How to Balance Chemical Equations Without Changing Subscripts
Learn why chemical subscripts must stay fixed when balancing equations, how to translate compound names into correct formulas, and how to adjust coefficients systematically. Includes worked examples, experimental evidence from water electrolysis, and five HSC-style practice questions with detailed solutions.
An electrolyser splits water into two gases. Suppose you collect 20.0 mL of gas at one electrode and 10.0 mL at the other. The gases are identified as hydrogen and oxygen.
A student notices that the oxygen volume is only half the hydrogen volume. They try changing the formula of water to make the numbers work.
That’s the wrong move. Changing a chemical subscript changes the substance itself. The correct equation is:
\[ \ce{2H2O(l) -> 2H2(g) + O2(g)} \]
The rule is straightforward: write the correct chemical formulas first, then balance the equation by changing coefficients in front of those formulas. Never change subscripts just to make the atom counts match.
The interesting part is understanding why. An experiment involving the decomposition of water gives us a useful way to see the difference between what we measure and what we infer about the particles involved.
01Why chemical subscripts cannot be changed
Imagine that you are balancing an equation involving water:
\[ \ce{H2 + O2 -> H2O} \]
Count the atoms.
| Element | Reactants | Products |
|---|---|---|
| Hydrogen | 2 | 2 |
| Oxygen | 2 | 1 |
Oxygen is unbalanced. You need two oxygen atoms on the product side.
It might seem sensible to change \(\ce{H2O}\) into \(\ce{H2O2}\). That gives you two oxygen atoms.
But \(\ce{H2O2}\) is hydrogen peroxide, not water.
You have made the atom count work by changing the reaction you were supposed to describe.
The correct approach is to keep the formula of water unchanged and put a coefficient of 2 in front of it. You then adjust the hydrogen coefficient as well:
\[ \ce{2H2 + O2 -> 2H2O} \]
Now both sides contain four hydrogen atoms and two oxygen atoms.
Coefficients and subscripts do different jobs
Consider the formula \(\ce{FeCl3}\), which represents iron(III) chloride.
There are two different ways to change the numbers in an equation involving this substance.
| Notation | What it means |
|---|---|
| \(\ce{2FeCl3}\) | Two formula units of iron(III) chloride |
| \(\ce{FeCl3}\) | One iron and three chlorine atoms per formula unit |
| \(\ce{FeCl2}\) | A different compound: iron(II) chloride |
A coefficient multiplies the entire chemical formula.
For example, \(\ce{2FeCl3}\) represents:
- 2 iron atoms
- 6 chlorine atoms
The subscript 3, however, belongs to the formula itself. It tells you the composition of each formula unit.
Think of a chemical formula as the ingredient list for a particular substance. A coefficient tells you how many copies of that substance are involved.
The analogy has one limitation: ionic compounds such as iron(III) chloride form extended crystal lattices rather than separate molecules. Their formulas describe the simplest whole-number ratio of ions, called a formula unit.
The important principle remains the same. The formula describes the substance. The coefficient describes how much of it participates in the reaction.
02What an experiment tells us about chemical formulas
Return to the electrolysis of water.
There is an important difference between observing a reaction and describing what happens to its particles.
An experiment might show bubbles forming at two electrodes. It might also allow us to measure the volume of each gas.
But you cannot look at an individual bubble and see the chemical formula of the gas inside it.
That requires further evidence and a particle model.
The apparatus and the original problem
A simple electrolysis experiment uses a container of water, two inert electrodes, and a direct-current power supply.
The negative terminal connects to the cathode, while the positive terminal connects to the anode.
A small amount of dilute sulfuric acid is added to improve the electrical conductivity of the water. Platinum electrodes can be used so that the electrodes do not participate significantly in the overall reaction.
When current passes through the solution:
- Gas forms at the cathode.
- Gas forms at the anode.
- The gases rise through the solution.
A basic beaker arrangement demonstrates that gases are produced, but it does not provide a reliable measurement of their relative volumes. Bubbles can escape, and the gases may not be collected separately.
To improve the experiment, use a gas-collection apparatus such as a Hofmann voltameter. It has separate graduated collection tubes above the electrodes, allowing each gas to be collected and measured independently.
What would you predict if water decomposes into hydrogen and oxygen?
The hydrogen volume should be twice the oxygen volume, provided both gases are measured at the same temperature and pressure.
This is what the experiment is designed to investigate.
