How to Check Reaction-Mass Answers Conserve Atoms

A student starts with 4.86 g of magnesium and 3.20 g of oxygen gas. They calculate that the reaction produces 8.06 g of magnesium oxide because the starting masses add to 8.06 g.

The answer looks convincing. But is adding the reactant masses enough to prove that the calculation conserves atoms?

No. You also need to check that the correct number of each type of atom has been accounted for. A calculation can have a perfectly reasonable total mass and still use an incorrect mole ratio.

To check whether a reaction-mass answer conserves atoms, use two checks:

1. Check the balanced equation. Count each type of atom on both sides. The numbers must match.
2. Check the reaction masses. Convert mass to moles, apply the coefficients as mole ratios, and convert back to mass. Account for every product and any unreacted reactants.

This is the mass-to-moles-to-moles-to-mass method. It is one of the most reliable ways to catch mistakes in HSC Chemistry stoichiometry.

01Start by checking the balanced equation

Consider the reaction between magnesium and oxygen:

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

Before calculating any masses, check the number of atoms on each side.

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

This is what conservation of atoms means: atoms are rearranged into new substances, but the number of atoms of each element does not change.

The equation also tells us something important about the calculation.

The coefficients give the mole ratio:

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

They do not give the mass ratio.

A useful way to picture this is to treat the equation like a recipe. The coefficients tell you how many portions of each substance react together. However, unlike identical portions of food, one mole of magnesium and one mole of oxygen gas have different masses.

You must convert to moles before using the ratio.

If an equation is not balanced, stop before doing the mass calculation. Its coefficients cannot give you a valid mole ratio.

For more background on why this works, see Conservation of Atoms in Chemical Reactions: HSC Chemistry Guide.

02Understand what a mass measurement actually tells you

There is a subtle difference between conserving mass and measuring the mass of the substances left in a container.

Imagine reacting calcium carbonate with hydrochloric acid:

\[ \ce{CaCO3 + 2HCl -> CaCl2 + CO2 + H2O} \]

One of the products is carbon dioxide gas.

Suppose you carry out the reaction in an open beaker sitting on a balance.

What should happen to the balance reading?

It will decrease as carbon dioxide bubbles out of the beaker. The balance is measuring the beaker and its contents, not the gas that has escaped into the surrounding air.

That does not mean atoms have disappeared. Carbon and oxygen atoms have left the beaker together in the form of carbon dioxide.

Improving the experiment

The open-beaker arrangement cannot measure the mass of the complete reacting system because gas escapes.

A revised arrangement can retain the gas:

• The reactants are kept inside a flask before they are mixed.
• A delivery tube connects the flask to a gas syringe.
• Carbon dioxide travels through the tube and collects in the syringe.
• The entire connected apparatus is weighed before and after the reaction.

The expected result is that the total mass remains unchanged, within the balance’s measurement uncertainty.

The gas has moved from the flask into the syringe, but it has not left the measured system.

A real gas-producing experiment requires appropriate teacher-approved apparatus and pressure control. An ordinary rigid container should not simply be sealed to contain a gas-producing reaction.

A quick mass estimate

Suppose 2.00 g of calcium carbonate reacts completely with excess hydrochloric acid.

The molar mass of calcium carbonate is 100.1 g mol⁻¹, while the molar mass of carbon dioxide is 44.0 g mol⁻¹.

First, calculate the amount of calcium carbonate:

\[ n(\ce{CaCO3})=\frac{2.00}{100.1} \]

\[ n(\ce{CaCO3})=0.0200\text{ mol} \]

The balanced equation has a 1:1 mole ratio between calcium carbonate and carbon dioxide.

Therefore:

\[ n(\ce{CO2})=0.0200\text{ mol} \]

Convert this amount into mass:

\[ m(\ce{CO2})=0.0200\times44.0 \]

\[ \boxed{m(\ce{CO2})=0.880\text{ g}} \]

If all the carbon dioxide escapes, the open beaker’s mass decreases by approximately 0.880 g.

The atoms are still accounted for. They are simply no longer inside the beaker.

This distinction matters whenever a reaction produces a gas, uses a gas from the surrounding air, or leaves an unreacted substance behind.

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

Now we can build a reliable calculation method.

The central idea is simple:

Mass tells you how much a substance weighs. Moles tell you how much of the substance is present. The balanced equation connects the amounts in moles.

There are two essential formulas:

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

and

\[ m=nM \]

where:

• \(n\) = amount of substance, in mol
• \(m\) = mass, in g
• \(M\) = molar mass, in g mol⁻¹

Follow these steps.

Step 1: Write and balance the chemical equation

Never begin with an unbalanced equation.

