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Mole Concept and Stoichiometry

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Step 1: Atoms, Molecules, and How We ‘Weigh’ Them

Everything around you—the air you breathe, the water you drink, the device you’re reading this on—is made of incredibly tiny particles called atoms. Think of an atom as the smallest possible Lego brick of an element, like Hydrogen (H), Carbon (C), or Oxygen (O).

When two or more atoms join together chemically, they form a molecule. It’s like clipping a few Lego bricks together. For example, a water molecule (H₂O) is made of two hydrogen atoms and one oxygen atom.

The Problem of Weighing an Atom

Atoms are unbelievably small and light. Using kilograms or grams to weigh them is like trying to weigh a single grain of sand with a scale meant for trucks! A single carbon atom, for example, weighs about 0.0000000000000000000000199 grams. This number is impossible to work with.

The Clever Solution: Relative Atomic Mass (RAM)

To solve this, scientists created a special “relative” scale. They didn’t use grams; they created a new, tiny unit.

  1. They took one specific atom, Carbon-12, and decided to fix its mass at exactly 12 units.
  2. This unit is called the atomic mass unit (amu).
  3. Then, they measured all other atoms relative to this standard.

Analogy: The Fruit Stall Imagine a fruit seller decides that a standard mango weighs exactly 1 kg. Now he can weigh all other fruits relative to his standard mango.

  • An apple might weigh 0.2 kg (it’s 0.2 times as heavy as the standard mango).
  • A watermelon might weigh 5 kg (it’s 5 times as heavy as the standard mango).

The Relative Atomic Mass (RAM) of an element tells us how heavy its atom is, on average, compared to 1/12th of a Carbon-12 atom.

For our chemistry, we can simply think of it as:

  • Hydrogen (H) has a RAM of 1 amu. It’s the lightest.
  • Carbon (C) has a RAM of 12 amu (our standard).
  • Oxygen (O) has a RAM of 16 amu (meaning an oxygen atom is roughly 16 times heavier than a hydrogen atom).

So, when you see C = 12, O = 16, you’re looking at the Relative Atomic Masses of those elements.

Practice Questions (Step 1):

  1. What is the standard element and its specific isotope used to define the atomic mass unit?
  2. If the RAM of Sodium (Na) is 23 amu, what does that mean in simple terms compared to Hydrogen (RAM = 1 amu)?
  3. What is the difference between an atom of hydrogen (H) and a molecule of hydrogen (H₂)?
  4. The RAM of Magnesium (Mg) is 24 and the RAM of Carbon (C) is 12. How many times heavier is a magnesium atom than a carbon atom?
  5. What do we call the tiny unit of mass used to express the weight of individual atoms and molecules?

Answers to Practice Questions (Step 1):

  1. The standard used is the Carbon-12
  2. It means that a single atom of Sodium is approximately 23 times heavier than a single atom of Hydrogen.
  3. An atom of hydrogen (H) is a single, individual particle of hydrogen. A molecule of hydrogen (H₂) consists of two hydrogen atoms chemically bonded together.
  4. A magnesium atom (24 amu) is twice as heavy as a carbon atom (12 amu), since 24 / 12 = 2.
  5. The unit is the atomic mass unit (amu).

Excellent work! You’ve grasped the fundamental concept of how we measure the mass of single atoms. Now, let’s move from individual atoms to the molecules they form.

Step 2: Calculating Molecular Mass

If we know the mass of each Lego brick, how do we find the mass of a structure we build? Simple — we just add up the masses of all the bricks we used! We do the exact same thing in chemistry.

The Relative Molecular Mass (RMM) of a compound is the sum of the relative atomic masses of all the atoms present in one molecule of that compound. It’s also measured in amu.

How to Calculate RMM

Let’s calculate the RMM for a water molecule, H₂O.

The formula H₂O tells us it contains:

  • 2 atoms of Hydrogen (H)
  • 1 atom of Oxygen (O)

Now we use the RAM values we learned: H = 1 amu, O = 16 amu.

