Why Equal Moles Can Have Different Masses in HSC Chemistry

Two identical 1.00 L gas bulbs sit on a laboratory bench. Both are at 25°C and 100 kPa. One contains helium gas; the other contains carbon dioxide.

Which bulb contains more gas particles? Which contains more mass?

The surprising result is that, to a good approximation, both bulbs contain the same number of gas particles, but the carbon dioxide bulb contains about 11 times as much gas mass.

This is why equal moles do not mean equal masses. A mole tells you how many specified particles you have. It does not tell you how much each particle weighs.

The key is to separate three quantities: the number of particles, the amount of substance, and the mass. Once you understand how they connect, the calculations become much easier.

## The experiment: same volume, same pressure, same temperature

Before looking at the mathematics, predict what happens in the two gas bulbs.

Both bulbs have:
- The same internal volume: \(1.00\ \mathrm{L}\).
- The same temperature: \(25^\circ\mathrm{C}\).
- The same absolute pressure: \(100\ \mathrm{kPa}\).

One contains helium, \(\ce{He}\), and the other contains carbon dioxide, \(\ce{CO2}\).

You might predict that equal volumes of gas have equal masses. After all, the bulbs are identical, and their gas conditions are identical.

That prediction is wrong. Equal volumes at the same temperature and pressure contain approximately equal numbers of gas particles, not equal masses.

[IMAGE: Two identical sealed laboratory gas bulbs side by side, one labelled helium (He) and the other carbon dioxide (CO2), each with a volume of 1.00 L, pressure of 100 kPa, and temperature of 25°C. Show a balance comparison indicating that the CO2 bulb has substantially more gas mass, while keeping the vessels themselves identical. Include a small inset showing the much lighter individual helium atoms compared with the heavier CO2 molecules.]

### How would you test the prediction?

A suitable experiment uses a rigid gas bulb connected to a vacuum pump, a gas supply, a pressure gauge, and a temperature sensor.

First, weigh the evacuated bulb. Then fill it with helium until it reaches 100 kPa absolute pressure at 25°C. Close the valve and record its mass. Next, evacuate the bulb again, fill it with carbon dioxide to the same pressure and temperature, and record its mass.

Subtract the empty bulb's mass from each measurement. The differences give the masses of the two gas samples.

The important controls are the bulb volume, gas temperature, and absolute pressure. If any of these differ, the samples may contain different amounts of gas, which would make the comparison unfair.

What should the experiment show? The carbon dioxide sample should be much heavier, even though both samples contain approximately the same amount of gas in moles.

### Why the original prediction fails

The reasoning becomes clearer if you picture the particles.

A helium sample contains separate helium atoms. Each helium atom has a relative atomic mass of approximately 4.00.

A carbon dioxide sample contains \(\ce{CO2}\) molecules. Each molecule contains one carbon atom and two oxygen atoms, giving a relative molecular mass of approximately:

\[
12.01+2(16.00)=44.01
\]

The carbon dioxide molecule is therefore about 11 times as massive as a helium atom.

Equal numbers of these particles cannot have equal total masses when the individual particles have different masses.

This is the central distinction: **equal particle counts do not require equal particle masses.**

Two identical 1.00 L gas bulbs sit on a laboratory bench. Both are at 25°C and 100 kPa. One contains helium gas; the other contains carbon dioxide.

Which bulb contains more gas particles? Which contains more mass?

The surprising result is that, to a good approximation, both bulbs contain the same number of gas particles, but the carbon dioxide bulb contains about 11 times as much gas mass.

This is why equal moles do not mean equal masses. A mole tells you how many specified particles you have. It does not tell you how much each particle weighs.

The key is to separate three quantities: the number of particles, the amount of substance, and the mass. Once you understand how they connect, the calculations become much easier.

## The experiment: same volume, same pressure, same temperature

Before looking at the mathematics, predict what happens in the two gas bulbs.

Both bulbs have:
- The same internal volume: \(1.00\ \mathrm{L}\).
- The same temperature: \(25^\circ\mathrm{C}\).
- The same absolute pressure: \(100\ \mathrm{kPa}\).

One contains helium, \(\ce{He}\), and the other contains carbon dioxide, \(\ce{CO2}\).

You might predict that equal volumes of gas have equal masses. After all, the bulbs are identical, and their gas conditions are identical.

That prediction is wrong. Equal volumes at the same temperature and pressure contain approximately equal numbers of gas particles, not equal masses.

[IMAGE: Two identical sealed laboratory gas bulbs side by side, one labelled helium (He) and the other carbon dioxide (CO2), each with a volume of 1.00 L, pressure of 100 kPa, and temperature of 25°C. Show a balance comparison indicating that the CO2 bulb has substantially more gas mass, while keeping the vessels themselves identical. Include a small inset showing individual helium atoms compared with heavier CO2 molecules.]

