How to Calculate Gas Mass Lost from an Open Reaction Vessel

An open flask contains a solution with 2.55 g of hydrogen peroxide. A catalyst makes the solution bubble, oxygen escapes, and the balance reading falls. If all the hydrogen peroxide reacts, should the flask lose 2.55 g?

No. Some of that mass stays behind as water. The theoretical mass of oxygen lost is only 1.20 g, assuming the reaction finishes, all the oxygen escapes, and nothing else enters or leaves the flask.

To calculate the mass of gas lost from an open reaction vessel, follow this route:

Mass of limiting reactant → moles of reactant → moles of gas → mass of gas.

The balanced chemical equation supplies the mole ratio. If you also know the initial balance reading, subtract the escaping gas mass to predict the final reading.

The important distinction is between the mass of gas produced by the reaction and the mass of gas that actually leaves the vessel.

01Why the balance loses less mass than the reactant

Picture an open conical flask sitting on an electronic balance.

At the bottom is hydrogen peroxide solution. A small amount of manganese dioxide catalyst is held separately inside the flask until the measurement begins.

When the catalyst contacts the solution, oxygen bubbles form. They rise through the liquid, enter the space above it, and escape through the open neck.

The catalyst remains in the flask. The water produced by the reaction remains in the solution. Only material that crosses the opening can reduce the mass of the weighed system.

Hydrogen peroxide decomposes according to:

\[ \ce{2H2O2(aq) -> 2H2O(l) + O2(g)} \]

The catalyst increases the reaction rate but is not consumed in the overall equation.

Before calculating, look at the coefficients. Two moles of hydrogen peroxide produce one mole of oxygen.

Does that mean 2 g of hydrogen peroxide produces 1 g of oxygen?

No. Coefficients represent amounts in moles, not masses in grams.

A mole is a counting unit for particles. Different chemicals have different masses per mole, so the coefficient ratio cannot usually be used directly as a mass ratio.

For this reaction:

• Two moles of hydrogen peroxide have a mass of \(2(34.0)=68.0\text{ g}\).
• Two moles of water have a mass of \(2(18.0)=36.0\text{ g}\).
• One mole of oxygen has a mass of \(32.0\text{ g}\).

The 68.0 g of hydrogen peroxide becomes 36.0 g of water and 32.0 g of oxygen.

If the oxygen escapes and the water stays, the balance loses 32.0 g, not 68.0 g.

The three calculations you need

Let:

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

First, convert the mass of the reacting substance into moles:

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

Next, use the coefficients in the balanced equation.

For hydrogen peroxide decomposition, the ratio is \(2:1\), so:

\[ n(\ce{O2})=\frac{n(\ce{H2O2})}{2} \]

Finally, convert the amount of oxygen into mass:

\[ m(\ce{O2})=n(\ce{O2})M(\ce{O2}) \]

If all the oxygen escapes and no other matter moves across the vessel boundary, this is the predicted mass lost.

02How the experiment tests the calculation

The balanced equation predicts the amount of oxygen that can form. A balance measures the net change in mass of the apparatus and its contents.

Those are related measurements, but they are not automatically identical.

What should happen as the reaction proceeds?

Imagine starting the reaction without removing any equipment from the balance. The catalyst is already inside the flask, held separately until it is tipped into the solution.

What should the mass reading do while oxygen bubbles escape?

It should generally decrease. If oxygen is the only material leaving, each decrease represents more oxygen crossing the flask’s open boundary.

As gas escape slows, the reading should start to level off.

Here is an illustrative set of readings for the 4.25 g hydrogen peroxide example. These values are invented to demonstrate how the data would be analysed, not records of a performed experiment.

The measured mass decrease is:

\[ \begin{aligned} m_{\text{lost}}&=165.30-163.32\\ &=1.98\text{ g} \end{aligned} \]

The theoretical oxygen mass was 2.00 g.

The observed decrease is therefore 0.02 g smaller than the prediction.