  1. Mass from Hydrogen: 2 atoms × 1 amu/atom = 2 amu
  2. Mass from Oxygen: 1 atom × 16 amu/atom = 16 amu
  3. Total Mass (RMM): 2 amu + 16 amu = 18 amu

So, the Relative Molecular Mass of one water molecule is 18 amu. It’s that easy!

Another Example: Sulphuric Acid (H₂SO₄)

This is a bigger molecule, but the method is the same. The formula H₂SO₄ tells us we have:

  • 2 atoms of Hydrogen (H)
  • 1 atom of Sulphur (S)
  • 4 atoms of Oxygen (O)

Given RAMs: H = 1, S = 32, O = 16.

  1. Mass from H: 2 × 1 = 2 amu
  2. Mass from S: 1 × 32 = 32 amu
  3. Mass from O: 4 × 16 = 64 amu
  4. Total Mass (RMM): 2 + 32 + 64 = 98 amu

A Quick Note: “Formula Mass”

Sometimes you’ll hear the term Formula Mass. It’s used for ionic compounds like table salt (NaCl), which form large crystal structures instead of individual molecules.

The good news: For all our calculations, “Formula Mass” is calculated in the exact same way as “Molecular Mass”. So, you don’t need to worry about any difference in the math.

  • Formula Mass of NaCl: (Na = 23, Cl = 35.5) = 23 + 35.5 = 58.5 amu.

Practice Questions (Step 2):

  1. Calculate the Relative Molecular Mass (RMM) of methane (CH₄). [C=12, H=1]
  2. Calculate the Formula Mass of calcium carbonate (CaCO₃), a compound found in chalk and marble. [Ca=40, C=12, O=16]
  3. What is the RMM of glucose (C₆H₁₂O₆)? [C=12, H=1, O=16]
  4. Calculate the RMM of ammonia (NH₃). [N=14, H=1]
  5. Find the Formula Mass of copper sulphate (CuSO₄). [Cu=64, S=32, O=16]

Answers to Practice Questions (Step 2):

  1. Methane (CH₄): (1 × 12) + (4 × 1) = 12 + 4 = 16 amu.
  2. Calcium Carbonate (CaCO₃): (1 × 40) + (1 × 12) + (3 × 16) = 40 + 12 + 48 = 100 amu.
  3. Glucose (C₆H₁₂O₆): (6 × 12) + (12 × 1) + (6 × 16) = 72 + 12 + 96 = 180 amu.
  4. Ammonia (NH₃): (1 × 14) + (3 × 1) = 14 + 3 = 17 amu.
  5. Copper Sulphate (CuSO₄): (1 × 64) + (1 × 32) + (4 × 16) = 64 + 32 + 64 = 160 amu.

Perfect! You can now find the mass of any molecule. Let’s use that skill to answer a very practical question: If you have a water molecule, what percentage of its weight is just from oxygen? This is called Percentage Composition.

Step 3: Calculating Percentage Composition

Percentage composition tells us what percentage of a compound’s total mass comes from each element within it.

Analogy: A Fruit Salad Imagine you have a fruit salad that weighs 500 grams.

  • It contains 300g of apples and 200g of bananas.
  • The percentage of apples is (300 / 500) × 100 = 60%.
  • The percentage of bananas is (200 / 500) × 100 = 40%.

We do the exact same thing with the masses of atoms in a molecule.

The formula is: % of an element = (Total mass of that element in the formula / Total formula mass) × 100

Example: Water (H₂O)

Let’s find the percentage composition of hydrogen and oxygen in water. (RAM: H=1, O=16)

Step A: Find the total formula mass (RMM) of H₂O.

  • RMM = (2 × 1) + (1 × 16) = 18 amu.

Step B: Find the percentage of each element.

  • Percentage of Hydrogen (H):
    • Total mass of H in the formula = 2 amu.
    • % H = (2 / 18) × 100 = 11.11%
  • Percentage of Oxygen (O):
    • Total mass of O in the formula = 16 amu.
    • % O = (16 / 18) × 100 = 88.89%

Verification Tip: The percentages should always add up to 100%. 11.11% + 88.89% = 100%. It works!