### How would you test the prediction?

A suitable experiment uses a rigid gas bulb connected to a vacuum pump, a gas supply, a pressure gauge, and a temperature sensor.

First, weigh the evacuated bulb. Then fill it with helium until it reaches 100 kPa absolute pressure at 25°C. Close the valve and record its mass.

Next, evacuate the bulb again, fill it with carbon dioxide to the same pressure and temperature, and record its mass.

Subtract the empty bulb's mass from each measurement. The differences give the masses of the two gas samples.

The important controls are the bulb volume, gas temperature, and absolute pressure. If any of these differ, the samples may contain different amounts of gas, which would make the comparison unfair.

An initial comparison using two gas containers without controlling temperature and pressure would not establish that the samples contain equal numbers of particles. The revised experiment controls all three conditions so that the comparison is meaningful.

### Why the original prediction fails

The reasoning becomes clearer if you picture the particles.

A helium sample contains separate helium atoms. Each helium atom has a relative atomic mass of approximately 4.00.

A carbon dioxide sample contains \(\ce{CO2}\) molecules. Each molecule contains one carbon atom and two oxygen atoms, giving a relative molecular mass of approximately:

\[
12.01+2(16.00)=44.01
\]

The carbon dioxide molecule is therefore about 11 times as massive as a helium atom.

Equal numbers of these particles cannot have equal total masses when the individual particles have different masses.

This is the central distinction: **equal particle counts do not require equal particle masses.**

### The historical clue: Avogadro's hypothesis

This idea has an important history.

In the early nineteenth century, scientists were investigating the volumes of gases involved in chemical reactions. Joseph Louis Gay-Lussac observed that reacting gases often combined in simple whole-number volume ratios.

For example, two volumes of hydrogen gas react with one volume of oxygen gas to form two volumes of water vapour, when measured under the same conditions.

But there was a problem. How could scientists connect measured gas volumes to the numbers of individual particles?

In 1811, Italian scientist Amedeo Avogadro proposed that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

His hypothesis provided a particle-level explanation for the observed gas-volume relationships. The original measurements were volume observations, not direct counts of molecules.

[IMAGE: Historical scientific concept illustration showing three graduated gas-volume tubes representing two equal volumes of hydrogen gas combining with one equal volume of oxygen gas to produce two equal volumes of water vapour. Show equal particle counts per equal volume, with correctly represented H2, O2, and H2O molecules. Label this as a conceptual reconstruction of the gas-volume reasoning, not an exact historical apparatus.]

The modern form of this relationship is called Avogadro's law:

\[
V\propto n
\]

At constant temperature and pressure, gas volume, \(V\), is directly proportional to amount of gas, \(n\), when the gases behave approximately ideally.

This explains why the two bulbs in our experiment contain approximately equal amounts of gas in moles.

It does not say that they have equal masses. The gas particles themselves have different masses.

You can read more about Avogadro's historical contribution in this [American Physical Society account](https://www.aps.org/apsnews/2016/02/amadeo-avogadro-enumerated-molecular-world).

### A quick calculation: how much gas is actually in each bulb?

We can estimate the amount of gas in the 1.00 L bulbs using the ideal gas equation:

\[
PV=nRT
\]

Where:

- \(P\) = absolute pressure, in kPa.
- \(V\) = gas volume, in L.
- \(n\) = amount of gas, in mol.
- \(R\) = gas constant, \(8.314\ \mathrm{kPa\,L\,mol^{-1}\,K^{-1}}\).
- \(T\) = absolute temperature, in K.

Convert the temperature:

\[
T=25+273=298\ \mathrm{K}
\]

Rearrange the equation:

\[
n=\frac{PV}{RT}
\]

Substitute the experimental conditions:

\[
\begin{aligned}
n&=\frac{100\times1.00}{8.314\times298}\\
&=0.0404\ \mathrm{mol}
\end{aligned}
\]

Each bulb therefore contains approximately \(0.0404\ \mathrm{mol}\) of gas.

Now calculate the mass of each sample.

For helium:

\[
\begin{aligned}
m&=nM\\
&=0.0404\times4.00\\
&=0.162\ \mathrm{g}
\end{aligned}
\]

For carbon dioxide:

\[
\begin{aligned}
m&=nM\\
&=0.0404\times44.01\\
&=1.78\ \mathrm{g}
\end{aligned}
\]

The carbon dioxide sample has approximately 11 times the mass of the helium sample.

This calculation assumes ideal-gas behaviour. Real gases do not behave perfectly ideally, but at these conditions the approximation is suitable for illustrating the relationship.

## What a mole actually counts

A mole is a counting unit, not a mass unit.

Think about a dozen. A dozen eggs contains 12 eggs. A dozen bricks contains 12 bricks.

The number is identical, but the total mass is very different.

A mole works in the same way, except that it represents an enormous number of particles.

One mole contains exactly:

\[
N_A=6.02214076\times10^{23}
\]

specified elementary entities.

