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New YorkChemistrySyllabus dot point

Why must a chemical equation be balanced, and how do coefficients express conservation of mass and charge?

Balancing equations and conservation of mass: balance chemical equations by adjusting coefficients so atoms and charge are conserved, and interpret the coefficients as mole ratios.

A focused Regents Chemistry answer on balancing chemical equations: why mass and charge are conserved, how to adjust coefficients (never subscripts), and how the balanced coefficients give the mole ratios used in all stoichiometry.

Generated by Claude Opus 4.89 min answer

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  1. What this topic is asking
  2. Why equations must balance
  3. Coefficients, not subscripts
  4. A reliable balancing order
  5. Coefficients are mole ratios
  6. Try this

What this topic is asking

The Core Curriculum asks you to balance chemical equations so that matter is conserved, and to interpret the balanced coefficients as the mole ratios that drive stoichiometry. The Regents tests this as a Part A "what is the coefficient" question and as a Part B-2 "balance the equation" task. The rule that you adjust coefficients but never subscripts is central.

Why equations must balance

This is why an unbalanced equation, even if the formulas are correct, is chemically incomplete: it would imply atoms appearing or disappearing. Conservation of mass also means the total mass of reactants equals the total mass of products, a fact tested directly in some Part B-2 questions.

Coefficients, not subscripts

You balance only by changing coefficients. Changing a subscript would turn the substance into a different one: writing H2O2\text{H}_2\text{O}_2 (hydrogen peroxide) in place of H2O\text{H}_2\text{O} (water) to "balance" oxygen is wrong because it changes the chemical. The smallest whole-number set of coefficients is the accepted answer.

A reliable balancing order

A method that works for most Regents equations:

  1. Balance metals and other single elements first.
  2. Balance polyatomic ions as whole units if they appear unchanged on both sides.
  3. Leave hydrogen and then oxygen until last, as they often appear in several substances.
  4. Reduce the coefficients to the smallest whole numbers.

Always finish by re-counting every element on both sides, because adjusting one coefficient can unbalance an element you set earlier.

Coefficients are mole ratios

Once balanced, the coefficients tell you the mole ratio in which substances react and form. In CH4+2 O2→CO2+2 H2O\text{CH}_4 + 2\,\text{O}_2 \rightarrow \text{CO}_2 + 2\,\text{H}_2\text{O}, one mole of methane reacts with two moles of oxygen to make one mole of carbon dioxide and two moles of water. This ratio is exactly what you use in the stoichiometric-calculations page to convert between amounts of different substances.

Try this

Q1. Balance N2+H2→NH3\text{N}_2 + \text{H}_2 \rightarrow \text{NH}_3 with the smallest whole-number coefficients. [1 point]

  • Cue. N2+3 H2→2 NH3\text{N}_2 + 3\,\text{H}_2 \rightarrow 2\,\text{NH}_3.

Q2. State what the coefficient 22 in front of H2O\text{H}_2\text{O} tells you about the reaction. [1 point]

  • Cue. That two moles of water are produced for the amounts shown by the other coefficients.

Exam-style practice questions

Practice questions written in the style of NYSED exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Regents (Part B-2 style)2 marksBalance the following equation using the smallest whole-number coefficients: ‾ C3H8+‾ O2→‾ CO2+‾ H2O\underline{\quad}\,\text{C}_3\text{H}_8 + \underline{\quad}\,\text{O}_2 \rightarrow \underline{\quad}\,\text{CO}_2 + \underline{\quad}\,\text{H}_2\text{O}
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A 2-point constructed-response item asking for coefficients only.

Balance carbon first: 33 carbons need 3 CO23\,\text{CO}_2. Balance hydrogen next: 88 hydrogens need 4 H2O4\,\text{H}_2\text{O}. Balance oxygen last: the right side now has 3(2)+4(1)=103(2) + 4(1) = 10 oxygen atoms, so the left needs 5 O25\,\text{O}_2.

The balanced equation is C3H8+5 O2→3 CO2+4 H2O\text{C}_3\text{H}_8 + 5\,\text{O}_2 \rightarrow 3\,\text{CO}_2 + 4\,\text{H}_2\text{O}. Markers reward the correct smallest whole-number coefficients with oxygen balanced last.

Regents (Part A style)1 marksWhen the equation ‾ Al+‾ O2→‾ Al2O3\underline{\quad}\,\text{Al} + \underline{\quad}\,\text{O}_2 \rightarrow \underline{\quad}\,\text{Al}_2\text{O}_3 is correctly balanced using smallest whole numbers, the coefficient of O2\text{O}_2 is (1) 1 (2) 2 (3) 3 (4) 4
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A 1-point Part A balancing item. The answer is (3) 3.

Balancing aluminum and oxygen: 4 Al+3 O2→2 Al2O34\,\text{Al} + 3\,\text{O}_2 \rightarrow 2\,\text{Al}_2\text{O}_3. Check: 44 aluminum atoms on each side; oxygen is 3×2=63 \times 2 = 6 on the left and 2×3=62 \times 3 = 6 on the right. The coefficient of O2\text{O}_2 is 33.

The trap is balancing aluminum but forgetting to recheck oxygen; the smallest whole-number set has O2=3\text{O}_2 = 3.

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