The observation and the inference
Consider the following illustrative experimental readings. These are teaching data, not measurements from a historical experiment.
| Electrode | Gas identified | Collected volume |
|---|---|---|
| Cathode (negative) | Hydrogen | 20.0 mL |
| Anode (positive) | Oxygen | 10.0 mL |
The volume ratio is:
\[ \frac{V_{\ce{H2}}}{V_{\ce{O2}}} = \frac{20.0}{10.0} =2.00 \]
The hydrogen-to-oxygen volume ratio is therefore \(2:1\).
At the same temperature and pressure, equal volumes of gases contain equal numbers of particles, assuming ideal gas behaviour.
This means the ratio of hydrogen molecules to oxygen molecules is also \(2:1\).
We can represent the result as:
\[ 2\text{ hydrogen molecules} : 1\text{ oxygen molecule} \]
The gas molecules are represented by \(\ce{H2}\) and \(\ce{O2}\), respectively.
We can now construct the balanced equation:
\[ \ce{2H2O(l) -> 2H2(g) + O2(g)} \]
What did we actually observe?
This distinction is important in experimental chemistry.
| Experimental observation | Chemical interpretation |
|---|---|
| Bubbles form at both electrodes. | Gaseous products are being produced. |
| The cathode gas is identified as hydrogen. | Hydrogen molecules are represented by \(\ce{H2}\). |
| The anode gas is identified as oxygen. | Oxygen molecules are represented by \(\ce{O2}\). |
| Hydrogen has twice the measured volume of oxygen. | At equal temperature and pressure, the gas particle ratio is \(2:1\). |
| The balanced equation is constructed. | Two water molecules correspond to two hydrogen molecules and one oxygen molecule in the overall reaction. |
The apparatus does not directly show individual water molecules breaking apart. The particle-level explanation is an interpretation supported by the gas identification, volume measurements, and chemical model.
There is another important limitation: the \(2:1\) gas-volume ratio alone does not prove the molecular formula of water. You also need to know that the collected gases are molecular hydrogen and molecular oxygen.
This is why balancing an equation cannot be used to invent the formulas of unknown substances. You need chemical evidence to establish the substances first.
03How to translate chemical names into formulas
Before balancing any equation, you need to know which substances are involved.
A name such as magnesium chloride does not tell you the formula merely by looking at the number of letters or atoms in the name. You need to understand how the substance is constructed.
The method depends on the type of substance.
Elemental molecules
Some elements exist naturally as molecules containing two atoms.
The seven common diatomic elements are:
| Element | Formula |
|---|---|
| Hydrogen | \(\ce{H2}\) |
| Nitrogen | \(\ce{N2}\) |
| Oxygen | \(\ce{O2}\) |
| Fluorine | \(\ce{F2}\) |
| Chlorine | \(\ce{Cl2}\) |
| Bromine | \(\ce{Br2}\) |
| Iodine | \(\ce{I2}\) |
For example, if a question says chlorine gas reacts with sodium, the reactant is \(\ce{Cl2}\), not \(\ce{Cl}\).
This matters when balancing because the coefficient counts whole chlorine molecules, each containing two chlorine atoms.
Ionic compounds
Ionic compounds are formed from positively charged ions (cations) and negatively charged ions (anions).
Their formulas must have an overall charge of zero.
Consider magnesium chloride.
Magnesium forms \(\mathrm{Mg}^{2+}\), while chloride forms \(\mathrm{Cl}^{-}\).
One magnesium ion needs two chloride ions to balance its charge:
\[ (+2)+2(-1)=0 \]
Therefore, magnesium chloride is:
\[ \ce{MgCl2} \]
You do not write \(\ce{MgCl}\), because that would give the formula unit an overall charge of \(+1\).
The same reasoning works for more complicated ionic compounds.
Polyatomic ions and parentheses
A polyatomic ion contains multiple atoms bonded together that carry an overall charge.
For example, nitrate is:
\[ \mathrm{NO_3}^{-} \]
Calcium forms \(\mathrm{Ca}^{2+}\).
Two nitrate ions are needed to balance one calcium ion, giving:
\[ \ce{Ca(NO3)2} \]
The parentheses are important.
The subscript 2 outside the parentheses multiplies every atom inside the nitrate group.
Therefore, one formula unit of calcium nitrate contains:
- 1 calcium atom
- 2 nitrogen atoms
- 6 oxygen atoms
Do not confuse the subscript 3 inside nitrate with the subscript 2 outside the brackets.
Roman numerals in compound names
Transition metals can form ions with different charges.