This number is called Avogadro's constant. It is the exact value used in the modern SI definition of the mole, established from 20 May 2019.

The [Bureau International des Poids et Mesures (BIPM)](https://www.bipm.org/en/si-base-units/mole) defines the mole as the SI unit of amount of substance.

The important word is *specified*. You must know what particles you are counting.

For example:

| Substance | Particle being counted | Entities in 1 mol |
|---|---|---:|
| Helium, \(\ce{He}\) | Helium atoms | \(6.022\times10^{23}\) |
| Hydrogen, \(\ce{H2}\) | Hydrogen molecules | \(6.022\times10^{23}\) |
| Water, \(\ce{H2O}\) | Water molecules | \(6.022\times10^{23}\) |
| Carbon dioxide, \(\ce{CO2}\) | Carbon dioxide molecules | \(6.022\times10^{23}\) |
| Sodium chloride, \(\ce{NaCl}\) | Formula units | \(6.022\times10^{23}\) |

Notice that sodium chloride is different from the molecular substances in the table.

It is an ionic compound with an extended lattice structure, so we describe its simplest repeating composition using formula units rather than molecules.

The mole counts the specified entities, regardless of whether they are atoms, molecules, or formula units.

[IMAGE: Scientific comparison of equal one-mole samples of helium atoms, water molecules, carbon dioxide molecules, and sodium chloride formula units. Represent each sample with the same number of particle symbols in a magnified conceptual illustration, while showing different total sample masses: 4.00 g, 18.02 g, 44.01 g, and 58.44 g respectively. Include accurate molecular structures and make clear that the particle drawings represent a tiny sample of the enormous actual count.]

### So where does mass come into it?

The missing quantity is molar mass.

Molar mass, represented by \(M\), is the mass of one mole of a substance.

Its unit is:

\[
\mathrm{g\,mol^{-1}}
\]

This unit means grams per mole.

For example:

- One mole of helium atoms has a mass of approximately \(4.00\ \mathrm{g}\).
- One mole of water molecules has a mass of approximately \(18.02\ \mathrm{g}\).
- One mole of carbon dioxide molecules has a mass of approximately \(44.01\ \mathrm{g}\).

Every sample contains the same number of specified entities, but each entity has a different mass.

That is why the molar masses differ.

To calculate molar mass, add the relative atomic masses of the atoms in the chemical formula, accounting for every subscript.

For example, carbon dioxide contains one carbon atom and two oxygen atoms:

\[
\begin{aligned}
M(\ce{CO2})&=12.01+2(16.00)\\
&=44.01\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

For a refresher on calculating molar mass from more complicated formulae, see [How to Calculate Molar Mass When a Formula Contains Brackets](https://tutorgum.com/learn/resources/hsc-chemistry-how-to-calculate-molar-mass-when-a-formula-contains-brackets/).

## The mathematical relationship between mass, moles, and particles

There are three quantities you need to keep separate.

| Quantity | Symbol | Unit | What it tells you |
|---|---|---|---|
| Mass | \(m\) | g | How much matter is present |
| Amount of substance | \(n\) | mol | How many specified entities are present |
| Number of entities | \(N\) | No units (a count) | The actual number of specified particles |

Two conversion relationships connect these quantities.

### Relationship 1: converting moles into particles

Avogadro's constant connects the amount of substance to the number of entities:

\[
N=nN_A
\]

Where:

- \(N\) = number of specified entities.
- \(n\) = amount of substance, in mol.
- \(N_A\) = Avogadro's constant, \(6.022\times10^{23}\ \mathrm{mol^{-1}}\).

For example, \(0.500\ \mathrm{mol}\) of water molecules contains:

\[
N=0.500\times6.022\times10^{23}
\]

\[
N=3.011\times10^{23}\text{ molecules}
\]

### Relationship 2: converting moles into mass

Molar mass connects the amount of substance to the sample's mass:

\[
m=nM
\]

Where:

- \(m\) = sample mass, in g.
- \(n\) = amount of substance, in mol.
- \(M\) = molar mass, in \(\mathrm{g\,mol^{-1}}\).

For example, \(0.500\ \mathrm{mol}\) of water has a mass of:

\[
m=0.500\times18.02
\]

\[
m=9.01\ \mathrm{g}
\]

The number of molecules is determined by the amount in moles. The mass depends on both the amount and the molar mass.

### Why the units matter

Look at the units in the mass equation:

\[
\mathrm{mol}\times\mathrm{g\,mol^{-1}}
=\mathrm{g}
\]

The mole units cancel, leaving grams.

For the particle equation:

\[
\mathrm{mol}\times\mathrm{entities\,mol^{-1}}
=\mathrm{entities}
\]

Again, the mole units cancel, leaving the number of entities.

This is dimensional analysis. It helps you check whether your equation is producing the type of answer the question actually asks for.

You can combine the two relationships:

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

Rearranging gives:

\[
m=\frac{N}{N_A}\times M
\]

This final equation shows exactly why equal particle counts do not necessarily produce equal masses.