Does the stable reading at 150 seconds prove that all the hydrogen peroxide reacted?

Not necessarily.

A stable reading tells us that the net mass of the weighed system is no longer changing detectably. Some oxygen might remain dissolved, the reaction might have slowed substantially, or the measurement might be affected by small experimental errors.

A balance cannot identify these possibilities by itself.

What the first apparatus cannot control

An uncovered reaction flask allows oxygen to escape, but it can also lose other material.

Bubbling may eject tiny liquid droplets. Water may evaporate, particularly if the reaction warms the solution.

Both effects make the measured mass loss larger, even though neither represents additional oxygen production.

The opposite problem is also possible: some oxygen may remain dissolved in the liquid when the final reading is taken, making the measured loss smaller than the mass of oxygen formed.

A more controlled arrangement could use a loose, non-airtight splash guard that remains part of the weighed apparatus. This reduces the chance of liquid droplets escaping while still allowing gas to leave.

A separate comparison run without oxygen production may help estimate ordinary evaporation. However, it is not automatically a perfect correction because the actual reaction can change the liquid temperature and bubbling behaviour.

For an ideal calculation, we normally make these assumptions:

1. The balanced chemical equation correctly describes the reaction.
2. The limiting reactant reacts completely.
3. All the gas being counted escapes from the weighed system.
4. No other significant material enters or leaves.

Under those assumptions:

\[ m_{\text{gas lost}} =m_{\text{initial}}-m_{\text{final}} \]

More generally, define the positive mass loss \(L\) as:

\[ L=m_{\text{initial}}-m_{\text{final}} \]

The signed mass change is \(\Delta m=m_{\text{final}}-m_{\text{initial}}\), so \(\Delta m=-L\) when the reading falls.

If matter also enters the system, the measured loss is the mass leaving minus the mass entering. A balance measures that net change, not the identity of the material involved.

For a fuller explanation of this measurement principle, see Why Mass Seems to Disappear When a Reaction Produces Gas.

03What if the reactant producing the gas is not the limiting reactant?

So far, we have assumed that all the stated hydrogen peroxide reacts.

That is a useful starting case. In many HSC Chemistry questions, however, two reactants are supplied, and one runs out first.

The amount of gas depends on whichever reactant limits the reaction.

Consider sodium carbonate reacting with hydrochloric acid:

\[ \ce{Na2CO3(s) + 2HCl(aq) -> 2NaCl(aq) + H2O(l) + CO2(g)} \]

One mole of sodium carbonate requires two moles of hydrochloric acid and produces one mole of carbon dioxide.

If you have plenty of carbonate but too little acid, you cannot calculate the actual gas production from all the carbonate initially present.

You must first determine how much can react.

04Choosing the correct calculation from the information given

The full mass-to-moles-to-moles-to-mass route works whenever a reactant mass is known. What changes between questions is how you obtain the initial number of moles.

One common mistake involves solution concentration.

Suppose you are given 10.0 g of a solution containing 8.0% hydrogen peroxide by mass.

The mass of hydrogen peroxide is:

\[ m(\ce{H2O2})=10.0(0.080) =0.800\text{ g} \]

It is 0.800 g, not 10.0 g, that must be converted into moles.

The rest of the solution still contributes to the initial balance reading, but it does not count as hydrogen peroxide available to react.

A quick diagnostic comparison

Two mass losses can look the same on a balance but mean different things chemically.

The useful rule is simple: calculate gas production with stoichiometry, and interpret measured mass loss by identifying what crossed the boundary of the weighed system.

You will need both ideas when experimental readings do not match the theoretical result.

05Questions and solutions

Question 1

An open beaker contains 5.30 g of sodium carbonate, which reacts completely with excess hydrochloric acid.

The initial balance reading, including the beaker and all reagents, is 182.40 g.