Trickier Example: Calcium Nitrate [Ca(NO₃)₂]

What is the percentage of nitrogen in calcium nitrate? (RAM: Ca=40, N=14, O=16)

Step A: Find the total formula mass.

  • Be careful with the brackets! The ₂ outside the bracket multiplies everything inside it.
  • Ca: 1 × 40 = 40
  • N: 2 × 14 = 28
  • O: 2 × 3 × 16 = 96
  • Total Mass = 40 + 28 + 96 = 164 amu.

Step B: Find the percentage of Nitrogen (N).

  • Total mass of N in the formula = 28 amu.
  • % N = (Total mass of N / Total formula mass) × 100
  • % N = (28 / 164) × 100 ≈ 17.07%.

Practice Questions (Step 3):

  1. Calculate the percentage by mass of carbon in carbon dioxide (CO₂). [C=12, O=16]
  2. Find the percentage composition of each element in ammonia (NH₃). [N=14, H=1]
  3. Which compound is a better nitrogenous fertilizer for crops (i.e., has a higher percentage of nitrogen): sodium nitrate (NaNO₃) or ammonium sulphate [(NH₄)₂SO₄]? [Na=23, N=14, O=16, S=32, H=1]
  4. Calculate the percentage of oxygen in magnesium nitrate crystals, Mg(NO₃)₂.6H₂O. [Mg=24, N=14, O=16, H=1]
  5. What is the percentage by mass of calcium in calcium phosphate [Ca₃(PO₄)₂]? [Ca=40, P=31, O=16]

Answers to Practice Questions (Step 3):

  1. Carbon in CO₂: The total mass of CO₂ is 12 + (2×16) = 44 amu. The percentage of Carbon is (12 / 44) × 100 = 27%.
  2. Ammonia (NH₃): The total mass is 14 + (3×1) = 17 amu.
    • % Nitrogen = (14 / 17) × 100 = 35%.
    • % Hydrogen = (3 / 17) × 100 = 65%.
  3. Better Fertilizer:
    • % N in NaNO₃ (mass 85) = (14 / 85) × 100 = 16.47%.
    • % N in (NH₄)₂SO₄ (mass 132) = (28 / 132) × 100 = 21.21%.
    • Ammonium sulphate [(NH₄)₂SO₄] is better because it has a higher percentage of nitrogen.
  4. Oxygen in Mg(NO₃)₂.6H₂O: The total mass is 256 amu. There are 12 oxygen atoms in total (6 from nitrate and 6 from water), with a combined mass of 12 × 16 = 192 amu.
    • % Oxygen = (192 / 256) × 100 = 75%.
  5. Calcium in Ca₃(PO₄)₂: The total mass is (3×40) + (2×31) + (8×16) = 310 amu. The mass of Calcium is 3 × 40 = 120 amu.
    • % Calcium = (120 / 310) × 100 = 71%.

You’re an expert at analyzing the composition of molecules now. Get ready for the next step, where we learn the single most important concept in all of chemistry. It’s a special number that acts as a bridge between the tiny world of atoms and the real world of grams that we can actually measure. This concept is called the mole.

Step 4: The Mole – A Chemist’s Dozen

In everyday life, we use words to represent numbers. For example:

  • A pair = 2 items
  • A dozen = 12 items
  • A century = 100 items

These words make it easier to talk about quantities. A chemist faces a similar problem. Atoms and molecules are so small that any sample we can see or weigh contains a gigantic number of them. Counting them would be impossible. So, chemists came up with their own counting word: the mole (abbreviated as mol).

A mole is simply a word that stands for a very, very big number: 602,200,000,000,000,000,000,000

In scientific notation, this is written as 6.022 × 10²³. This massive number is called Avogadro’s Number (or Avogadro’s Constant), in honour of the scientist Amedeo Avogadro.

So, the mole is the chemist’s dozen. It’s a counting unit.