If two samples contain the same number of particles, their values of \(N\) are equal. However, their molar masses, \(M\), can be different.

Therefore, their masses can be different.

[IMAGE: Annotated scientific relationship diagram showing three physical quantities connected by equations: a balance measuring sample mass in grams, a labelled amount of substance in moles, and a magnified collection of individual particles. Show the conversion moles to mass using multiplication by molar mass and moles to particle count using Avogadro's constant. Include the units on both conversion paths and show that the two paths answer different questions.]

### Worked example: equal moles of water and carbon dioxide

You have \(0.250\ \mathrm{mol}\) of water molecules and \(0.250\ \mathrm{mol}\) of carbon dioxide molecules. Calculate the mass of each sample and compare their number of molecules.

**Step 1**: Calculate the molar mass of each substance.

For water:

\[
\begin{aligned}
M(\ce{H2O})&=2(1.008)+16.00\\
&=18.02\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

For carbon dioxide:

\[
\begin{aligned}
M(\ce{CO2})&=12.01+2(16.00)\\
&=44.01\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

**Step 2**: Calculate the mass of each sample.

Use:

\[
m=nM
\]

For water:

\[
m=0.250\times18.02
\]

\[
m=4.50\ \mathrm{g}
\]

For carbon dioxide:

\[
m=0.250\times44.01
\]

\[
m=11.0\ \mathrm{g}
\]

**Step 3**: Calculate the number of molecules in each sample.

Both samples contain \(0.250\ \mathrm{mol}\) of molecules.

Therefore:

\[
\begin{aligned}
N&=0.250\times6.022\times10^{23}\\
&=1.51\times10^{23}\text{ molecules}
\end{aligned}
\]

Both samples contain exactly the same number of molecules.

**Step 4**: Compare their masses.

\[
\frac{m(\ce{CO2})}{m(\ce{H2O})}
=\frac{44.01}{18.02}
\]

\[
=2.44
\]

The carbon dioxide sample has approximately 2.44 times the mass of the water sample, despite containing the same number of molecules.

[IMAGE: Side-by-side particle illustration of 0.250 mol water and 0.250 mol carbon dioxide. Show equal numbers of representative molecules, with accurate H2O bent molecular geometry and CO2 linear molecular geometry. Include mass labels of 4.50 g and 11.0 g respectively, and a clear visual comparison of the different molecular masses.]

### Equal masses do not mean equal moles either

Now reverse the situation.

Suppose you have two samples, each with a mass of \(10.0\ \mathrm{g}\). One contains ammonia, \(\ce{NH3}\), and the other contains sulfur dioxide, \(\ce{SO2}\).

Would they contain the same number of molecules?

No. The ammonia molecules are much lighter, so more of them fit into the same total mass.

### Worked example: equal masses of ammonia and sulfur dioxide

A laboratory has \(10.0\ \mathrm{g}\) of ammonia and \(10.0\ \mathrm{g}\) of sulfur dioxide. Calculate the amount and number of molecules in each sample. Determine which sample contains more molecules and by what factor.

**Step 1**: Calculate the molar masses.

For ammonia:

\[
\begin{aligned}
M(\ce{NH3})&=14.01+3(1.008)\\
&=17.03\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

For sulfur dioxide:

\[
\begin{aligned}
M(\ce{SO2})&=32.06+2(16.00)\\
&=64.06\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

**Step 2**: Calculate the amount of each substance.

Rearrange the mass equation:

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

For ammonia:

\[
n=\frac{10.0}{17.03}
=0.587\ \mathrm{mol}
\]

For sulfur dioxide:

\[
n=\frac{10.0}{64.06}
=0.156\ \mathrm{mol}
\]

**Step 3**: Calculate the number of molecules.

For ammonia:

\[
\begin{aligned}
N&=0.587\times6.022\times10^{23}\\
&=3.53\times10^{23}\text{ molecules}
\end{aligned}
\]

For sulfur dioxide:

\[
\begin{aligned}
N&=0.156\times6.022\times10^{23}\\
&=9.40\times10^{22}\text{ molecules}
\end{aligned}
\]

**Step 4**: Compare the particle counts.

\[
\frac{N(\ce{NH3})}{N(\ce{SO2})}
=\frac{0.587}{0.156}
\]

\[
=3.76
\]

The ammonia sample contains approximately 3.76 times as many molecules as the sulfur dioxide sample.

Both samples have the same mass, but the ammonia sample has a much greater amount of substance.

[IMAGE: Two equal-mass laboratory sample containers, each labelled 10.0 g, one containing ammonia molecules and the other sulfur dioxide molecules. Show noticeably more NH3 molecules than SO2 molecules in the equal-mass samples. Use accurate molecular structures: trigonal pyramidal NH3 and bent SO2. Do not use arbitrary particle sizes to represent molecular mass.]