The reaction is:

\[ \ce{Na2CO3(s) + 2HCl(aq) -> 2NaCl(aq) + H2O(l) + CO2(g)} \]

Calculate:

a. The amount of carbon dioxide produced, in moles.

b. The theoretical mass lost from the beaker.

c. The final balance reading.

Assume all carbon dioxide escapes and no other matter enters or leaves. Use \(M(\ce{Na2CO3})=106.0\text{ g mol}^{-1}\) and \(M(\ce{CO2})=44.0\text{ g mol}^{-1}\).

Solution 1

The reaction produces 0.0500 mol of carbon dioxide, giving a theoretical mass loss of 2.20 g and a final balance reading of 180.20 g.

a. Calculate carbon dioxide moles.

First, convert sodium carbonate into moles:

\[ \begin{aligned} n(\ce{Na2CO3})&=\frac{5.30}{106.0}\\ &=0.0500\text{ mol} \end{aligned} \]

The equation gives a \(1:1\) mole ratio between sodium carbonate and carbon dioxide.

Therefore:

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

b. Calculate the gas mass lost.

\[ \begin{aligned} m(\ce{CO2})&=nM\\ &=0.0500(44.0)\\ &=2.20\text{ g} \end{aligned} \]

c. Calculate the final reading.

\[ \begin{aligned} m_{\text{final}}&=182.40-2.20\\ &=180.20\text{ g} \end{aligned} \]

The 5.30 g of sodium carbonate is not the mass lost. Its atoms become several products, and only the escaping carbon dioxide leaves the beaker under the stated assumptions.

Question 2

An open beaker contains 3.18 g of sodium carbonate. A separate cup contains 50.0 mL of \(1.00\text{ mol L}^{-1}\) hydrochloric acid.

Both containers are initially on the same balance, reading 210.00 g.

The acid is poured into the beaker without removing anything from the balance. Bubbling eventually stops, and the reading stabilises at 208.90 g.

The student then adds another 10.0 mL of the same hydrochloric acid from outside the balance. This added solution has an independently measured mass of 10.10 g.

The reaction is:

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

a. Identify the limiting reactant during the first stage.

b. Calculate the mass of sodium carbonate remaining when the first reading stabilises.

c. Predict the final balance reading after the second addition and reaction.

d. Explain why the first stable reading did not prove that all the sodium carbonate had reacted.

Assume each reaction continues until a reactant is exhausted, all carbon dioxide escapes, and there are no other mass transfers. Use \(M(\ce{Na2CO3})=106.0\text{ g mol}^{-1}\) and \(M(\ce{CO2})=44.0\text{ g mol}^{-1}\).

Solution 2

Hydrochloric acid is limiting in the first stage, leaving 0.530 g of sodium carbonate. The second addition consumes this remaining carbonate, giving a final balance reading of 218.78 g.

a. Identify the first limiting reactant.

The initial amount of sodium carbonate is:

\[ \begin{aligned} n(\ce{Na2CO3})&=\frac{3.18}{106.0}\\ &=0.0300\text{ mol} \end{aligned} \]

The initial amount of HCl is:

\[ \begin{aligned} n(\ce{HCl})&=1.00(0.0500)\\ &=0.0500\text{ mol} \end{aligned} \]

Reacting all the carbonate would require:

\[ 2(0.0300)=0.0600\text{ mol HCl} \]

Only 0.0500 mol is available. HCl is limiting.

b. Find how much carbonate remains.

From the \(2:1\) HCl-to-carbonate ratio:

\[ \begin{aligned} n(\text{carbonate reacted}) &=\frac{0.0500}{2}\\ &=0.0250\text{ mol} \end{aligned} \]

Therefore:

\[ \begin{aligned} n(\text{remaining}) &=0.0300-0.0250\\ &=0.00500\text{ mol} \end{aligned} \]

Its mass is:

\[ m=0.00500(106.0)=0.530\text{ g} \]

The first stage produces 0.0250 mol of carbon dioxide, with mass:

\[ m(\ce{CO2})=0.0250(44.0) =1.10\text{ g} \]

This agrees with the measured decrease from 210.00 g to 208.90 g.

c. Predict the final reading.