  • 1 dozen eggs is 12 eggs.
  • 1 mole of carbon atoms is 6.022 × 10²³ carbon atoms.
  • 1 mole of water molecules is 6.022 × 10²³ water molecules.
  • 2 moles of iron atoms is 2 × (6.022 × 10²³) iron atoms.

That’s it! For now, just think of “mole” as a word for this specific, giant number. It’s the standard way we count particles in chemistry.

(Why this strange number? It wasn’t chosen randomly! It was cleverly defined as the number of atoms present in exactly 12 grams of Carbon-12. This creates a magical link between the tiny ‘amu’ and the real-world ‘gram’, which we will explore in the very next step!)

Practice Questions (Step 4):

  1. What name is given to the number 6.022 × 10²³?
  2. How many molecules are present in 1 mole of carbon dioxide (CO₂)?
  3. How many atoms are present in 3 moles of helium (He) gas?
  4. A dozen is to 12 as a mole is to what? (Write the number in scientific notation).
  5. How many molecules are in 0.5 moles of water (H₂O)? (You can use 6 × 10²³ for Avogadro’s number for this calculation).

Answers to Practice Questions (Step 4):

  1. The number 6.022 × 10²³ is called Avogadro’s Number or Avogadro’s Constant.
  2. There are 022 × 10²³ molecules in 1 mole of carbon dioxide.
  3. There are 3 × (6.022 × 10²³) = 066 × 10²³ atoms in 3 moles of helium.
  4. A dozen is to 12 as a mole is to 022 × 10²³.
  5. There are 0.5 × (6 × 10²³) = 3 × 10²³ molecules in 0.5 moles of water.

You’ve correctly understood that a mole is just a specific, giant number. Now for the magical part. Why did scientists pick such a specific, strange number? They did it to create a perfect bridge between the tiny world of amu and the real world of grams. This bridge is called Molar Mass.

Step 5: Molar Mass – Connecting Worlds

We know that weighing a single atom in grams is impractical. But we can easily weigh a few grams of a substance in a lab. The mole is the concept that connects these two scales.

Molar Mass is defined as the mass of one mole of a substance. The unit for Molar Mass is grams per mole (g/mol).

The Magical Connection

Here is the most useful fact in all of basic chemistry:

The Relative Molecular Mass (RMM) of a substance in amu is numerically the same as its Molar Mass in grams/mole.

This means all the hard work we did in Step 2 to calculate RMM now has a real-world use!

Let’s see it in action:

  • For Carbon (C):
    • The mass of 1 single atom of Carbon is 12 amu.
    • The mass of 1 mole (6.022 × 10²³ atoms) of Carbon is 12 grams.
    • The Molar Mass of Carbon is 12 g/mol.
  • For Water (H₂O):
    • The mass of 1 single molecule of water is 18 amu.
    • The mass of 1 mole (6.022 × 10²³ molecules) of water is 18 grams.
    • The Molar Mass of water is 18 g/mol.
  • For Sulphuric Acid (H₂SO₄):
    • The mass of 1 single molecule of H₂SO₄ is 98 amu.
    • The mass of 1 mole (6.022 × 10²³ molecules) of H₂SO₄ is 98 grams.
    • The Molar Mass of H₂SO₄ is 98 g/mol.

So, to find the Molar Mass of any compound, you do the exact same calculation as for RMM, you just change the units from ‘amu’ to ‘g/mol’.

Analogy: Think of a dozen eggs. A dozen small eggs might weigh 600g, while a dozen large eggs might weigh 800g. The number of eggs (12) is the same, but the total mass is different. Similarly, one mole of any substance always has 6.022 × 10²³ particles, but its weight in grams (the Molar Mass) depends on how heavy each individual particle is.