The Roman numeral in a compound name tells you which charge to use.
For example:
| Compound name | Relevant ions | Formula |
|---|---|---|
| Iron(II) chloride | \(\mathrm{Fe}^{2+}\), \(\mathrm{Cl}^{-}\) | \(\ce{FeCl2}\) |
| Iron(III) chloride | \(\mathrm{Fe}^{3+}\), \(\mathrm{Cl}^{-}\) | \(\ce{FeCl3}\) |
| Copper(II) oxide | \(\mathrm{Cu}^{2+}\), \(\mathrm{O}^{2-}\) | \(\ce{CuO}\) |
The Roman numeral is not a coefficient or a subscript. It identifies the charge of the metal ion.
Molecular compounds
For many covalent compounds, prefixes in the name indicate the number of atoms.
For example:
- Carbon monoxide: \(\ce{CO}\)
- Carbon dioxide: \(\ce{CO2}\)
- Dinitrogen monoxide: \(\ce{N2O}\)
- Dinitrogen tetroxide: \(\ce{N2O4}\)
These subscripts are part of the chemical formula. They cannot be adjusted during balancing.
One final detail: state symbols also matter when interpreting a reaction. For example, hydrogen chloride gas is \(\ce{HCl(g)}\), while hydrochloric acid is \(\ce{HCl(aq)}\). The formula is the same, but the state symbol tells you how the substance is present.
04Worked examples
Now apply the distinction between writing formulas and balancing coefficients.
Worked example: Formation of iron(III) chloride
Solid iron reacts with chlorine gas to produce iron(III) chloride. Write the balanced chemical equation.
Step 1
Iron is represented by \(\ce{Fe}\).
Chlorine is diatomic in its elemental form, so chlorine gas is \(\ce{Cl2}\).
The product is iron(III) chloride. Iron(III) has a charge of \(+3\), while chloride has a charge of \(-1\).
Therefore, the product formula is \(\ce{FeCl3}\).
Our unbalanced equation is:
\[ \ce{Fe(s) + Cl2(g) -> FeCl3(s)} \]
Step 2
There are two chlorine atoms on the reactant side and three on the product side.
We need a common number that can be made from groups of two and three.
The smallest common number is six.
Place a coefficient of 3 before chlorine gas and a coefficient of 2 before iron(III) chloride:
\[ \ce{Fe + 3Cl2 -> 2FeCl3} \]
Step 3
There are now two iron atoms on the product side.
Place a coefficient of 2 before iron:
\[ \ce{2Fe + 3Cl2 -> 2FeCl3} \]
Step 4
| Element | Reactants | Products |
|---|---|---|
| Fe | 2 | 2 |
| Cl | 6 | 6 |
The final balanced equation is:
\[ \boxed{\ce{2Fe(s) + 3Cl2(g) -> 2FeCl3(s)}} \]
Notice that we never changed \(\ce{FeCl3}\) into \(\ce{FeCl2}\).
Doing so would change the product from iron(III) chloride into iron(II) chloride. The coefficient is what needed adjusting.
Worked example: Lead(II) nitrate reacting with potassium iodide
Aqueous lead(II) nitrate reacts with aqueous potassium iodide. A yellow precipitate of lead(II) iodide forms, while potassium nitrate remains dissolved.
Write the balanced molecular equation, including state symbols.
Step 1
Lead(II) has a charge of \(+2\), and nitrate has a charge of \(-1\).
Therefore, lead(II) nitrate is:
\[ \ce{Pb(NO3)2} \]
Potassium has a charge of \(+1\), while iodide has a charge of \(-1\). Potassium iodide is:
\[ \ce{KI} \]
For the products, lead(II) iodide is \(\ce{PbI2}\), and potassium nitrate is \(\ce{KNO3}\).
Our initial equation is:
\[ \ce{Pb(NO3)2(aq) + KI(aq) -> PbI2(s) + KNO3(aq)} \]
Step 2
There are two nitrate groups on the reactant side, but only one on the product side.
Place a coefficient of 2 before potassium nitrate:
\[ \ce{Pb(NO3)2 + KI -> PbI2 + 2KNO3} \]
Step 3
There are now two potassium atoms on the product side.
Place a coefficient of 2 before potassium iodide:
\[ \ce{Pb(NO3)2 + 2KI -> PbI2 + 2KNO3} \]
This also gives us two iodine atoms, which balances the lead(II) iodide.