The second addition supplies:

\[ \begin{aligned} n(\ce{HCl})&=1.00(0.0100)\\ &=0.0100\text{ mol} \end{aligned} \]

This is exactly enough to react with the remaining 0.00500 mol of sodium carbonate.

The additional carbon dioxide formed is therefore:

\[ n(\ce{CO2})=0.00500\text{ mol} \]

Its mass is:

\[ m(\ce{CO2})=0.00500(44.0) =0.220\text{ g} \]

Adding the second acid portion initially increases the balance reading:

\[ 208.90+10.10=219.00\text{ g} \]

The additional carbon dioxide then escapes:

\[ \begin{aligned} m_{\text{final}}&=219.00-0.220\\ &=218.78\text{ g} \end{aligned} \]

d. Explain the first plateau.

The first reading stabilised because the HCl had been exhausted, not because every reactant had disappeared.

A tempting mistake is to interpret a flat mass-versus-time graph as proof that the original solid was completely consumed. The second acid addition reveals the unreacted carbonate by producing more carbon dioxide.

The increase in balance mass after the addition is also important. You cannot simply subtract all gas produced across both stages from the original 210.00 g without including the 10.10 g of added acid solution.

Question 3

An open reaction vessel initially contains a solution with 5.10 g of hydrogen peroxide.

The hydrogen peroxide decomposes according to:

\[ \ce{2H2O2(aq) -> 2H2O(l) + O2(g)} \]

By the time the balance reading becomes stable, the measured mass decrease is 2.10 g.

A matched comparison experiment without oxygen production loses 0.30 g through evaporation.

For this question, assume that:

• The actual reaction loses exactly 0.30 g of water through evaporation.
• All oxygen produced escapes before the final reading.
• No other matter enters or leaves the vessel.

a. Calculate the actual mass of oxygen that escaped.

b. Determine the mass of hydrogen peroxide that reacted and the percentage of the original hydrogen peroxide consumed.

c. If some oxygen may instead remain dissolved, explain which numerical conclusion becomes uncertain and what minimum percentage reacted can still be established.

Use \(M(\ce{H2O2})=34.0\text{ g mol}^{-1}\) and \(M(\ce{O2})=32.0\text{ g mol}^{-1}\).

Solution 3

Under the stated assumptions, 1.80 g of oxygen escaped, 3.825 g of hydrogen peroxide reacted, and 75.0% of the original hydrogen peroxide was consumed. If some oxygen was retained, 75.0% becomes a minimum rather than an exact conversion.

a. Separate oxygen loss from evaporation.

The observed decrease includes two substances leaving the vessel:

\[ \begin{aligned} m(\ce{O2})&=2.10-0.30\\ &=1.80\text{ g} \end{aligned} \]

The 2.10 g decrease is not entirely oxygen.

b. Work backwards from oxygen to hydrogen peroxide.

Calculate oxygen moles:

\[ \begin{aligned} n(\ce{O2})&=\frac{1.80}{32.0}\\ &=0.05625\text{ mol} \end{aligned} \]

The balanced equation requires two moles of hydrogen peroxide for every mole of oxygen.

\[ \begin{aligned} n(\ce{H2O2})&=2(0.05625)\\ &=0.1125\text{ mol} \end{aligned} \]

Convert to mass:

\[ \begin{aligned} m(\ce{H2O2})&=0.1125(34.0)\\ &=3.825\text{ g} \end{aligned} \]

The percentage consumed is:

\[ \begin{aligned} \%\text{ reacted} &=\frac{3.825}{5.10}\times100\\ &=75.0\% \end{aligned} \]

The remaining hydrogen peroxide mass is:

\[ 5.10-3.825=1.275\text{ g} \]

This result depends on the assumption that all oxygen produced escaped.

c. What changes if oxygen remains dissolved?

