Why You Must Balance Equations Before Calculating Reacting Masses

Suppose 8.00 g of oxygen reacts with 1.00 g of hydrogen to produce water. You write the equation:

\[ \ce{H2 + O2 -> H2O} \]

The equation looks reasonable. But if you use its apparent 1:1 mole ratio, you predict that 0.250 mol of oxygen produces 0.250 mol of water, or 4.50 g.

There is a problem. The correctly balanced equation is:

\[ \ce{2H2 + O2 -> 2H2O} \]

This predicts 0.500 mol of water, or 9.00 g. That’s twice the previous answer, using exactly the same starting materials.

Why the difference? The unbalanced equation has two oxygen atoms on the reactant side but only one on the product side. It has effectively lost half the oxygen atoms from the calculation.

You must balance a chemical equation before calculating reacting masses because its coefficients determine the mole ratios between reactants and products. An unbalanced equation cannot provide valid mole ratios, so any mass calculation based on its coefficients can be wrong.

This is not a minor correction to make at the end. Balancing is what makes the calculation possible in the first place.

01Why balancing matters: where the idea came from

The reason balanced equations matter becomes clearer when you consider an old experimental problem.

When a metal is heated in air, it can become heavier. That seems strange at first. If heating is changing the metal, where does the extra mass come from?

In the eighteenth century, this was an important question in chemistry. Different explanations were proposed for the increase in mass during a process called calcination, in which a metal reacts with air to form a solid oxide.

Antoine Lavoisier investigated this problem using carefully controlled measurements.

Lavoisier’s experiment with tin

Imagine a glass retort containing a measured quantity of tin.

A retort is a glass vessel with a rounded body and a narrow neck. The tin sits inside the rounded section, while the narrow neck allows the vessel to be sealed.

The apparatus was arranged as follows:

1. Tin was placed inside the glass retort.
2. The tin was heated until it melted.
3. The narrow glass neck was sealed, trapping the air inside.
4. The complete sealed apparatus was weighed.
5. The sealed retort was heated again to allow the tin to calcine.
6. After cooling, the complete apparatus was weighed again.
7. Finally, the narrow neck was broken, allowing outside air to enter.

The important measurement was the mass of the entire retort and its contents, not just the tin.

What should the experiment show?

Consider two possible explanations.

• If some material from the fire passed through the glass and joined the tin, the total mass of the sealed apparatus should increase.
• If the tin gained mass by combining with air already trapped inside the retort, the tin would become heavier, but the total mass of the sealed apparatus would remain unchanged.

The second prediction was supported by the observations.

The sealed retort had the same total mass before and after heating. However, when its neck was broken, air rushed inside and the total mass increased.

The increase in the mass of the calcined tin was accounted for by the mass of air that entered when the vessel was opened.

Lavoisier concluded that part of the air had combined with the tin. Later work established that oxygen was the component responsible.

His experiments were described in his work on the calcination of tin in closed vessels. A historical account of the apparatus and measurements is available in M. M. Pattison Muir’s Heroes of Science: Chemists.

Why was the revised apparatus important?

An open-air experiment that measures only the metal cannot track everything entering and leaving the system.

The sealed retort addressed this problem by keeping the reacting materials inside a defined boundary. The balance could then measure the total mass of the system before and after the reaction.

This helped distinguish two different observations:

• The metal gained mass because material from the air combined with it.
• The total mass of the closed system did not change because that material was already inside the system.

There is an important historical qualification here. Lavoisier did not invent the idea that matter is conserved. Earlier thinkers, including Jean Rey, had considered the possibility that air contributed to the increased mass of calcined metals. Lavoisier’s contribution included developing a particularly careful quantitative approach and establishing the significance of oxygen in chemical reactions. Carmen Giunta’s historical teaching notes discuss this distinction.

A small calculation using the modern interpretation

We can represent the formation of tin(IV) oxide using the modern balanced equation:

\[ \ce{Sn + O2 -> SnO2} \]

Assume that pure tin(IV) oxide forms completely.

The equation tells us that:

\[ 1\text{ mol Sn}:1\text{ mol O}_2:1\text{ mol SnO}_2 \]

Suppose 0.0100 mol of tin reacts.

The amount of oxygen required is:

\[ n(\ce{O2})=0.0100\text{ mol} \]

Using the molar mass of oxygen gas, \(32.0\ \mathrm{g\,mol^{-1}}\):

\[ \begin{aligned} m(\ce{O2})&=nM\\ &=0.0100\times32.0\\ &=0.320\text{ g} \end{aligned} \]

The tin gains 0.320 g as oxygen combines with it.