Step 4
| Element | Reactants | Products |
|---|---|---|
| Pb | 1 | 1 |
| K | 2 | 2 |
| N | 2 | 2 |
| O | 6 | 6 |
| I | 2 | 2 |
The final balanced equation is:
\[ \boxed{ \ce{ Pb(NO3)2(aq) + 2KI(aq) -> PbI2(s) + 2KNO3(aq) } } \]
The yellow solid is an experimental observation. Identifying it as lead(II) iodide is the chemical interpretation based on the known reactants and product chemistry.
Notice that the nitrate group remained unchanged in the overall equation. Keeping track of it as a group makes balancing much easier.
05A reliable decision process
When you are given a chemical equation to balance, use this sequence.
| Stage | What you should do | What to avoid |
|---|---|---|
| 1. Identify substances | Translate every name into its correct formula. | Guessing formulas to make the equation balance. |
| 2. Check the formulas | Verify ionic charges, subscripts, and elemental molecules. | Changing a formula because an atom count looks wrong. |
| 3. Write the equation | Place the correct reactants and products on their respective sides. | Trying to balance before identifying the products. |
| 4. Adjust coefficients | Change the numbers in front of formulas. | Editing subscripts or removing atoms. |
| 5. Check the result | Count every element on both sides. | Checking only the elements you changed. |
| 6. Simplify | Use the smallest whole-number coefficient ratio. | Leaving unnecessary common factors. |
A useful final check is to ask yourself two separate questions:
- Do the formulas represent the actual substances involved?
- Are the numbers of each type of atom equal on both sides?
Both conditions must be satisfied.
An equation can have perfectly balanced atoms and still describe the wrong chemical reaction.
06Questions and solutions
Question 1
Sodium metal reacts with chlorine gas to produce solid sodium chloride.
Write the correct formulas and balance the chemical equation.
Solution 1
The balanced equation is:
\[ \ce{2Na(s) + Cl2(g) -> 2NaCl(s)} \]
Sodium forms \(\mathrm{Na}^{+}\), while chloride forms \(\mathrm{Cl}^{-}\). Their charges cancel in a \(1:1\) ratio, so sodium chloride has the formula \(\ce{NaCl}\).
Chlorine exists as a diatomic molecule, so its elemental formula is \(\ce{Cl2}\).
There are two chlorine atoms on the reactant side. Therefore, we need two formula units of sodium chloride:
\[ \ce{Na + Cl2 -> 2NaCl} \]
This gives two sodium atoms on the product side, so we also place a coefficient of 2 before sodium.
The final equation contains two sodium atoms and two chlorine atoms on each side.
The important trap is writing chlorine as \(\ce{Cl}\) simply because sodium chloride contains one chlorine atom per formula unit. The elemental reactant and the ionic product have different formulas.
Question 2
Aqueous calcium nitrate reacts with aqueous sodium carbonate to produce solid calcium carbonate and aqueous sodium nitrate.
(a) Write the correct formula for each substance.
(b) Write the balanced equation, including state symbols.
(c) Explain why the formula of calcium nitrate contains parentheses.
Solution 2
The correct formulas are \(\ce{Ca(NO3)2}\), \(\ce{Na2CO3}\), \(\ce{CaCO3}\), and \(\ce{NaNO3}\).
The balanced equation is:
\[ \ce{ Ca(NO3)2(aq) + Na2CO3(aq) -> CaCO3(s) + 2NaNO3(aq) } \]
First, construct the formulas using ionic charges.
Calcium forms \(\mathrm{Ca}^{2+}\), and nitrate is \(\mathrm{NO_3}^{-}\). Two nitrate ions are needed to balance one calcium ion.
Therefore:
\[ \ce{Ca(NO3)2} \]
Sodium forms \(\mathrm{Na}^{+}\), while carbonate is \(\mathrm{CO_3}^{2-}\). Two sodium ions are needed to balance one carbonate ion:
\[ \ce{Na2CO3} \]
For the products, calcium carbonate has the formula \(\ce{CaCO3}\), and sodium nitrate is \(\ce{NaNO3}\).
Now balance the equation.
There are two nitrate groups on the reactant side. Place a coefficient of 2 before sodium nitrate:
\[ \ce{Ca(NO3)2 + Na2CO3 -> CaCO3 + 2NaNO3} \]
This also balances sodium because there are two sodium atoms on each side.
Calcium and carbonate are already balanced.
The parentheses in calcium nitrate are necessary because the subscript 2 applies to the entire nitrate group. Without them, the formula would not correctly represent two nitrate ions.