## Common misconceptions to avoid

There are a few mistakes worth catching before they become habits.

| Tempting assumption | What is actually correct |
|---|---|
| Equal moles mean equal masses. | Equal moles mean equal numbers of specified entities. Mass also depends on molar mass. |
| Equal masses mean equal moles. | The amount depends on the mass divided by the molar mass. |
| One mole always means one mole of molecules. | A mole can count atoms, molecules, ions, or formula units. |
| Equal gas volumes always contain equal amounts. | This is approximately true only when temperature and pressure are also equal and ideal-gas behaviour is a suitable model. |
| One mole of a compound contains one mole of atoms. | The number of moles of atoms depends on the compound's formula. |

For example, \(1.00\ \mathrm{mol}\) of water molecules contains \(2.00\ \mathrm{mol}\) of hydrogen atoms and \(1.00\ \mathrm{mol}\) of oxygen atoms.

The number of moles changes depending on which entity you are counting.

If you want to practise that distinction further, see [How to Convert Moles of a Compound into Moles of Its Elements](https://tutorgum.com/learn/resources/hsc-chemistry-how-to-convert-between-moles-of-compound-and-moles-of-an-element-in-it/).

Also remember that the mole and molar mass answer different questions:

- The mole tells you how many specified particles are present.
- Molar mass tells you the mass of one mole of those particles.
- Mass tells you how much matter is present in the sample.

Keeping these three ideas separate is the key to solving mole calculations correctly.

## Questions and solutions

### Question 1

A sample contains \(0.250\ \mathrm{mol}\) of calcium carbonate, \(\ce{CaCO3}\), formula units.

Calculate the mass of the sample.

Use:

- Ca = 40.08
- C = 12.01
- O = 16.00

### Solution 1

The mass of the calcium carbonate sample is **\(25.0\ \mathrm{g}\)**.

**Step 1**: Calculate the molar mass.

\[
\begin{aligned}
M(\ce{CaCO3})
&=40.08+12.01+3(16.00)\\
&=100.09\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

**Step 2**: Apply the mass equation.

\[
m=nM
\]

Substitute:

\[
\begin{aligned}
m&=0.250\times100.09\\
&=25.0225\ \mathrm{g}
\end{aligned}
\]

To three significant figures:

\[
\boxed{m=25.0\ \mathrm{g}}
\]

The sample contains \(0.250\ \mathrm{mol}\) of calcium carbonate formula units, each with the same chemical composition and molar mass.

### Question 2

A student is given two gas samples:

- Sample A contains \(3.00\ \mathrm{g}\) of hydrogen gas, \(\ce{H2}\).
- Sample B contains \(3.00\ \mathrm{g}\) of helium gas, \(\ce{He}\).

Calculate:

a) The amount of each gas in moles.

b) The number of gas particles in each sample.

c) The total number of atoms in each sample.

d) Explain why the answers to parts (b) and (c) give different comparisons.

Use:

- \(M(\ce{H2})=2.016\ \mathrm{g\,mol^{-1}}\)
- \(M(\ce{He})=4.00\ \mathrm{g\,mol^{-1}}\)

[IMAGE: Two equal-mass gas samples in separate labelled containers, one containing hydrogen molecules represented as pairs of bonded hydrogen atoms and the other containing individual helium atoms. Label both samples 3.00 g. Show the distinction between a molecule as one gas particle and the individual atoms that make up that molecule, without showing the final particle counts.]

### Solution 2

The hydrogen sample contains approximately twice as many gas particles and four times as many atoms as the helium sample.

The difference comes from what counts as one gas particle.

**Step 1**: Calculate the amount of hydrogen gas.

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

\[
\begin{aligned}
n(\ce{H2})&=\frac{3.00}{2.016}\\
&=1.49\ \mathrm{mol}
\end{aligned}
\]

**Step 2**: Calculate the amount of helium gas.

\[
\begin{aligned}
n(\ce{He})&=\frac{3.00}{4.00}\\
&=0.750\ \mathrm{mol}
\end{aligned}
\]

**Step 3**: Calculate the number of gas particles.

Hydrogen gas consists of \(\ce{H2}\) molecules:

\[
\begin{aligned}
N(\ce{H2})
&=1.49\times6.022\times10^{23}\\
&=8.96\times10^{23}\text{ molecules}
\end{aligned}
\]

Helium gas consists of individual atoms:

\[
\begin{aligned}
N(\ce{He})
&=0.750\times6.022\times10^{23}\\
&=4.52\times10^{23}\text{ atoms}
\end{aligned}
\]

The hydrogen sample contains approximately twice as many gas particles.

**Step 4**: Calculate the total number of atoms.

Each hydrogen molecule contains two hydrogen atoms.