The sealed apparatus, however, has not gained any mass. The oxygen has simply moved from the trapped air into the solid product.

This is an illustrative modern calculation, not a measurement from Lavoisier’s historical experiment.

The connection to modern stoichiometry is straightforward: a balanced equation keeps track of how atoms are rearranged without creating or destroying them.

02What the coefficients actually tell you

Consider the reaction between magnesium and oxygen:

\[ \ce{2Mg + O2 -> 2MgO} \]

Count the atoms on each side.

Both elements have the same number of atoms before and after the reaction.

The coefficients provide the proportions in which the substances react:

\[ 2\text{ mol Mg}:1\text{ mol O}_2:2\text{ mol MgO} \]

These are mole ratios, not mass ratios.

A coefficient tells you how many particles of a substance participate in the reaction. Because a mole contains a fixed number of particles, the same coefficient ratio applies when working with moles.

Coefficients are not subscripts

This is another important distinction.

A coefficient changes how many particles of a substance are involved. A subscript changes the chemical formula itself.

For example:

• \(\ce{2H2O}\) means two water molecules.
• \(\ce{H2O2}\) means one hydrogen peroxide molecule.

These are different substances.

When balancing an equation, you change the coefficients in front of formulas. You do not change the subscripts to make the atom counts match.

Changing \(\ce{H2O}\) into \(\ce{H2O2}\) does not balance the formation of water. It changes the product.

Why coefficients do not give you masses

Think of the coefficients as the quantities in a recipe. They tell you how many portions of each ingredient are needed.

But knowing that a recipe uses two portions of one ingredient and one portion of another does not tell you that their masses are in a 2:1 ratio.

The same applies to chemical equations.

For example:

\[ \ce{2H2 + O2 -> 2H2O} \]

The coefficients give a mole ratio of:

\[ 2:1:2 \]

However, the corresponding masses are:

The masses follow the relationship:

\[ 4.0+32.0=36.0\text{ g} \]

The coefficients give the number of moles, while the molar masses tell us how much those moles weigh.

The recipe analogy stops here: chemical particles have specific compositions and masses, and reactions rearrange their atoms rather than simply combining identical portions.

This is why you cannot normally read a balanced equation and immediately use its coefficients as a mass ratio.

03The mass-to-moles-to-moles-to-mass method

Once the equation is balanced, you can calculate the mass of a substance produced or consumed.

The method follows four stages:

Mass → moles → mole ratio → moles → mass

You need two equations:

To convert mass into moles:

\[ n=\frac{m}{M} \]

To convert moles into mass:

\[ m=nM \]

Where:

• \(n\) is the amount of substance, measured in moles (mol).
• \(m\) is the mass, measured in grams (g).
• \(M\) is the molar mass, measured in grams per mole (\(\mathrm{g\,mol^{-1}}\)).

The balanced equation supplies the conversion between the two amounts in moles.

For a known substance A and a required substance B:

\[ n_B=n_A\times\frac{\nu_B}{\nu_A} \]

Here, \(\nu_A\) and \(\nu_B\) are the coefficients of substances A and B in the balanced equation.

The complete calculation can be organised like this:

There is one additional decision if the question supplies two or more reactants.

You must identify the limiting reactant: the reactant that runs out first and therefore determines how much product can form.

The balanced equation tells you the required proportions, but it does not guarantee that all the starting reactants will be consumed.

04What balancing cannot tell you

A balanced equation gives the theoretical stoichiometric relationship between substances.

It does not automatically tell you the actual mass collected in an experiment.

For a straightforward HSC calculation, you will often assume:

• The reactants are pure.
• The equation describes the only reaction occurring.
• The limiting reactant reacts completely.
• No product is lost during collection.
• There are no significant side reactions.

If these assumptions do not hold, the actual product mass may differ from the theoretical prediction.

For example, an incomplete reaction produces less product than the theoretical amount. A side reaction may consume some of the starting material. A gas may escape before it can be collected.

The coefficients still describe the specified reaction, but you must account for what actually happens to the substances.

All numerical data in the following practice questions are illustrative. Use the molar masses supplied in each question and assume the stated reaction conditions apply.

05Questions and solutions

Question 1

Calcium oxide reacts with carbon dioxide to form calcium carbonate.

\[ \ce{CaO + CO2 -> CaCO3} \]

A student reacts 5.60 g of calcium oxide with excess carbon dioxide.