The coefficients must represent the correct number of each type of atom on both sides.

Step 2: Identify the known substance and the substance you need

For example, a question might give you the mass of magnesium and ask for the mass of magnesium oxide.

You know the starting mass, but you need the product mass.

Step 3: Convert the known mass into moles

Use:

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

Make sure you use the molar mass of the correct substance.

Step 4: Apply the mole ratio

Read the coefficients from the balanced equation.

For a known substance A and a target substance B:

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

Here, \(\nu_A\) and \(\nu_B\) are the coefficients of A and B.

This is where the balanced equation becomes useful. It tells you how many moles of product form from the amount of reactant available.

If there are two or more reactants, you must also identify the limiting reagent. The limiting reagent is the reactant that runs out first and determines how much product can form.

Step 5: Convert the product’s moles into mass

Use:

\[ m=nM \]

Use the molar mass of the product, not the reactant.

Step 6: Check the complete mass balance

Ask yourself:

• Have I accounted for every element?
• Have I included every product?
• Is any reactant left over?
• Has a gas escaped from the measured system?
• Does the total mass of all substances before the reaction equal the total mass afterwards?

A reaction-mass answer should account for the entire system, not just the product you were asked to calculate.

If you are checking an answer that someone else has already calculated, you can also work backwards: convert the proposed product mass into moles, use the equation to calculate the reactant amounts required, and compare those amounts with what was originally available.

05The common mistake: Treating coefficients as mass ratios

Let’s return to the magnesium reaction.

A student sees:

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

They assume the ratio 2:1 means 2 g of magnesium reacts with 1 g of oxygen.

That sounds reasonable if you have not yet connected coefficients to moles. Unfortunately, it is wrong.

For the 4.86 g magnesium sample, this shortcut predicts that only 2.43 g of oxygen is required.

But 2.43 g of oxygen gas represents:

\[ n(\ce{O2})=\frac{2.43}{32.0} \]

\[ n(\ce{O2})=0.0759\text{ mol} \]

This provides only:

\[ 2\times0.0759=0.1518\text{ mol O atoms} \]

However, the magnesium sample contains 0.200 mol of magnesium atoms.

Since each \(\ce{MgO}\) unit requires one magnesium atom and one oxygen atom, there is not enough oxygen to convert all the magnesium into magnesium oxide.

The correct oxygen requirement is 3.20 g, not 2.43 g.

The coefficient ratio describes particles or moles, not grams. The different molar masses determine the actual mass ratio.

This is why the conversion to moles is essential.

07A quick diagnostic for reaction-mass answers

When checking your working, look for these common problems.

There is one final distinction worth remembering.

A balanced equation tells you that atoms are conserved. A correct stoichiometric calculation tells you how much of each substance is involved. A mass balance checks that the complete system has been accounted for.

You need all three ideas working together.

08Questions and solutions

Question 1

Calcium carbonate reacts with excess hydrochloric acid according to the equation:

\[ \ce{CaCO3 + 2HCl -> CaCl2 + CO2 + H2O} \]

A sample of 5.00 g of calcium carbonate reacts completely.

Calculate the mass of carbon dioxide produced. Explain why the mass of carbon dioxide is not equal to the starting mass of calcium carbonate.

Solution 1

The reaction produces 2.20 g of carbon dioxide. The remaining atoms are accounted for in calcium chloride and water.

Step 1: Calculate the molar mass of calcium carbonate.

\[ M(\ce{CaCO3})=40.1+12.0+3(16.0) \]

\[ M(\ce{CaCO3})=100.1\text{ g mol}^{-1} \]

Step 2: Convert the calcium carbonate mass into moles.

\[ n(\ce{CaCO3})=\frac{5.00}{100.1} \]

\[ n(\ce{CaCO3})=0.04995\text{ mol} \]

Step 3: Apply the mole ratio.

The equation gives a 1:1 ratio between calcium carbonate and carbon dioxide.

\[ n(\ce{CO2})=0.04995\text{ mol} \]

Step 4: Calculate the carbon dioxide mass.

\[ m(\ce{CO2})=0.04995\times44.0 \]

\[ \boxed{m(\ce{CO2})=2.20\text{ g}} \]

The carbon dioxide is only one of the products.

Calcium atoms are present in calcium chloride, while the remaining oxygen and hydrogen atoms are present in carbon dioxide, calcium chloride, and water as required by the balanced equation.

The carbon dioxide mass is therefore smaller than the starting calcium carbonate mass because the other atoms have been rearranged into different products.