If some produced oxygen remains in the liquid, the amount of oxygen formed could be greater than 0.05625 mol.

That means more hydrogen peroxide could have reacted than the balance loss suggests.

Under the other stated assumptions, at least 75.0% of the initial hydrogen peroxide reacted, but the exact conversion cannot be established from the mass decrease alone. It could be as high as 100%.

This is why a corrected mass loss is not automatically an exact measurement of reaction completion.

In a real experiment, the comparison run may also fail to reproduce evaporation perfectly. The reaction can warm the liquid or change its exposed surface, so that assumption needs experimental justification.

Question 4

A dry 11.40 g sample contains sodium hydrogencarbonate and inert sand.

The sample is heated in an open vessel until all the sodium hydrogencarbonate decomposes:

\[ \ce{2NaHCO3(s) -> Na2CO3(s) + CO2(g) + H2O(g)} \]

After cooling under dry conditions, the solid residue has a mass of 9.85 g.

The entire residue is then reacted with excess hydrochloric acid in a separate open vessel:

\[ \ce{Na2CO3(s) + 2HCl(aq) -> 2NaCl(aq) + H2O(l) + CO2(g)} \]

During this second reaction, the mass lost specifically as carbon dioxide is 1.10 g.

Assume both reactions are complete, the sand is chemically unchanged, and the heating stage loses only carbon dioxide and water vapour. During the acid stage, assume all carbon dioxide escapes with no other mass transfers.

a. Determine the original mass of sodium hydrogencarbonate and the mass of sand.

b. Calculate the separate masses of carbon dioxide and water vapour lost during heating.

c. Show that the heating mass loss agrees with the measured solid residue.

d. Explain why dividing the heating mass loss by the molar mass of carbon dioxide would give an incorrect amount of carbon dioxide.

Use \(M(\ce{NaHCO3})=84.0\text{ g mol}^{-1}\), \(M(\ce{Na2CO3})=106.0\text{ g mol}^{-1}\), \(M(\ce{CO2})=44.0\text{ g mol}^{-1}\), and \(M(\ce{H2O})=18.0\text{ g mol}^{-1}\).

Solution 4

The original sample contained 4.20 g of sodium hydrogencarbonate and 7.20 g of sand. Heating released 1.10 g of carbon dioxide and 0.450 g of water vapour, giving the measured 1.55 g mass decrease.

a. Use the second reaction to reconstruct the starting sample.

The second reaction releases 1.10 g of carbon dioxide.

\[ \begin{aligned} n(\ce{CO2})&=\frac{1.10}{44.0}\\ &=0.0250\text{ mol} \end{aligned} \]

Each mole of sodium carbonate produces one mole of carbon dioxide when reacted with sufficient acid.

Therefore, the cooled residue contained:

\[ n(\ce{Na2CO3})=0.0250\text{ mol} \]

Now use the first equation.

Two moles of sodium hydrogencarbonate produce one mole of sodium carbonate.

\[ \begin{aligned} n(\ce{NaHCO3})&=2(0.0250)\\ &=0.0500\text{ mol} \end{aligned} \]

Its original mass was:

\[ \begin{aligned} m(\ce{NaHCO3})&=0.0500(84.0)\\ &=4.20\text{ g} \end{aligned} \]

The original sample also contained inert sand.

\[ m(\text{sand})=11.40-4.20 =7.20\text{ g} \]

b. Calculate both gases released during heating.

From the first balanced equation, two moles of sodium hydrogencarbonate release one mole of carbon dioxide and one mole of water vapour.

For 0.0500 mol of sodium hydrogencarbonate:

\[ n(\ce{CO2})=n(\ce{H2O}) =0.0250\text{ mol} \]

Carbon dioxide mass:

\[ m(\ce{CO2})=0.0250(44.0) =1.10\text{ g} \]

Water vapour mass:

\[ m(\ce{H2O})=0.0250(18.0) =0.450\text{ g} \]

Total heating loss:

\[ m_{\text{lost}}=1.10+0.450 =1.55\text{ g} \]

c. Check the solid residue.

