Junior High Chemistry | Equation Coefficients
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Chemical equation
Balancing coefficients

This is not coefficient guessing. First use particle models to understand atom conservation, then remove the model: choose a single-path element → start from the larger atom count → lock balanced elements → follow the chain. If the chain gets stuck, introduce variables.

① Particle models② Observation method③ Chain balancing④ x–y algebra
1
See it first
Atoms are rearranged, not lost.
2
Remove the particle model
Count atoms directly from formulas.
3
Lock one element at a time
Once balanced, do not change it casually.
4
Use variables only when needed
Use observation first; algebra only when stuck.
01

Particle models: rule out two common mistakes

First recognize molecular forms and atom conservation, then move to coefficients.

Wrong ① | Not a valid molecular representation×
H+O
H + O
→
HOH
H₂O
Hydrogen and oxygen are not normally represented as isolated H and O atoms; recognize H₂ and O₂ first.
Wrong ② | Atoms are not conserved×
HH
+
OO
H₂ + O₂
→
HOH
H₂O
There are 2 O atoms on the left but only 1 on the right. Do not change subscripts; change coefficients.
Correct | Add one H₂ molecule✓
HH
HH
+
OO
2H₂ + O₂
→
HOH
HOH
2H₂O
Both sides have H = 4 and O = 2; atom types and counts match.
Next stage:The particle model makes conservation visible. In actual problems, remove the particles and count each element directly under the equation.
Open particle-model practice →
02

Remove the particle model: the observation sequence

Students often get stuck not on arithmetic, but on where to start.

① Find a single-path element

Choose an element present on both sides that appears in only one formula on each side.

② Start from the larger atom count

Compare its atom counts. Use the larger side as the reference; equal counts are also fine. The goal is to avoid repeated changes.

③ Lock it after balancing

Once balanced, leave that element alone. Follow the compound you just changed to the next connected element.

1
Find a single-path pair on both sides For C₂H₆ + O₂ → CO₂ + H₂O, C appears only in C₂H₆ on the left and only in CO₂ on the right, so C is a good starting element.
2
Start from the side with the larger atom countC₂H₆ has 2 C, CO₂ has 1 C, sofirstRight 2CO₂.
3
Lock C, then follow the H connected through C₂H₆Left 6 H, thereforeRightneed 3H₂O.at this point C、H then.
4
Finish with ORighthas 4 + 3 = 7 O, soLeft O₂ is 7/2; 2, smallest integer ratio.
03

Demo: equation on top, atom counts below

Watch the reasoning one action at a time. Yellow marks the current focus; the target element stays highlighted until balanced.

Ready
Look at the full equation firstPress “Next step” to reveal one reasoning move at a time.
1C2H6
+
7/2O2
→
2CO2
+
3H2O
—
+
—
=
—
+
—
—
+
—
=
—
+
—
—
+
—
=
—
+
—
How to read it:Under each compound, record only how many atoms of the current element it contributes; the arrow column always shows an equals sign.
Thinking nowBalanced / locked
0 / 16
04

Practice without hints

Choose the first target element yourself. Use a hint only when stuck.

Basic

Enter the smallest integer coefficients

Question 1 / 8
05

Too many free coefficients? Use x and y

Do not assign variables to every coefficient at the start. Use observation first; introduce x or y only where the next coefficient cannot be determined uniquely.

Example 1 | After Fe is balanced, C is still free

　Fe2O3+　CO→　CO2+　Fe
1
Use the observation method first: start with FeWrite the Fe₂O₃ coefficient 1, Lefthas 2 Fe, soRight Fe coefficientis 2.Fe balancelock it first.
2
Next comes C: both sides must match, but the amount is still unknownCO CO₂ has only 1 C, so“has x CO, has x CO₂”.at this point coefficient x.
1Fe2O3+xCO→xCO2+2Fe
ElementLeftRight
Fe2=2
Cx=x
O3 + x=2x
Only one linear equation remains:3 + x = 2x, so x = 3. This gives Fe₂O₃ + 3CO → 3CO₂ + 2Fe.

Example 2 | Set H₂O = 1, then use x and y

　Cu+　HNO3→　Cu(NO3)2+　NO2↑+1H2O
1
Start with H: set the H₂O coefficient to 1Right H₂O has 2 H, soLeft HNO₃ coefficientfirst 2.at this pointLefthas N = 2、O = 6.
　Cu+2HNO3→　Cu(NO3)2+　NO2↑+1H2O
2
Tie the Cu pair together with xhas x Cu, has x Cu(NO₃)₂.Righttherefore 2x N、6x O.
3
NO₂ is still free: introduce yy NO₂ contributes y N、2y O.now N、O equal on both sides x、y.
xCu+2HNO3→xCu(NO3)2+yNO2↑+1H2O
ElementLeftRight
H2=2
Cux=x
N2=2x + y
O6=6x + 2y + 1
Solve the final two equations:N: 2 = 2x + y; O: 6 = 6x + 2y + 1. Solving gives x = 1/2 and y = 1. The equation is ½Cu + 2HNO₃ → ½Cu(NO₃)₂ + NO₂↑ + H₂O, then multiply every coefficient by 2 to get Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂↑ + 2H₂O.
Key idea:Do not set every coefficient to a, b, c, d at the start. Use the observation method as far as possible; introduce x only for a truly free coefficient, and y only if one more degree of freedom remains. Solve them by keeping each element conserved on both sides.
06

Further practice

Use particle models to confirm atom conservation, then return here to balance without them.

Particle-model practice｜G82C11

Good for practicing the visual idea that atom types and counts are the same before and after a reaction.

Open particle-model question bank Open teaching video
A

Animated solution library (Engine A)

This section is data-driven: one animation engine can replay many equations. More question-bank reactions can be added without rebuilding the UI.

Ready
Choose an equationUse the buttons above to switch among equations from the question bank.
How to read it:Highlight the target element first, then the relevant compounds; coefficients and atom counts appear one step at a time in the teaching order.
Thinking nowBalanced / locked
0 / 0

📚 Want more chemistry practice?

Materials are tools. First learn the reasoning process here; choose extra paper practice only if you need it.

Junior-high science practice workbookSuitable for junior-high science practice and transition work. Advanced junior-high chemistry & physicsSuitable for students who want more intermediate problems and concept practice. Visual ChemistryExtends chemistry concepts through visual explanations.

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