Question 3
An electrolyser decomposes water into hydrogen and oxygen.
The following illustrative measurements are recorded after correcting for water vapour. Both gases are measured at the same temperature and pressure.
| Gas | Collected volume |
|---|---|
| Hydrogen | 18.0 mL |
| Oxygen | 9.00 mL |
(a) Calculate the hydrogen-to-oxygen volume ratio.
(b) Write the balanced chemical equation for the decomposition of water.
(c) If 15.0 mL of oxygen is produced under the same conditions, calculate the expected hydrogen volume and the total volume of dry gas collected.
(d) Explain why the measured volume ratio does not, by itself, establish the molecular formula of water.
Solution 3
(a) The hydrogen-to-oxygen volume ratio is \(2:1\).
Calculate the ratio:
\[ \frac{V_{\ce{H2}}}{V_{\ce{O2}}} = \frac{18.0}{9.00} =2.00 \]
Therefore:
\[ \ce{H2:O2}=2:1 \]
At constant temperature and pressure, gas volume is proportional to the amount of gas in moles.
This means the mole ratio of hydrogen to oxygen is also \(2:1\).
(b) The balanced chemical equation is:
\[ \ce{2H2O(l) -> 2H2(g) + O2(g)} \]
Two water molecules contain four hydrogen atoms and two oxygen atoms.
The products contain the same totals:
- Two \(\ce{H2}\) molecules contain four hydrogen atoms.
- One \(\ce{O2}\) molecule contains two oxygen atoms.
Both elements are conserved.
(c) The expected hydrogen volume is 30.0 mL, giving a total dry gas volume of 45.0 mL.
The volume ratio is \(2:1\).
Therefore:
\[ V_{\ce{H2}}=2(15.0) \]
\[ V_{\ce{H2}}=30.0\text{ mL} \]
The total volume is:
\[ V_{\text{total}}=30.0+15.0 \]
\[ \boxed{V_{\text{total}}=45.0\text{ mL}} \]
This assumes complete collection, no gas losses, no competing reactions, and equal temperature and pressure.
(d) The volume ratio establishes the relative amounts of the two gaseous products, not the molecular formula of the original liquid by itself.
To construct the equation, we also need to identify the gases as molecular hydrogen, \(\ce{H2}\), and molecular oxygen, \(\ce{O2}\).
The molecular formula of water is established through chemical evidence about its composition and structure. The balanced equation then describes how those particles are rearranged.
The tempting mistake is to treat a measured gas ratio as direct proof of every subscript in the reactant formula. Measurements support a chemical model, but they do not replace the evidence needed to identify the substances.
Question 4
A student mixes aqueous iron(III) chloride with aqueous sodium hydroxide.
A reddish-brown precipitate forms, while sodium chloride remains dissolved in the solution.
(a) Write the correct formula for each reactant and product.
(b) Write the balanced chemical equation, including state symbols.
(c) Another student suggests changing iron(III) hydroxide from \(\ce{Fe(OH)3}\) to \(\ce{Fe(OH)2}\) to make the equation easier to balance. Explain why this is incorrect.
Solution 4
The correct balanced equation is:
\[ \ce{ FeCl3(aq) + 3NaOH(aq) -> Fe(OH)3(s) + 3NaCl(aq) } \]
The formula of iron(III) chloride is \(\ce{FeCl3}\), because iron(III) has a \(+3\) charge and chloride has a \(-1\) charge.
Sodium hydroxide is \(\ce{NaOH}\), and iron(III) hydroxide is \(\ce{Fe(OH)3}\).
The hydroxide ion has a charge of \(-1\). Three hydroxide ions are therefore needed to balance one iron(III) ion.
This fixes the product formula before balancing begins.
Next, count the chlorine atoms. There are three chlorine atoms on the reactant side, so place a coefficient of 3 before sodium chloride.
This gives three sodium atoms on the product side, requiring a coefficient of 3 before sodium hydroxide.
The completed equation is:
\[ \ce{ FeCl3(aq) + 3NaOH(aq) -> Fe(OH)3(s) + 3NaCl(aq) } \]
The atom counts are:
| Element | Reactants | Products |
|---|---|---|
| Fe | 1 | 1 |
| Cl | 3 | 3 |
| Na | 3 | 3 |
| O | 3 | 3 |
| H | 3 | 3 |
Changing \(\ce{Fe(OH)3}\) to \(\ce{Fe(OH)2}\) would change the product into a different compound.