Question 2

Iron reacts with copper(II) sulfate according to:

\[ \ce{Fe + CuSO4 -> FeSO4 + Cu} \]

A 5.58 g iron strip is placed into a solution containing 20.0 g of anhydrous copper(II) sulfate.

Assume the reaction goes to completion, the water is a spectator, and the stated masses refer to pure solutes.

Calculate:

a) The mass of copper produced.

b) The mass of copper(II) sulfate remaining unreacted.

c) Whether the complete reaction conserves mass.

Solution 2

Iron is the limiting reagent. The reaction produces 6.35 g of copper, and 4.04 g of copper(II) sulfate remains unreacted. The complete mass balance is conserved.

Step 1: Convert the iron mass into moles.

The molar mass of iron is 55.8 g mol⁻¹.

\[ n(\ce{Fe})=\frac{5.58}{55.8} \]

\[ n(\ce{Fe})=0.100\text{ mol} \]

Step 2: Convert the copper(II) sulfate mass into moles.

Its molar mass is 159.6 g mol⁻¹.

\[ n(\ce{CuSO4})=\frac{20.0}{159.6} \]

\[ n(\ce{CuSO4})=0.1253\text{ mol} \]

Step 3: Identify the limiting reagent.

The equation has a 1:1 mole ratio.

Since there are fewer moles of iron, iron is the limiting reagent.

Therefore, 0.100 mol of copper(II) sulfate reacts.

Step 4: Calculate the mass of copper produced.

The equation gives a 1:1 ratio between iron and copper.

\[ n(\ce{Cu})=0.100\text{ mol} \]

The molar mass of copper is 63.5 g mol⁻¹.

\[ m(\ce{Cu})=0.100\times63.5 \]

\[ \boxed{m(\ce{Cu})=6.35\text{ g}} \]

Step 5: Calculate the excess copper(II) sulfate.

\[ n_{\text{remaining}}=0.1253-0.100 \]

\[ n_{\text{remaining}}=0.0253\text{ mol} \]

Convert this amount into mass:

\[ m=0.0253\times159.6 \]

\[ \boxed{m_{\text{remaining}}=4.04\text{ g}} \]

Step 6: Check conservation of mass.

The products are iron(II) sulfate and copper.

The molar mass of iron(II) sulfate is 151.9 g mol⁻¹.

\[ m(\ce{FeSO4})=0.100\times151.9 \]

\[ m(\ce{FeSO4})=15.19\text{ g} \]

Total product mass:

\[ 15.19+6.35=21.54\text{ g} \]

Add the unreacted copper(II) sulfate:

\[ 21.54+4.04=25.58\text{ g} \]

Initial reactant mass:

\[ 5.58+20.0=25.58\text{ g} \]

The masses agree.

The deposited copper alone is not expected to equal the total starting mass. Iron and sulfate atoms are also present in iron(II) sulfate, while some copper(II) sulfate remains unreacted.

Question 3

Sulfuric acid reacts with sodium hydroxide according to:

\[ \ce{H2SO4 + 2NaOH -> Na2SO4 + 2H2O} \]

A reaction uses 4.90 g of pure sulfuric acid and 5.00 g of pure sodium hydroxide.

A student claims that 9.90 g of products must form because the starting reactant masses add to 9.90 g.

Calculate the theoretical mass of each product and determine whether the student’s claim is correct.

Also calculate the mass of any unreacted reactant.

Solution 3

The student is incorrect. The reaction produces 7.10 g of sodium sulfate and 1.80 g of water, leaving 1.00 g of sodium hydroxide unreacted.

The total mass is conserved, but not all the starting reactants become products.

Step 1: Convert sulfuric acid into moles.

The molar mass of sulfuric acid is 98.0 g mol⁻¹.

\[ n(\ce{H2SO4})=\frac{4.90}{98.0} \]

\[ n(\ce{H2SO4})=0.0500\text{ mol} \]

Step 2: Convert sodium hydroxide into moles.

The molar mass of sodium hydroxide is 40.0 g mol⁻¹.

\[ n(\ce{NaOH})=\frac{5.00}{40.0} \]

\[ n(\ce{NaOH})=0.125\text{ mol} \]

Step 3: Identify the limiting reagent.

The balanced equation requires:

\[ 1\text{ mol H}_2\text{SO}_4:2\text{ mol NaOH} \]

For 0.0500 mol of sulfuric acid, the required sodium hydroxide is:

\[ n(\ce{NaOH})=2\times0.0500 \]

\[ n(\ce{NaOH})=0.100\text{ mol} \]

There are 0.125 mol available.

Sulfuric acid is the limiting reagent, and sodium hydroxide is in excess.

Step 4: Calculate the sodium sulfate mass.