Calculate the theoretical mass of calcium carbonate produced.

Use:

• \(M(\ce{CaO})=56.1\ \mathrm{g\,mol^{-1}}\)
• \(M(\ce{CaCO3})=100.1\ \mathrm{g\,mol^{-1}}\)

Assume the reaction proceeds completely.

Solution 1

The theoretical mass of calcium carbonate is 9.99 g.

Step 1: Identify the mole ratio.

The equation is already balanced:

\[ \ce{CaO + CO2 -> CaCO3} \]

The ratio of calcium oxide to calcium carbonate is 1:1.

Step 2: Convert calcium oxide into moles.

\[ \begin{aligned} n(\ce{CaO})&=\frac{5.60}{56.1}\\ &=0.0998\text{ mol} \end{aligned} \]

Step 3: Calculate the moles of calcium carbonate.

Using the 1:1 ratio:

\[ \begin{aligned} n(\ce{CaCO3}) &=0.0998\times\frac{1}{1}\\ &=0.0998\text{ mol} \end{aligned} \]

Step 4: Convert the product amount into mass.

\[ \begin{aligned} m(\ce{CaCO3})&=nM\\ &=0.0998\times100.1\\ &=9.99\text{ g} \end{aligned} \]

The answer is 9.99 g of calcium carbonate.

The product has a greater mass than the starting calcium oxide because carbon dioxide supplies additional carbon and oxygen atoms.

Question 2

An impure sample has a total mass of 8.00 g. It contains potassium chlorate and an inert solid that does not react.

When potassium chlorate is heated, it decomposes according to:

\[ \ce{2KClO3 -> 2KCl + 3O2} \]

The reaction proceeds to completion, and 1.92 g of oxygen gas is collected.

Calculate:

(a) The mass of pure potassium chlorate in the original sample.

(b) The percentage purity of the original sample.

(c) The mass of potassium chloride produced.

(d) The mass of inert solid remaining.

Use:

• \(M(\ce{KClO3})=122.6\ \mathrm{g\,mol^{-1}}\)
• \(M(\ce{KCl})=74.6\ \mathrm{g\,mol^{-1}}\)
• \(M(\ce{O2})=32.0\ \mathrm{g\,mol^{-1}}\)

Assume all oxygen is collected and no potassium chloride or inert solid is lost.

Solution 2

The original sample contains 4.90 g of potassium chlorate, giving a purity of 61.3%. It produces 2.98 g of potassium chloride, with 3.10 g of inert solid remaining.

Step 1: Calculate the moles of oxygen produced.

\[ \begin{aligned} n(\ce{O2})&=\frac{m}{M}\\ &=\frac{1.92}{32.0}\\ &=0.0600\text{ mol} \end{aligned} \]

Step 2: Use the balanced equation to find the amount of potassium chlorate that decomposed.

The equation gives:

\[ 2\text{ mol }\ce{KClO3}:3\text{ mol }\ce{O2} \]

Therefore:

\[ \begin{aligned} n(\ce{KClO3}) &=0.0600\times\frac{2}{3}\\ &=0.0400\text{ mol} \end{aligned} \]

Step 3: Convert the potassium chlorate amount into mass.

\[ \begin{aligned} m(\ce{KClO3})&=nM\\ &=0.0400\times122.6\\ &=4.904\text{ g} \end{aligned} \]

The original sample contained 4.90 g of pure potassium chlorate.

Step 4: Calculate the percentage purity.

Percentage purity is the mass of the pure substance divided by the total sample mass, multiplied by 100.

\[ \begin{aligned} \text{Purity} &=\frac{4.904}{8.00}\times100\\ &=61.3\% \end{aligned} \]

Step 5: Calculate the mass of potassium chloride produced.

The equation gives a 2:2 ratio between potassium chlorate and potassium chloride.

\[ \begin{aligned} n(\ce{KCl})&=0.0400\text{ mol}\\ m(\ce{KCl})&=0.0400\times74.6\\ &=2.984\text{ g} \end{aligned} \]

The potassium chloride mass is 2.98 g.

Step 6: Calculate the inert solid remaining.

The inert solid does not participate in the reaction.

\[ \begin{aligned} m_{\text{inert}} &=8.00-4.904\\ &=3.096\text{ g} \end{aligned} \]

The inert solid mass is 3.10 g.

A final mass check confirms the result:

\[ 2.984+3.096+1.920=8.000\text{ g} \]

The potassium chloride, inert solid, and released oxygen account for the entire original sample.