It would also fail to represent the charge-balanced combination of \(\mathrm{Fe}^{3+}\) and \(\mathrm{OH}^{-}\).
The reddish-brown precipitate is an observation. Its identification as iron(III) hydroxide is a chemical interpretation based on the reaction and the known ions.
You cannot change the formula simply because a different formula might make the arithmetic easier.
Question 5
A hypothetical pure molecular compound contains only nitrogen and oxygen.
The compound is completely decomposed into nitrogen gas and oxygen gas. No other products form, and all the gases are collected.
At the same temperature and pressure, the measured gas volumes are:
| Product gas | Volume |
|---|---|
| Nitrogen, \(\ce{N2}\) | 40.0 mL |
| Oxygen, \(\ce{O2}\) | 30.0 mL |
A student makes the following claims:
- The empirical formula of the original compound is \(\ce{N4O3}\).
- The molecular formula must also be \(\ce{N4O3}\), because that formula gives the balanced equation with the smallest whole-number coefficients.
- Changing a subscript is acceptable whenever it makes an equation balance.
Assess all three claims.
Determine the empirical formula, write a balanced decomposition equation using that empirical formula, and explain whether the molecular formula can be established from the information provided.
Solution 5
The empirical formula is \(\ce{N4O3}\), but the molecular formula cannot be determined uniquely from the gas-volume data alone.
The student’s first claim is correct. The second and third claims are incorrect.
Step 1: Determine the ratio of product gas molecules.
At the same temperature and pressure, gas volumes are proportional to the number of gas particles.
Therefore:
\[ \ce{N2:O2}=40.0:30.0 \]
Simplifying:
\[ \ce{N2:O2}=4:3 \]
There are four nitrogen molecules for every three oxygen molecules in the collected products.
Step 2: Convert the molecular ratio into an atomic ratio.
Each nitrogen molecule contains two nitrogen atoms.
Each oxygen molecule contains two oxygen atoms.
Therefore, the relative numbers of atoms are:
\[ N:O=(4\times2):(3\times2) \]
\[ N:O=8:6 \]
Simplifying gives:
\[ N:O=4:3 \]
The empirical formula is therefore:
\[ \boxed{\ce{N4O3}} \]
The empirical formula represents the simplest whole-number ratio of atoms in the compound.
Step 3: Write the balanced decomposition equation.
Using the empirical formula as the starting composition:
\[ \ce{N4O3 -> N2 + O2} \]
We need four nitrogen atoms and three oxygen atoms on the product side for each empirical formula unit.
This gives:
\[ \boxed{\ce{2N4O3 -> 4N2 + 3O2}} \]
Check the atom counts:
| Element | Reactants | Products |
|---|---|---|
| N | 8 | 8 |
| O | 6 | 6 |
The equation is balanced.
Step 4: Decide whether the molecular formula is known.
This is where the student’s second claim fails.
An empirical formula gives the simplest atom ratio, but a molecular formula gives the actual number of each atom in one molecule.
For a molecular compound, the molecular formula must be a whole-number multiple of the empirical formula.
For example, the possible relationship is:
\[ (\ce{N4O3})_n \]
where \(n\) is a positive whole number.
The gas-volume measurements establish the \(4:3\) atomic ratio. They do not establish how many empirical formula units are contained in one molecule.
To determine the molecular formula, we would need additional information, such as the compound’s molar mass.
Step 5: Resolve the apparent contradiction about changing subscripts.
The student’s third claim confuses two different stages of a chemistry problem.
When determining an unknown compound’s empirical formula, you use experimental evidence to infer the appropriate atom ratio. That is how we obtained \(\ce{N4O3}\).
Once the identity and formula of a substance have been established, however, you cannot change its subscripts simply to balance a reaction.
The formula must remain fixed while coefficients are adjusted.
This is the important distinction: experimental evidence can establish a formula, but balancing an equation cannot establish or alter the identity of a substance.
A balanced equation must describe the chemistry that actually occurs, not merely an arrangement of numbers that makes the atom totals agree.
The next useful step is learning to use balanced coefficients as mole ratios. Once the formulas are correct and the equation is balanced, you can calculate how much reactant is needed, predict product quantities, and identify the limiting reactant.
For a broader treatment of identifying products and balancing ordinary chemical reactions, see How to Predict Products and Balance Chemical Equations.
Remember that nuclear equations use a different system of notation and conservation rules. See How to Balance Nuclear Equations for HSC Chemistry for that separate method.