The equation gives a 1:1 ratio between sulfuric acid and sodium sulfate.

\[ n(\ce{Na2SO4})=0.0500\text{ mol} \]

The molar mass of sodium sulfate is 142.0 g mol⁻¹.

\[ m(\ce{Na2SO4})=0.0500\times142.0 \]

\[ \boxed{m(\ce{Na2SO4})=7.10\text{ g}} \]

Step 5: Calculate the water mass.

The equation gives a 1:2 ratio between sulfuric acid and water.

\[ n(\ce{H2O})=2\times0.0500 \]

\[ n(\ce{H2O})=0.100\text{ mol} \]

The molar mass of water is 18.0 g mol⁻¹.

\[ m(\ce{H2O})=0.100\times18.0 \]

\[ \boxed{m(\ce{H2O})=1.80\text{ g}} \]

Step 6: Calculate the excess sodium hydroxide.

\[ n_{\text{remaining}}=0.125-0.100 \]

\[ n_{\text{remaining}}=0.0250\text{ mol} \]

\[ m=0.0250\times40.0 \]

\[ \boxed{m_{\text{remaining}}=1.00\text{ g}} \]

Step 7: Check the mass balance.

Total product mass:

\[ 7.10+1.80=8.90\text{ g} \]

Include the excess sodium hydroxide:

\[ 8.90+1.00=9.90\text{ g} \]

This matches the initial reactant mass:

\[ 4.90+5.00=9.90\text{ g} \]

The student’s mistake was assuming every reactant must be consumed.

Conservation of mass includes unreacted substances. It does not mean the mass of the products alone must equal the mass of all the starting reactants.

Question 4

Sodium hydrogen carbonate decomposes when heated according to:

\[ \ce{2NaHCO3 -> Na2CO3 + CO2 + H2O} \]

A student heats 8.40 g of sodium hydrogen carbonate in an open crucible until the reaction is complete.

Assume all carbon dioxide and water vapour escape from the crucible.

Calculate:

a) The mass of sodium carbonate remaining.

b) The mass of carbon dioxide produced.

c) The mass of water produced.

d) The total mass lost from the crucible.

A student argues that the decrease in the balance reading proves that atoms have been lost. Explain why this conclusion is incorrect.

Solution 4

The crucible contains 5.30 g of sodium carbonate after heating, while 3.10 g of gaseous products escape. The atoms are conserved because the carbon dioxide and water vapour leave the weighed crucible.

Step 1: Calculate the initial amount of sodium hydrogen carbonate.

Its molar mass is 84.0 g mol⁻¹.

\[ n(\ce{NaHCO3})=\frac{8.40}{84.0} \]

\[ n(\ce{NaHCO3})=0.100\text{ mol} \]

Step 2: Apply the mole ratios.

The balanced equation is:

\[ \ce{2NaHCO3 -> Na2CO3 + CO2 + H2O} \]

The ratio is 2:1:1:1.

Therefore:

\[ n(\ce{Na2CO3})=0.0500\text{ mol} \]

\[ n(\ce{CO2})=0.0500\text{ mol} \]

\[ n(\ce{H2O})=0.0500\text{ mol} \]

Step 3: Calculate the sodium carbonate mass.

Its molar mass is 106.0 g mol⁻¹.

\[ m(\ce{Na2CO3})=0.0500\times106.0 \]

\[ \boxed{m(\ce{Na2CO3})=5.30\text{ g}} \]

Step 4: Calculate the carbon dioxide mass.

\[ m(\ce{CO2})=0.0500\times44.0 \]

\[ \boxed{m(\ce{CO2})=2.20\text{ g}} \]

Step 5: Calculate the water mass.

\[ m(\ce{H2O})=0.0500\times18.0 \]

\[ \boxed{m(\ce{H2O})=0.900\text{ g}} \]

Step 6: Calculate the total mass lost.

Both carbon dioxide and water vapour leave the crucible.

\[ m_{\text{lost}}=2.20+0.900 \]

\[ \boxed{m_{\text{lost}}=3.10\text{ g}} \]

The final crucible mass is:

\[ 8.40-3.10=5.30\text{ g} \]

This matches the calculated sodium carbonate mass.

The complete mass balance is:

\[ 5.30+2.20+0.900=8.40\text{ g} \]

The apparent mass loss is explained by the movement of gaseous products out of the measured system.

If all the products were captured and included in the measurement, the total mass would remain 8.40 g.

Question 5

A 10.00 g sample contains only calcium carbonate and magnesium carbonate.

When heated sufficiently, both substances decompose completely:

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

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

After heating:

• The total mass of solid residue is 5.40 g.
• The total mass of carbon dioxide collected is 4.60 g.
• No other products are formed, and no material is lost.