The important detail is that the equation calculates the amount of potassium chlorate that reacted. It does not tell you how much of the original sample was inert. You need the initial total mass to calculate purity.

Question 3

Carbon monoxide reacts with oxygen to form carbon dioxide.

\[ \ce{2CO + O2 -> 2CO2} \]

A reaction vessel initially contains:

• 12.0 g of carbon monoxide
• 16.0 g of oxygen gas

Calculate:

(a) Which reactant is limiting.

(b) The theoretical mass of carbon dioxide produced.

(c) The mass of oxygen remaining after the reaction.

Use:

• \(M(\ce{CO})=28.0\ \mathrm{g\,mol^{-1}}\)
• \(M(\ce{O2})=32.0\ \mathrm{g\,mol^{-1}}\)
• \(M(\ce{CO2})=44.0\ \mathrm{g\,mol^{-1}}\)

Assume the reaction proceeds completely and no products are lost.

Solution 3

Carbon monoxide is the limiting reactant. The reaction produces 18.9 g of carbon dioxide, leaving 9.14 g of oxygen unreacted.

Step 1: Convert both reactants into moles.

For carbon monoxide:

\[ \begin{aligned} n(\ce{CO})&=\frac{12.0}{28.0}\\ &=0.4286\text{ mol} \end{aligned} \]

For oxygen:

\[ \begin{aligned} n(\ce{O2})&=\frac{16.0}{32.0}\\ &=0.500\text{ mol} \end{aligned} \]

Step 2: Determine the limiting reactant.

The balanced equation requires two moles of carbon monoxide for every mole of oxygen.

The oxygen needed to react with all the available carbon monoxide is:

\[ \begin{aligned} n(\ce{O2})_{\text{required}} &=0.4286\times\frac{1}{2}\\ &=0.2143\text{ mol} \end{aligned} \]

We have 0.500 mol of oxygen available.

Since 0.500 mol is greater than 0.2143 mol, oxygen is in excess.

Carbon monoxide is the limiting reactant.

Step 3: Calculate the amount of carbon dioxide produced.

The coefficient ratio between carbon monoxide and carbon dioxide is 2:2, or 1:1.

\[ \begin{aligned} n(\ce{CO2}) &=0.4286\times\frac{2}{2}\\ &=0.4286\text{ mol} \end{aligned} \]

Step 4: Calculate the mass of carbon dioxide.

\[ \begin{aligned} m(\ce{CO2})&=nM\\ &=0.4286\times44.0\\ &=18.9\text{ g} \end{aligned} \]

Step 5: Calculate the oxygen remaining.

First, subtract the oxygen consumed from the initial oxygen amount:

\[ \begin{aligned} n(\ce{O2})_{\text{left}} &=0.500-0.2143\\ &=0.2857\text{ mol} \end{aligned} \]

Convert the remaining oxygen into mass:

\[ \begin{aligned} m(\ce{O2})_{\text{left}} &=0.2857\times32.0\\ &=9.14\text{ g} \end{aligned} \]

The final results are:

• Carbon dioxide produced: 18.9 g.
• Oxygen remaining: 9.14 g.

A useful mass check is:

\[ 18.9+9.14=28.0\text{ g} \]

This matches the initial total mass:

\[ 12.0+16.0=28.0\text{ g} \]

The tempting mistake is to calculate the product using all the oxygen initially present. But the balanced equation tells us that carbon monoxide runs out first. Only the oxygen required to react with that carbon monoxide is consumed.

Question 4

Aluminium reacts with chlorine gas to produce aluminium chloride:

\[ \ce{2Al + 3Cl2 -> 2AlCl3} \]

A student is given 5.40 g of aluminium and claims that 8.10 g of chlorine is required.

Their reasoning is:

“The equation has a 2:3 coefficient ratio, so 2 g of aluminium reacts with 3 g of chlorine.”

Explain why this reasoning is incorrect.

Then calculate:

(a) The correct mass of chlorine required to react completely with the aluminium.

(b) The theoretical mass of aluminium chloride produced.

Use:

• \(M(\ce{Al})=27.0\ \mathrm{g\,mol^{-1}}\)
• \(M(\ce{Cl2})=71.0\ \mathrm{g\,mol^{-1}}\)
• \(M(\ce{AlCl3})=133.5\ \mathrm{g\,mol^{-1}}\)

Assume chlorine is supplied in exactly the required amount.

Solution 4

The correct chlorine mass is 21.3 g, not 8.10 g. The theoretical mass of aluminium chloride is 26.7 g.

