as above. The layouts look different, but the reasoning and the answers are the same.

This lesson uses the atomic masses on the AP periodic table: H 1.008, C 12.01, O 16.00, Cl 35.45, K 39.10, Fe 55.85, and Zn 65.38.

Pick a reaction, a given substance, and a substance to find, then enter a mass and watch the three steps. The last line checks the answer with the mass-ratio method from Stage 1:

Try it: mass → moles → moles → mass
Given:
Find:

Example 1: zinc and hydrochloric acid

6.50 g of zinc reacts with excess hydrochloric acid. How many moles and how many grams of hydrogen gas form?

Zn(s)+2 HCl(aq)→ZnClX2(aq)+HX2(g)\ce{Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g)}
Step-by-step solution

  1. Decide what controls the amount. The acid is in excess, so all the zinc reacts and the zinc decides.

  2. Mass → moles.

    n(Zn)=6.50 g65.38 g/mol=0.09942 mol\begin{aligned} n(\ce{Zn}) &= \frac{6.50\ \text{g}}{65.38\ \text{g/mol}} \\ &= 0.09942\ \text{mol} \end{aligned}

  3. Mole ratio from the coefficients. Zn and H₂ both have a coefficient of 1:

    n(HX2)=0.09942×11=0.0994 mol\begin{aligned} n(\ce{H2}) &= 0.09942 \times \frac{1}{1} \\ &= \boxed{0.0994\ \text{mol}} \end{aligned}

  4. Moles → mass. M(H₂) = 2.016 g/mol:

    m(HX2)=0.09942×2.016=0.200 g\begin{aligned} m(\ce{H2}) &= 0.09942 \times 2.016 \\ &= \boxed{0.200\ \text{g}} \end{aligned}

  5. A common mistake. Applying the coefficient ratio to grams (6.50 g of zinc makes 6.50 g of hydrogen). Coefficients give mole ratios, not mass ratios.

Example 2: oxygen from potassium chlorate

24.5 g of potassium chlorate decomposes completely, with MnO₂ as a catalyst. What mass of oxygen gas forms?

2 KClOX3(s)→ΔMnOX22 KCl(s)+3 OX2(g)\ce{2KClO3(s) ->[\ce{MnO2}][\Delta] 2KCl(s) + 3O2(g)}
Step-by-step solution

  1. Molar mass. KClO₃: 39.10 + 35.45 + 3(16.00) = 122.55 g/mol.

  2. Mass → moles.

    n(KClOX3)=24.5122.55=0.1999 mol\begin{aligned} n(\ce{KClO3}) &= \frac{24.5}{122.55} \\ &= 0.1999\ \text{mol} \end{aligned}

  3. Mole ratio. O₂ and KClO₃ are in a 3 : 2 ratio:

    n(OX2)=0.1999×32=0.2999 mol\begin{aligned} n(\ce{O2}) &= 0.1999 \times \frac{3}{2} \\ &= 0.2999\ \text{mol} \end{aligned}

  4. Moles → mass.

    m(OX2)=0.2999×32.00=9.60 g\begin{aligned} m(\ce{O2}) &= 0.2999 \times 32.00 \\ &= \boxed{9.60\ \text{g}} \end{aligned}

  5. Check with mass ratios. 2 × 122.55 = 245.1 g of KClO₃ gives 3 × 32.00 = 96.00 g of O₂, and 24.5 g is (almost exactly) one-tenth of 245.1 g, so it gives one-tenth of 96.00 g, or 9.60 g.

Practice: for C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l), how many moles of the requested substance?
  1. O₂ needed to burn 1 mol C₃H₈
  2. CO₂ formed from 1 mol C₃H₈
  3. H₂O formed from 1 mol C₃H₈
  4. CO₂ formed from 2 mol C₃H₈
  5. C₃H₈ burned by 10 mol O₂
  6. H₂O formed along with 3 mol CO₂
  7. CO₂ formed from 0.5 mol C₃H₈
  8. C₃H₈ burned by 2.5 mol O₂

Common mistakes

Mistake 1: “2H₂ + O₂ means 2 g of hydrogen react with 1 g of oxygen.” Coefficients give ratios of particles and moles, not grams. 2 mol H₂ (4.032 g) reacts with 1 mol O₂ (32.00 g), a mass ratio of about 1 : 8.

Mistake 2: “6.50 g of zinc makes 6.50 g of hydrogen, since the ratio is 1 : 1.” 1 : 1 is a mole ratio. Convert 6.50 g Zn to 0.09942 mol first, and then back to 0.200 g H₂. Never multiply grams by a mole ratio.

Mistake 3: “Any equation will do.” Coefficients are mole ratios only in a balanced equation. Before you calculate, check that each kind of atom is balanced (1.6.3), then read off the mole ratio.

Quick check

For 2H₂(g) + O₂(g) → 2H₂O(l), how many moles of O₂ react completely with 4 mol H₂?

8.0 g of CH₄ burns completely: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l). What mass of water forms?

Summary

  • In a balanced equation, the coefficients give the ratio of particles and of moles, not of masses.
  • Stoichiometry in three steps: grams A → moles A → moles B → grams B. Only the middle step uses the equation.
  • The US conversion-factor chain, the Chinese proportion layout, and the mass-ratio method of Stage 1 rest on the same idea and give the same answer.
On to AP (for later)

Stoichiometry is the core of AP Chemistry Unit 4 and shows up in calculations on nearly every exam. AP problems enter the mole bridge from different sides: mass (÷ M), gases (PV = nRT, 2.8.6), and solutions (molarity × volume, 2.9.4), but the middle step is always the mole ratio. AP also uses particle diagrams and asks you to draw the products in the right ratio.

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

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