Balancing Chemical Equations配平化学方程式

Big Idea · Transformations (TRA)AP link · Unit 4~20 min

Learning goals

  • Explain why a chemical equation must be balanced
  • Check whether an equation is balanced by counting atoms
  • Balance simple equations by inspection and with the least common multiple

Before you start

  • What a Chemical Formula Means (1.5.1)
  • Conservation of Mass (1.6.1)

Why balance equations?

In a chemical reaction, atoms are never created or destroyed. They are only rearranged into new substances. This is the law of conservation of mass. So in a correct chemical equation, every kind of atom must appear in equal numbers on the left (the reactants反应物) and on the right (the products生成物).

Take hydrogen burning to form water. If we write only the formulas:

HX2+OX2→HX2O\ce{H2 + O2 -> H2O}

there are 2 oxygen atoms on the left but only 1 on the right. An oxygen atom has “vanished,” so this equation is not balanced yet.

The key rule: change coefficients, never subscripts

Where it goesExampleCan you change it when balancing?
Subscript下标Lower right of an element symbolthe 2 in HX2O\ce{H2O}No. Changing it makes a different substance
Coefficient系数In front of a formulathe 2 in 2 HX2O\ce{2H2O}Yes. It’s the only thing you change

2 HX2O\ce{2H2O} means 2 water molecules: 2 × 2 = 4 H atoms and 2 × 1 = 2 O atoms in total.

The basic process:

  1. Write the formulas of the reactants and products
  2. Count each kind of atom on both sides
  3. Add coefficients in front of formulas until every kind of atom matches
  4. Check that the coefficients are in the lowest whole-number ratio

Example 1: Burning hydrogen

HX2+OX2→HX2O\ce{H2 + O2 -> H2O}

Try it yourself with the tool below first, then read the step-by-step solution. Each circle is one atom, so counting them shows you what’s unbalanced.

Try it: use − / + to change the coefficients
ElementLeft (reactants)Right (products)

Step-by-step solution
  1. Count atoms. Left: H = 2, O = 2. Right: H = 2, O = 1. O doesn’t match.
  2. Balance O. Put a 2 in front of HX2O\ce{H2O}. The right side now has 2 O, matching the left, but its H has also become 4.
  3. Balance H. Put a 2 in front of HX2\ce{H2}, so the left side also has 4 H.
  4. Check. H: 4 = 4, O: 2 = 2, and 2, 1, 2 is already the lowest ratio. Answer: 2 HX2+OX2→2 HX2O\ce{2H2 + O2 -> 2H2O}

Example 2: Burning methane

Methane, the main component of natural gas, burns to form carbon dioxide and water:

CHX4+OX2→COX2+HX2O\ce{CH4 + O2 -> CO2 + H2O}

Tip: Start with elements that appear in only one substance on each side (here, C and H). Save free elements like OX2\ce{O2} for last, because changing their coefficient doesn’t affect any other element.

Try it: use − / + to change the coefficients
ElementLeft (reactants)Right (products)

Step-by-step solution
  1. Balance C. Left C = 1, right C = 1. Already equal.
  2. Balance H. The left has 4 H, but HX2O\ce{H2O} has only 2, so put a 2 in front of HX2O\ce{H2O}. Now the right side has 4 H.
  3. Balance O last. Right O = 2 (from COX2\ce{CO2}) + 2 × 1 (from 2 HX2O\ce{2H2O}) = 4. Each OX2\ce{O2} has 2 O, so put a 2 in front of OX2\ce{O2}.
  4. Check. C: 1 = 1, H: 4 = 4, O: 4 = 4. Answer: CHX4+2 OX2→COX2+2 HX2O\ce{CH4 + 2O2 -> CO2 + 2H2O}

Example 3: Oxidizing aluminum

Aluminum reacts with oxygen in the air, forming a thin protective layer of aluminum oxide:

Al+OX2→AlX2OX3\ce{Al + O2 -> Al2O3}

This time O comes in groups of 2 on the left and groups of 3 on the right, which is hard to match by eye. Use the least common multiple method.

Try it: use − / + to change the coefficients
ElementLeft (reactants)Right (products)

Step-by-step solution
  1. Find the hardest element. O comes in groups of 2 on the left and groups of 3 on the right.
  2. Find the least common multiple. The LCM of 2 and 3 is 6, so aim for 6 O on each side: put 6 ÷ 2 = 3 in front of OX2\ce{O2} and 6 ÷ 3 = 2 in front of AlX2OX3\ce{Al2O3}.
  3. Balance Al. The right side has 2 × 2 = 4 Al, so put a 4 in front of Al\ce{Al}.
  4. Check. Al: 4 = 4, O: 6 = 6. Answer: 4 Al+3 OX2→2 AlX2OX3\ce{4Al + 3O2 -> 2Al2O3}

Common mistakes

Mistake 1: Changing subscripts. Turning HX2+OX2→HX2O\ce{H2 + O2 -> H2O} into HX2+OX2→HX2OX2\ce{H2 + O2 -> H2O2} does make the atoms match, but HX2OX2\ce{H2O2} is hydrogen peroxide, not water. It’s a different reaction.

Mistake 2: Not reducing the ratio. 4 HX2+2 OX2→4 HX2O\ce{4H2 + 2O2 -> 4H2O} has matching atoms, but every coefficient can be divided by 2. Write 2 HX2+OX2→2 HX2O\ce{2H2 + O2 -> 2H2O} instead.

Mistake 3: Forgetting to multiply the coefficient by the subscript. 3 OX2\ce{3O2} contains 3 × 2 = 6 O atoms, not 3 and not 5.

Quick check

Which is the correct, lowest-ratio balanced equation for burning hydrogen?

After balancing NX2+HX2→NHX3\ce{N2 + H2 -> NH3}, what is the coefficient of HX2\ce{H2}?

Summary

  • Balancing is based on conservation of mass: the number of each kind of atom doesn’t change in a reaction.
  • Change coefficients, never subscripts.
  • Use inspection for simple equations and balance free elements last. When that doesn’t work, use the least common multiple.
  • Finally, check that the coefficients are in the lowest whole-number ratio.
On to AP (for later)

In AP Unit 4 (Chemical Reactions), balancing is the first step of every stoichiometry calculation. Later you’ll also balance net ionic equations and redox equations.

Practice this chapter: Module 6 · Chemical Equations · 12 questions

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