The student’s mistake is treating a mole ratio as a mass ratio.

Step 1: Convert the aluminium mass into moles.

\[ \begin{aligned} n(\ce{Al})&=\frac{5.40}{27.0}\\ &=0.200\text{ mol} \end{aligned} \]

Step 2: Use the balanced equation to calculate the chlorine required.

The equation gives:

\[ 2\text{ mol Al}:3\text{ mol }\ce{Cl2} \]

Therefore:

\[ \begin{aligned} n(\ce{Cl2}) &=0.200\times\frac{3}{2}\\ &=0.300\text{ mol} \end{aligned} \]

Step 3: Convert chlorine moles into mass.

\[ \begin{aligned} m(\ce{Cl2})&=nM\\ &=0.300\times71.0\\ &=21.3\text{ g} \end{aligned} \]

This is the correct chlorine requirement.

Why is it so different from the student’s 8.10 g?

The coefficients represent moles, and one mole of aluminium has a different mass from one mole of chlorine gas.

In fact:

• 2 mol of aluminium has a mass of \(2\times27.0=54.0\text{ g}\).
• 3 mol of chlorine has a mass of \(3\times71.0=213\text{ g}\).

The actual mass ratio is 54.0:213, not 2:3.

Step 4: Calculate the moles of aluminium chloride produced.

The balanced equation gives a 2:2 ratio between aluminium and aluminium chloride.

\[ \begin{aligned} n(\ce{AlCl3}) &=0.200\times\frac{2}{2}\\ &=0.200\text{ mol} \end{aligned} \]

Step 5: Convert the product amount into mass.

\[ \begin{aligned} m(\ce{AlCl3})&=nM\\ &=0.200\times133.5\\ &=26.7\text{ g} \end{aligned} \]

The product mass is 26.7 g.

Notice that:

\[ 5.40+21.3=26.7\text{ g} \]

The mass of the aluminium and chlorine consumed matches the mass of aluminium chloride formed.

The coefficients tell you how many moles react. You must use molar masses to determine the corresponding masses.

Question 5

A mixture contains only calcium carbonate and magnesium carbonate.

Both substances decompose when heated:

\[ \ce{CaCO3 -> CaO + CO2} \]

\[ \ce{MgCO3 -> MgO + CO2} \]

A 9.20 g sample of the mixture is heated until both carbonates have completely decomposed.

The following measurements are obtained:

• Initial mixture mass: 9.20 g
• Final dry solid residue mass: 4.80 g
• Carbon dioxide collected in a gas collector: 3.52 g

The gas collector is known to capture only 80.0% of the carbon dioxide produced. The remaining carbon dioxide escapes collection.

Calculate the original mass of calcium carbonate and magnesium carbonate in the mixture.

Then explain why using 3.52 g as the total mass of carbon dioxide produced gives an impossible result.

Assume:

• Both carbonates decompose completely.
• Calcium oxide and magnesium oxide remain in the solid residue.
• Carbon dioxide is the only gas produced.
• No solid material is lost.
• The collector captures exactly 80.0% of the carbon dioxide produced.

Use:

Solution 5

The original mixture contains 4.88 g of calcium carbonate and 4.32 g of magnesium carbonate.

The key is that the collector does not measure all the carbon dioxide produced. We must account for the missing gas before using the balanced equations.

This is the part that makes the question more difficult than a standard mass-to-moles calculation.

Step 1: Calculate the total mass of carbon dioxide produced.

The collector captures 80.0% of the carbon dioxide.

Therefore:

\[ \begin{aligned} m(\ce{CO2})_{\text{produced}} &=\frac{3.52}{0.800}\\ &=4.40\text{ g} \end{aligned} \]

There is an independent way to confirm this result.

The starting mixture has a mass of 9.20 g, while the dry solid residue has a mass of 4.80 g.

Since carbon dioxide is the only gas produced and no solid is lost:

\[ \begin{aligned} m(\ce{CO2})_{\text{produced}} &=9.20-4.80\\ &=4.40\text{ g} \end{aligned} \]

Both measurements agree.

The collector has captured 3.52 g, while 0.88 g has escaped collection.

Step 2: Convert the total carbon dioxide mass into moles.

\[ \begin{aligned} n(\ce{CO2}) &=\frac{4.40}{44.0}\\ &=0.100\text{ mol} \end{aligned} \]

Step 3: Connect the carbon dioxide amount to the two original carbonates.