Therefore, the amount of hydrogen atoms is:

\[
n(\text{H atoms})=2(1.49)
\]

\[
=2.98\ \mathrm{mol\ of\ atoms}
\]

The number of hydrogen atoms is:

\[
\begin{aligned}
N(\text{H atoms})
&=2.98\times6.022\times10^{23}\\
&=1.80\times10^{24}\text{ atoms}
\end{aligned}
\]

Helium is monatomic, meaning each gas particle is already one atom.

Therefore, the number of helium atoms is:

\[
N(\text{He atoms})=4.52\times10^{23}
\]

**Step 5**: Compare the total numbers of atoms.

\[
\frac{1.80\times10^{24}}{4.52\times10^{23}}
=3.98
\]

The hydrogen sample contains approximately four times as many atoms.

The trap is treating the number of gas particles as identical to the number of atoms. That works for helium, but not for hydrogen gas because each hydrogen molecule contains two atoms.

### Question 3

An unknown gas is placed inside a rigid \(2.00\ \mathrm{L}\) bulb.

The gas is measured under the following conditions:

- Absolute pressure: \(100\ \mathrm{kPa}\)
- Temperature: \(27^\circ\mathrm{C}\)
- Mass of gas: \(3.53\ \mathrm{g}\)

Assume ideal-gas behaviour.

a) Calculate the amount of gas in moles.

b) Calculate its molar mass.

c) The calculated molar mass is close to that of carbon dioxide. Does this prove that the gas is \(\ce{CO2}\)? Explain.

Use:

\[
R=8.314\ \mathrm{kPa\,L\,mol^{-1}\,K^{-1}}
\]

[IMAGE: A rigid 2.00 L laboratory gas bulb connected to a pressure gauge, temperature sensor, gas inlet valve, and vacuum line. Show the bulb being weighed on a laboratory balance after filling. Include labels for 100 kPa absolute pressure, 27°C, and 3.53 g gas mass. Do not show the calculated amount or molar mass.]

### Solution 3

The gas has an amount of approximately \(0.0802\ \mathrm{mol}\) and a molar mass of approximately \(44.0\ \mathrm{g\,mol^{-1}}\). This is consistent with carbon dioxide, but it does not prove the gas identity.

**Step 1**: Convert the temperature into kelvin.

\[
T=27+273
\]

\[
T=300\ \mathrm{K}
\]

**Step 2**: Calculate the amount of gas.

Use the ideal gas equation:

\[
PV=nRT
\]

Rearrange:

\[
n=\frac{PV}{RT}
\]

Substitute:

\[
\begin{aligned}
n&=\frac{100\times2.00}{8.314\times300}\\
&=\frac{200}{2494.2}\\
&=0.0802\ \mathrm{mol}
\end{aligned}
\]

**Step 3**: Calculate the molar mass.

Use:

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

Substitute:

\[
\begin{aligned}
M&=\frac{3.53}{0.0802}\\
&=44.0\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

**Step 4**: Evaluate the identification.

Carbon dioxide has a molar mass of approximately \(44.01\ \mathrm{g\,mol^{-1}}\).

The measured value is consistent with carbon dioxide.

However, other gases can have similar molar masses. For example, nitrous oxide, \(\ce{N2O}\), also has a molar mass close to \(44.01\ \mathrm{g\,mol^{-1}}\).

Therefore, the mass and gas conditions allow us to determine the molar mass, but additional chemical evidence would be needed to establish the identity.

The important distinction is that the volume, pressure, and temperature determine the amount of gas under the ideal-gas model. The measured mass then allows us to calculate its molar mass.

### Question 4

Two identical rigid gas bulbs, A and B, each have a volume of \(1.00\ \mathrm{L}\).

Both are maintained at \(25^\circ\mathrm{C}\) and an absolute pressure of \(100\ \mathrm{kPa}\).

- Bulb A contains pure carbon dioxide.
- Bulb B contains a mixture of helium and carbon dioxide.

The measured mass of gas in bulb B is \(60.0\%\) of the gas mass in bulb A.

Assume ideal-gas behaviour.

Calculate:

a) The average molar mass of the gas mixture in bulb B.

b) The mole fraction of helium in bulb B.

c) The amount of helium and carbon dioxide in bulb B.

Use:

- \(M(\ce{He})=4.00\ \mathrm{g\,mol^{-1}}\)
- \(M(\ce{CO2})=44.01\ \mathrm{g\,mol^{-1}}\)

[IMAGE: Two identical sealed 1.00 L gas bulbs at the same temperature and pressure. Bulb A contains only CO2 molecules. Bulb B contains a mixture of He atoms and CO2 molecules, with the proportions deliberately unspecified. Include a balance comparison indicating that bulb B has 60.0% of the gas mass of bulb A. Show the identical bulb dimensions and condition labels without revealing the mixture composition.]

### Solution 4

The helium mole fraction is **0.440**, meaning helium makes up \(44.0\%\) of the gas particles in bulb B.

The key is that equal volumes at the same temperature and pressure contain equal total amounts of gas, but their average molar masses can differ.