Practice Questions (Step 5):

  1. What is the Molar Mass of nitrogen gas (N₂)? [N=14]
  2. If the Relative Molecular Mass of a substance is 44 amu, what is the mass of one mole of that substance in grams?
  3. What are the standard units used to express Molar Mass?
  4. Calculate the Molar Mass of glucose (C₆H₁₂O₆). [C=12, H=1, O=16]
  5. Based on what you’ve learned, what is the mass of 6.022 × 10²³ molecules of ammonia (NH₃)? [N=14, H=1]

Answers to Practice Questions (Step 5):

  1. Molar Mass of N₂: The mass is 2 × 14 = 28. The units for Molar Mass are g/mol. So, the Molar Mass is 28 g/mol.
  2. Mass of one mole: The mass of one mole of a substance is its Molar Mass. If the RMM is 44 amu, the Molar Mass is 44 g/mol, meaning one mole has a mass of 44 grams.
  3. Units of Molar Mass: The standard units are grams per mole (g/mol).
  4. Molar Mass of Glucose (C₆H₁₂O₆): The calculation is (6 × 12) + (12 × 1) + (6 × 16) = 72 + 12 + 96 = 180. The Molar Mass is 180 g/mol.
  5. Mass of 6.022 × 10²³ molecules of NH₃: Since 6.022 × 10²³ is the number of molecules in one mole, the question is simply asking for the Molar Mass of NH₃. Molar Mass = 14 + (3 × 1) = 17 grams.

You’ve made the crucial connection between the atomic and the real world! Now, let’s put this powerful tool to use. This step is about the single most common and important calculation in all of chemistry: converting between the mass of a substance and the number of moles it contains.

Step 6: Mole-Mass Conversions

We now have the key conversion factor: the Molar Mass (M). Think of it as a bridge that lets you travel between the land of ‘Grams’ and the land of ‘Moles’.

There are two simple formulas that form the foundation of almost all chemical calculations.

The Two Key Formulas

  1. To find the number of moles (n) when you know the mass: Number of Moles = Mass (in grams) / Molar Mass (in g/mol)
  2. To find the mass (in grams) when you know the number of moles: Mass = Number of Moles × Molar Mass

Let’s see them in action.

Example 1: Converting Grams to Moles

Problem: How many moles of Carbon Dioxide (CO₂) are there in 11 grams of the gas? (C=12, O=16)

Step A: Find the Molar Mass of CO₂.

  • Molar Mass (M) = 12 + (2 × 16) = 12 + 32 = 44 g/mol.

Step B: Use the formula.

  • Number of Moles (n) = Mass / Molar Mass
  • n = 11 g / 44 g/mol = 0.25 moles.

So, 11 grams of CO₂ is a quarter of a mole.

Example 2: Converting Moles to Grams

Problem: What is the mass of 3 moles of Sodium Hydroxide (NaOH)? (Na=23, O=16, H=1)

Step A: Find the Molar Mass of NaOH.

  • Molar Mass (M) = 23 + 16 + 1 = 40 g/mol.

Step B: Use the formula.

  • Mass = Number of Moles × Molar Mass
  • Mass = 3 mol × 40 g/mol = 120 grams.

So, to get 3 moles of NaOH for an experiment, you would need to weigh out 120 grams.

Practice Questions (Step 6):

  1. Calculate the number of moles in 9 grams of water (H₂O). [H=1, O=16]
  2. What is the mass of 0.5 moles of Sulphuric Acid (H₂SO₄)? [H=1, S=32, O=16]
  3. How many moles of pure gold (Au) are in a 394-gram gold bar? [Au=197]
  4. Find the mass of 1.5 moles of methane gas (CH₄). [C=12, H=1]
  5. Which contains more moles: 32 grams of Methane (CH₄) or 32 grams of Oxygen (O₂)?

Answers to Practice Questions (Step 6):

  1. Moles in 9g of H₂O: The Molar Mass of water (H₂O) is 18 g/mol. Number of Moles = 9 g / 18 g/mol = 5 moles.
  2. Mass of 0.5 moles of H₂SO₄: The Molar Mass of sulphuric acid (H₂SO₄) is 98 g/mol. Mass = 0.5 mol × 98 g/mol = 49 grams.
  3. Moles in 394g of Au: The Molar Mass of gold (Au) is 197 g/mol. Number of Moles = 394 g / 197 g/mol = 2 moles.
  4. Mass of 1.5 moles of CH₄: The Molar Mass of methane (CH₄) is 16 g/mol. Mass = 1.5 mol × 16 g/mol = 24 grams.
  5. Which has more moles?:
    • Moles of CH₄ = 32 g / 16 g/mol = 2 moles.
    • Moles of O₂ = 32 g / 32 g/mol = 1 mole.
    • 32 grams of Methane (CH₄) contains more moles.