From the measured values:

\[ 11.40-9.85=1.55\text{ g} \]

This matches the predicted loss.

An independent check is to calculate the mass remaining after heating.

The sodium carbonate residue has mass:

\[ m(\ce{Na2CO3})=0.0250(106.0) =2.65\text{ g} \]

Adding the unchanged sand:

\[ m_{\text{residue}}=2.65+7.20 =9.85\text{ g} \]

Both methods give the same result.

d. Explain the misconception.

The tempting calculation is \(1.55/44.0\), treating the entire heating loss as carbon dioxide.

That is incorrect because heating creates two gaseous products: carbon dioxide and water vapour.

The first-stage mass loss is their combined mass.

There is also a useful carbon-accounting result here. The heating reaction releases half of the original carbon atoms as carbon dioxide and leaves the other half in sodium carbonate. The subsequent acid reaction releases that remaining carbon as more carbon dioxide.

The 1.10 g of carbon dioxide released in the second stage is additional gas, not the same gas that escaped during heating.

Question 5

An impure 5.30 g sample contains sodium carbonate and inert sand.

The sample reacts completely with excess hydrochloric acid inside a reaction flask. Carbon dioxide passes through a delivery tube into a separate absorption flask containing a solution with 1.60 g of sodium hydroxide.

Both flasks and their connecting apparatus sit on the same electronic balance. The absorption flask has an open outlet to the surrounding air.

The reactions are:

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

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

For this idealised problem, assume:

• All carbon dioxide produced in the reaction flask reaches the absorber.
• The sodium hydroxide captures all incoming carbon dioxide until it is exhausted.
• Once the sodium hydroxide is exhausted, any additional carbon dioxide leaves through the open outlet.
• All captured gas and reaction products remain in the absorption flask.
• The sand is inert, and there are no other mass transfers.
• Effects from displacement of initially trapped air, gas buoyancy, and tubing movement are negligible.

At first, the balance remains at 250.00 g even though carbon dioxide bubbles are visibly forming.

After a period of time, the balance begins to fall. When the reaction finishes, the final reading is 249.56 g.

a. Explain how the balance can initially remain constant while carbon dioxide is being produced.

b. Calculate the total mass of carbon dioxide produced, the mass absorbed by sodium hydroxide, and the mass escaping into the surrounding air.

c. Determine the percentage by mass of sodium carbonate in the original sample.

d. Imagine repeating the experiment with the same reactant amounts, but now only the reaction flask sits on the balance. The absorber remains connected by a flexible, independently supported tube but sits completely outside the weighed system. The initial reading of the reaction flask alone is 200.00 g.

Predict its final reading.

e. Explain why the two arrangements can give different balance decreases even though the same total amount of carbon dioxide is produced and the same amount reaches the surrounding air.

Use \(M(\ce{NaOH})=40.0\text{ g mol}^{-1}\), \(M(\ce{Na2CO3})=106.0\text{ g mol}^{-1}\), and \(M(\ce{CO2})=44.0\text{ g mol}^{-1}\).

Solution 5

The reaction produces 1.32 g of carbon dioxide, but only 0.440 g escapes from the combined apparatus. The absorber retains 0.880 g, and the original sample is 60.0% sodium carbonate by mass. When only the reaction flask is weighed, its predicted mass decrease is the full 1.32 g.

The central insight is that gas can leave the reaction flask without leaving the entire system being weighed.

a. Explain the initially constant reading.

At first, the reaction produces carbon dioxide, which leaves the reaction flask and travels through the tube.

But the gas is absorbed by the sodium hydroxide solution.

Both flasks sit on the balance, so the carbon dioxide has merely moved from one part of the weighed apparatus to another.

There is no net transfer of that carbon dioxide out of the complete weighed system.