**Step 1**: Establish that both bulbs contain the same amount of gas.

Since both bulbs have identical volume, pressure, and temperature, the ideal gas equation gives the same amount for each.

\[
n=\frac{PV}{RT}
\]

\[
\begin{aligned}
n&=\frac{100\times1.00}{8.314\times298}\\
&=0.0404\ \mathrm{mol}
\end{aligned}
\]

Therefore:

\[
n_A=n_B=0.0404\ \mathrm{mol}
\]

**Step 2**: Calculate the mass of gas in bulb A.

Bulb A contains pure carbon dioxide.

\[
m_A=nM
\]

\[
\begin{aligned}
m_A&=0.0404\times44.01\\
&=1.78\ \mathrm{g}
\end{aligned}
\]

**Step 3**: Calculate the mass of gas in bulb B.

The mass in B is \(60.0\%\) of the mass in A.

\[
m_B=0.600\times1.78
\]

\[
m_B=1.07\ \mathrm{g}
\]

**Step 4**: Calculate the average molar mass of the mixture.

Use:

\[
M_{\text{average}}=\frac{m_B}{n_B}
\]

\[
\begin{aligned}
M_{\text{average}}
&=\frac{1.07}{0.0404}\\
&=26.4\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

This is the average mass per mole of gas particles in bulb B.

**Step 5**: Calculate the mole fraction of helium.

Let \(x\) represent the mole fraction of helium.

The remaining fraction, \(1-x\), represents carbon dioxide.

The average molar mass is the weighted average of the two molar masses:

\[
26.4=4.00x+44.01(1-x)
\]

Expand:

\[
26.4=4.00x+44.01-44.01x
\]

Rearrange:

\[
44.01-26.4=40.01x
\]

\[
x=\frac{17.61}{40.01}
\]

\[
x=0.440
\]

Therefore:

\[
\text{Helium mole fraction}=0.440
\]

\[
\text{CO2 mole fraction}=1-0.440=0.560
\]

**Step 6**: Calculate the amount of each gas.

For helium:

\[
\begin{aligned}
n(\ce{He})&=0.440\times0.0404\\
&=0.0178\ \mathrm{mol}
\end{aligned}
\]

For carbon dioxide:

\[
\begin{aligned}
n(\ce{CO2})&=0.560\times0.0404\\
&=0.0226\ \mathrm{mol}
\end{aligned}
\]

The amounts add to the total:

\[
0.0178+0.0226=0.0404\ \mathrm{mol}
\]

This is an important result. Helium makes up \(44.0\%\) of the particles but contributes only about \(6.7\%\) of the mixture's mass.

The tempting mistake is to interpret \(60.0\%\) of the mass as \(60.0\%\) of the particles. That does not work because helium and carbon dioxide have very different molar masses.

### Question 5

A rigid, sealed vessel initially contains:

- \(1.00\ \mathrm{mol}\) of carbon monoxide, \(\ce{CO}\).
- \(0.500\ \mathrm{mol}\) of oxygen gas, \(\ce{O2}\).

The gases react completely when ignited according to:

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

Assume:
- The reaction proceeds to completion.
- No gas escapes from the vessel.
- All gases behave ideally.
- All substances remain gaseous.
- The vessel returns to its original temperature after the reaction.

A student argues that because mass is conserved, the final pressure must equal the initial pressure.

Evaluate this claim.

Calculate:

a) The total initial mass of gas.

b) The total final mass of gas.

c) The total amount of gas particles before and after the reaction.

d) The final pressure as a fraction of the initial pressure.

e) Explain why the mass remains constant even though the number of gas particles changes.

Use:

- \(M(\ce{CO})=28.01\ \mathrm{g\,mol^{-1}}\)
- \(M(\ce{O2})=32.00\ \mathrm{g\,mol^{-1}}\)
- \(M(\ce{CO2})=44.01\ \mathrm{g\,mol^{-1}}\)

[IMAGE: A rigid, sealed reaction vessel containing an initial mixture of CO molecules and O2 molecules, with an ignition source shown outside the vessel. Clearly show the fixed vessel volume, sealed walls, and initial gas composition. Do not show the final products or the final particle count in the question illustration.]

### Solution 5

The student's claim is incorrect. **The final mass is unchanged, but the final pressure is two-thirds of the initial pressure.**

This is the key conceptual distinction: mass is conserved during a chemical reaction, but the total number of gas molecules does not have to be conserved.

**Step 1**: Calculate the initial mass.

The initial carbon monoxide mass is:

\[
\begin{aligned}
m(\ce{CO})&=1.00\times28.01\\
&=28.01\ \mathrm{g}
\end{aligned}
\]

The initial oxygen mass is:

\[
\begin{aligned}
m(\ce{O2})&=0.500\times32.00\\
&=16.00\ \mathrm{g}
\end{aligned}
\]

Therefore, the total initial mass is:

\[
m_{\text{initial}}=28.01+16.00
\]

\[
m_{\text{initial}}=44.01\ \mathrm{g}
\]

**Step 2**: Determine the amount of carbon dioxide produced.

The balanced equation is:

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

The mole ratio is:

\[
2:1:2
\]

For every two moles of carbon monoxide, one mole of oxygen produces two moles of carbon dioxide.