Look at the two balanced equations:

\[ \ce{CaCO3 -> CaO + CO2} \]

\[ \ce{MgCO3 -> MgO + CO2} \]

Both have a 1:1 mole ratio between carbonate and carbon dioxide.

Therefore:

• 1 mol of calcium carbonate produces 1 mol of carbon dioxide.
• 1 mol of magnesium carbonate produces 1 mol of carbon dioxide.

Let:

• \(x\) = initial moles of \(\ce{CaCO3}\).
• \(y\) = initial moles of \(\ce{MgCO3}\).

The total carbon dioxide produced is 0.100 mol.

Therefore:

\[ x+y=0.100 \]

This is our first equation.

Step 4: Use the initial sample mass to create a second equation.

The original mixture contained 9.20 g of both carbonates.

The calcium carbonate contributes \(100.1x\) grams, while the magnesium carbonate contributes \(84.3y\) grams.

Therefore:

\[ 100.1x+84.3y=9.20 \]

We now have two equations:

\[ x+y=0.100 \]

\[ 100.1x+84.3y=9.20 \]

Both measurements are needed to distinguish the two carbonates.

Step 5: Solve the simultaneous equations.

Rearrange the first equation:

\[ y=0.100-x \]

Substitute into the mass equation:

\[ 100.1x+84.3(0.100-x)=9.20 \]

Expand:

\[ 100.1x+8.43-84.3x=9.20 \]

Collect like terms:

\[ 15.8x=0.770 \]

Therefore:

\[ x=\frac{0.770}{15.8} \]

\[ x=0.0487\text{ mol} \]

Substitute this into the first equation:

\[ \begin{aligned} y&=0.100-0.0487\\ &=0.0513\text{ mol} \end{aligned} \]

The original mixture contained:

• 0.0487 mol of calcium carbonate.
• 0.0513 mol of magnesium carbonate.

Step 6: Convert both amounts into their original masses.

For calcium carbonate:

\[ \begin{aligned} m(\ce{CaCO3}) &=0.0487\times100.1\\ &=4.88\text{ g} \end{aligned} \]

For magnesium carbonate:

\[ \begin{aligned} m(\ce{MgCO3}) &=0.0513\times84.3\\ &=4.32\text{ g} \end{aligned} \]

The original mixture therefore contained:

Step 7: Check the final solid residue.

The calcium carbonate produces calcium oxide in a 1:1 mole ratio.

\[ \begin{aligned} m(\ce{CaO}) &=0.0487\times56.1\\ &=2.73\text{ g} \end{aligned} \]

The magnesium carbonate produces magnesium oxide, also in a 1:1 mole ratio.

\[ \begin{aligned} m(\ce{MgO}) &=0.0513\times40.3\\ &=2.07\text{ g} \end{aligned} \]

The predicted solid residue is:

\[ 2.73+2.07=4.80\text{ g} \]

This matches the measured residue.

The carbon dioxide mass also checks:

\[ 3.52+0.88=4.40\text{ g} \]

All the supplied measurements are consistent.

Why the tempting calculation fails

Suppose you ignored the collection efficiency and treated 3.52 g as the total carbon dioxide produced.

You would calculate:

\[ n(\ce{CO2})=\frac{3.52}{44.0} =0.0800\text{ mol} \]

Since both carbonates produce carbon dioxide in a 1:1 mole ratio, you might conclude that the original mixture contained only 0.0800 mol of carbonate formula units.

But even if every one of those moles were calcium carbonate, the maximum possible starting mass would be:

\[ 0.0800\times100.1=8.01\text{ g} \]

The actual sample weighed 9.20 g.

That is impossible.

The missing constraint is that the gas collector captures only part of the carbon dioxide. The mass lost by the original sample and the stated collection efficiency reveal the actual amount produced.

This is why a balanced equation must be used alongside the experimental information. Correct coefficients give you the right chemical proportions, but you must still identify what was produced, what was collected, and what remained unreacted.

06Where to go next

Balancing an equation is the starting point for every reliable reacting-mass calculation. Once the coefficients are correct, your next decisions are whether a reactant is limiting, whether the reaction is complete, and whether any products escape the measured system.

For a deeper method of checking whether all atoms and reaction masses have been accounted for, read How to Check Reaction-Mass Answers Conserve Atoms.

If the question involves gas escaping from an open vessel, continue with How to Calculate Gas Mass Lost from an Open Reaction Vessel.

The essential order stays the same: balance first, convert mass into moles, use the coefficients, and only then convert back into mass.