The following molar masses apply:

Determine the original mass of each carbonate.

Then explain why knowing that 4.60 g of carbon dioxide was produced is not, by itself, enough to identify the composition of the original mixture.

Finally, verify that the calculated composition accounts for the complete reaction mass.

Solution 5

The original mixture contained approximately 7.52 g of calcium carbonate and 2.48 g of magnesium carbonate.

The key is that both carbonates produce one mole of carbon dioxide per mole of carbonate, but their different molar masses mean that they contribute different masses to the final oxide residue.

The gas measurement gives the total amount of carbonate that decomposed. The original sample mass provides the additional information needed to determine the proportions.

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

The molar mass of carbon dioxide is 44.0 g mol⁻¹.

\[ n(\ce{CO2})=\frac{4.60}{44.0} \]

\[ n(\ce{CO2})=0.1045\text{ mol} \]

Each carbonate produces one mole of carbon dioxide.

Therefore, if \(x\) represents the initial moles of calcium carbonate and \(y\) represents the initial moles of magnesium carbonate:

\[ x+y=0.1045 \]

This is our first relationship.

Step 2: Use the original sample mass.

The original sample contains 10.00 g of carbonate.

Using the molar masses:

\[ 100.1x+84.3y=10.00 \]

This is our second relationship.

We now have two equations:

\[ x+y=0.1045 \]

\[ 100.1x+84.3y=10.00 \]

Step 3: Solve for the individual amounts.

Rearrange the first equation:

\[ y=0.1045-x \]

Substitute this into the second equation:

\[ 100.1x+84.3(0.1045-x)=10.00 \]

Expand:

\[ 100.1x+8.813-84.3x=10.00 \]

Collect like terms:

\[ 15.8x=1.187 \]

\[ x=0.0751\text{ mol} \]

Substitute this value into the first equation:

\[ y=0.1045-0.0751 \]

\[ y=0.0294\text{ mol} \]

Step 4: Calculate the original calcium carbonate mass.

\[ m(\ce{CaCO3})=0.0751\times100.1 \]

\[ \boxed{m(\ce{CaCO3})=7.52\text{ g}} \]

Step 5: Calculate the original magnesium carbonate mass.

\[ m(\ce{MgCO3})=0.0294\times84.3 \]

\[ \boxed{m(\ce{MgCO3})=2.48\text{ g}} \]

The original mixture therefore contained approximately:

• 7.52 g calcium carbonate
• 2.48 g magnesium carbonate

Step 6: Verify the solid residue.

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

\[ m(\ce{CaO})=0.0751\times56.1 \]

\[ m(\ce{CaO})=4.21\text{ g} \]

Magnesium carbonate produces magnesium oxide in a 1:1 mole ratio.

\[ m(\ce{MgO})=0.0294\times40.3 \]

\[ m(\ce{MgO})=1.19\text{ g} \]

The total solid residue is:

\[ 4.21+1.19=5.40\text{ g} \]

This agrees with the measured residue.

Step 7: Complete the mass balance.

The total mass of the solid residue is 5.40 g.

The carbon dioxide mass is 4.60 g.

\[ m_{\text{final}}=5.40+4.60 \]

\[ \boxed{m_{\text{final}}=10.00\text{ g}} \]

This matches the original sample mass.

The atom accounting also works. Calcium and magnesium remain in their respective oxides, while carbon leaves in carbon dioxide. Oxygen atoms are distributed between the oxide residues and the carbon dioxide.

The interesting part is why the gas measurement alone is insufficient.

Both carbonates produce one mole of carbon dioxide per mole of carbonate. Therefore, 0.1045 mol of carbon dioxide tells us that the original mixture contained 0.1045 mol of carbonates in total.

It does not tell us how much of that amount was calcium carbonate and how much was magnesium carbonate.

The original 10.00 g sample supplies the second independent constraint. Because the two carbonates have different molar masses, their relative amounts can be determined.

The measured 5.40 g residue then provides an independent check on the calculated result, alongside the complete mass balance.

This is a useful way to approach more demanding stoichiometry questions: one measurement may establish the total amount, while another establishes how that amount is distributed.

09Where to go next

Once a reaction-mass calculation passes the atom and mass checks, the next useful distinction is between theoretical yield and actual yield.

The theoretical yield is the maximum product mass predicted by the balanced equation and the available reactants. The actual yield is what is collected experimentally.

A reaction can conserve every atom and still produce less isolated product than the theoretical prediction because of incomplete reaction, side reactions, or losses during separation.

That is why conservation of atoms, limiting reagents, and percentage yield belong together in reaction-mass calculations.