The initial amounts are exactly in the required stoichiometric ratio.

Therefore, the reaction produces:

\[
n(\ce{CO2})=1.00\ \mathrm{mol}
\]

**Step 3**: Calculate the final mass.

\[
\begin{aligned}
m_{\text{final}}
&=1.00\times44.01\\
&=44.01\ \mathrm{g}
\end{aligned}
\]

The initial and final masses are equal:

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

\[
44.01\ \mathrm{g}=44.01\ \mathrm{g}
\]

Mass is conserved.

**Step 4**: Compare the total amounts of gas particles.

Before the reaction:

\[
\begin{aligned}
n_{\text{initial}}
&=1.00+0.500\\
&=1.50\ \mathrm{mol}
\end{aligned}
\]

After the reaction:

\[
n_{\text{final}}=1.00\ \mathrm{mol}
\]

The total amount of gas particles has decreased.

Before the reaction, the vessel contained:

\[
N_{\text{initial}}
=1.50\times6.022\times10^{23}
\]

\[
N_{\text{initial}}=9.03\times10^{23}
\]

After the reaction:

\[
N_{\text{final}}
=1.00\times6.022\times10^{23}
\]

\[
N_{\text{final}}=6.02\times10^{23}
\]

There are fewer gas molecules after the reaction, even though the total mass has not changed.

**Step 5**: Calculate the final pressure.

The ideal gas equation is:

\[
PV=nRT
\]

Rearranging:

\[
P=\frac{nRT}{V}
\]

The vessel has the same volume and is returned to the same temperature.

Therefore, \(R\), \(T\), and \(V\) are unchanged, meaning pressure is directly proportional to the total amount of gas.

\[
\frac{P_f}{P_i}=\frac{n_f}{n_i}
\]

Substitute:

\[
\begin{aligned}
\frac{P_f}{P_i}
&=\frac{1.00}{1.50}\\
&=\frac{2}{3}
\end{aligned}
\]

Therefore:

\[
P_f=\frac{2}{3}P_i
\]

The final pressure is two-thirds of the initial pressure.

[IMAGE: Explanatory before-and-after particle diagram of the sealed rigid vessel. Before: two CO molecules for every one O2 molecule, representing the 1.00 mol CO and 0.500 mol O2 mixture. After: two CO2 molecules formed from the original three gas molecules, showing the 2:1 change in total gas particle count. Use accurate molecular structures and keep the vessel volume identical in both states. Label the total particle amounts as 1.50 mol before and 1.00 mol after, and show that the same total number of each type of atom is present.]

**Step 6**: Explain why the mass is conserved but the number of molecules changes.

A chemical reaction rearranges atoms. It does not create or destroy the atoms themselves.

Before the reaction:

- \(1.00\ \mathrm{mol}\) of \(\ce{CO}\) contains \(1.00\ \mathrm{mol}\) of carbon atoms and \(1.00\ \mathrm{mol}\) of oxygen atoms.
- \(0.500\ \mathrm{mol}\) of \(\ce{O2}\) contains another \(1.00\ \mathrm{mol}\) of oxygen atoms.

The total is:

\[
1.00\ \mathrm{mol\ C\ atoms}
\]

\[
2.00\ \mathrm{mol\ O\ atoms}
\]

After the reaction, \(1.00\ \mathrm{mol}\) of \(\ce{CO2}\) contains exactly:

- \(1.00\ \mathrm{mol}\) of carbon atoms.
- \(2.00\ \mathrm{mol}\) of oxygen atoms.

The total number of atoms is unchanged.

However, the atoms have been rearranged into a different number of molecules.

Initially, there were \(1.50\ \mathrm{mol}\) of gas molecules. Afterwards, there is only \(1.00\ \mathrm{mol}\) of gas molecules.

This is why the pressure changes even though the mass remains constant.

The tempting assumption was that unchanged mass must mean unchanged pressure. But pressure in this situation depends on the amount of gas particles, as well as temperature and volume, not directly on the total mass.

## Where this connects to stoichiometry

The difference between mass, moles, and particle numbers becomes essential when you start solving more complicated stoichiometry problems.

A balanced chemical equation gives the mole ratios between reactants and products. It does not give their mass ratios directly.

To move between them, you need the molar mass of each substance:

\[
\text{Mass}\longleftrightarrow
\text{Moles}\longleftrightarrow
\text{Particles}
\]

When a question gives you a mass, convert it into moles before using the balanced equation. When it asks for a particle count, specify whether you need atoms, molecules, ions, or formula units.

That is the decision that prevents most equal-mole and unequal-mass mistakes in HSC Chemistry.