Once the sodium hydroxide is exhausted, additional carbon dioxide passes through the absorber and escapes into the surrounding air. The balance then begins to fall.

The initial flat section of the graph does not indicate that the reaction has stopped or failed to begin.

b. Calculate how much carbon dioxide was absorbed, produced, and lost.

Start with the sodium hydroxide.

\[ \begin{aligned} n(\ce{NaOH})&=\frac{1.60}{40.0}\\ &=0.0400\text{ mol} \end{aligned} \]

The absorption equation shows that two moles of NaOH react with one mole of carbon dioxide.

Therefore, the maximum amount of carbon dioxide absorbed is:

\[ \begin{aligned} n(\ce{CO2,absorbed}) &=\frac{0.0400}{2}\\ &=0.0200\text{ mol} \end{aligned} \]

Its mass is:

\[ \begin{aligned} m(\ce{CO2,absorbed}) &=0.0200(44.0)\\ &=0.880\text{ g} \end{aligned} \]

The balance eventually falls, so under the stated assumptions the sodium hydroxide must have become exhausted. It captured its full 0.880 g capacity before additional gas escaped.

The combined apparatus loses:

\[ \begin{aligned} L&=250.00-249.56\\ &=0.440\text{ g} \end{aligned} \]

This is the mass of carbon dioxide that escaped into the surrounding air.

The total carbon dioxide produced must therefore be:

\[ \begin{aligned} m(\ce{CO2,total}) &=0.880+0.440\\ &=1.320\text{ g} \end{aligned} \]

The absorption reaction produces water, but that water remains inside the absorption flask. It does not create an extra mass loss.

c. Determine the sample purity.

Convert total carbon dioxide mass into moles:

\[ \begin{aligned} n(\ce{CO2})&=\frac{1.320}{44.0}\\ &=0.0300\text{ mol} \end{aligned} \]

The carbonate reaction gives a \(1:1\) mole ratio:

\[ n(\ce{Na2CO3})=0.0300\text{ mol} \]

Therefore:

\[ \begin{aligned} m(\ce{Na2CO3})&=0.0300(106.0)\\ &=3.18\text{ g} \end{aligned} \]

The original sample had a total mass of 5.30 g.

\[ \begin{aligned} \%\ce{Na2CO3} &=\frac{3.18}{5.30}\times100\\ &=60.0\% \end{aligned} \]

The remaining 2.12 g is inert sand.

d. Predict the reading when only the reaction flask is weighed.

In the new arrangement, every gram of carbon dioxide travelling out of the reaction flask crosses the boundary of the weighed system.

It does not matter that much of that carbon dioxide is absorbed afterwards. The absorber is not on the balance.

The reaction flask therefore loses the full 1.32 g of carbon dioxide.

\[ \begin{aligned} m_{\text{final}}&=200.00-1.32\\ &=198.68\text{ g} \end{aligned} \]

The predicted final reading is 198.68 g.

e. Reconcile the different measurements.

In the first arrangement, the balance supports the reaction flask and the absorber together.

Of the 1.32 g of carbon dioxide produced, 0.880 g is transferred within the weighed system. Only 0.440 g crosses the outer boundary and escapes.

In the second arrangement, the balance supports only the reaction flask. All 1.32 g of carbon dioxide leaves that weighed system, even though 0.880 g is subsequently captured elsewhere.

The tempting conclusion is that the smaller decrease in the first experiment means less carbon dioxide was produced.

But both experiments have identical chemistry. The difference is where the weighing boundary sits.

This also explains why a balance alone cannot always reveal how much gas a reaction has produced. You need to know where the gas travels, whether it is retained or absorbed, and what the instrument actually weighs.

If the next experiment measures gas volume using a gas syringe instead of mass loss, the balanced-equation mole ratio still works. The final step changes from \(m=nM\) to a volume calculation using molar volume or the gas equation.

That is the connection developed in Gas Volumes in Chemical Reactions for HSC Chemistry.