Fantastic! You’ve mastered the link between moles and mass. Now, let’s connect the mole back to its original meaning: a giant counting number. This step is all about converting between moles and the actual number of atoms or molecules.

Step 7: Mole-Particle Conversions

Remember that the mole is just a word for a number: Avogadro’s Number (Nₐ), which is 6.022 × 10²³. We can use this number as a conversion factor to figure out exactly how many particles we’re dealing with. A “particle” can be an atom, a molecule, an ion, etc.

The Two Key Formulas

  1. To find the number of particles when you know the moles: Number of Particles = Number of Moles × Avogadro’s Number
  2. To find the number of moles when you know the number of particles: Number of Moles = Number of Particles / Avogadro’s Number

Example 1: Moles to Molecules

Problem: How many molecules are in 2.5 moles of ammonia (NH₃)?

Step A: Use the formula.

  • Number of Molecules = Moles × Nₐ
  • Number of Molecules = 2.5 × (6.022 × 10²³)
  • Number of Molecules = 15.055 × 10²³ molecules (or 1.5055 × 10²⁴ molecules).

A Deeper Level: Finding the Number of Atoms

This is a very important skill that builds on the last example.

Problem: How many atoms are there in total in 2.5 moles of ammonia (NH₃)?

Step A: First, find the number of molecules (we just did this).

  • Number of NH₃ molecules = 15.055 × 10²³.

Step B: Find how many atoms are in ONE molecule.

  • One molecule of NH₃ contains 1 Nitrogen atom + 3 Hydrogen atoms = 4 atoms in total.

Step C: Multiply to find the total number of atoms.

  • Total Atoms = (Number of molecules) × (atoms per molecule)
  • Total Atoms = (15.055 × 10²³) × 4 = 60.22 × 10²³ atoms (or 6.022 × 10²⁴ atoms).

So you see, you first find the number of molecules, and then multiply by the number of atoms inside each molecule.

Practice Questions (Step 7):

  1. How many molecules of water are there in 5 moles of water (H₂O)? (Use Nₐ = 6 × 10²³)
  2. If a sample of iron (Fe) contains 1.8 × 10²⁴ atoms, how many moles of iron are in the sample? (Use Nₐ = 6 × 10²³)
  3. How many atoms of oxygen are present in 2 moles of oxygen molecules (O₂)?
  4. Calculate the total number of atoms in 0.5 moles of methane (CH₄).
  5. Challenge: How many atoms of hydrogen are present in 18 grams of water (H₂O)? [H=1, O=16] (Hint: First convert grams to moles!)

Answers to Practice Questions (Step 7):

  1. Molecules in 5 moles of H₂O: Number of Molecules = 5 mol × (6 × 10²³) = 30 × 10²³ molecules (or 3 × 10²⁴ molecules).
  2. Moles in 1.8 × 10²⁴ atoms of Fe: Number of Moles = (1.8 × 10²⁴) / (6 × 10²³) = 3 moles.
  3. Atoms in 2 moles of O₂:
    • Number of O₂ molecules = 2 × (6.022 × 10²³) = 12.044 × 10²³ molecules.
    • Total O atoms = (12.044 × 10²³) × 2 = 088 × 10²³ atoms.
  4. Total atoms in 0.5 moles of CH₄:
    • Number of CH₄ molecules = 0.5 × (6.022 × 10²³) = 3.011 × 10²³ molecules.
    • One molecule has 1 C + 4 H = 5 atoms.
    • Total Atoms = (3.011 × 10²³) × 5 = 055 × 10²³ atoms.
  5. Challenge Answer:
    • First, convert mass to moles: Molar Mass of H₂O is 18 g/mol. Moles = 18 g / 18 g/mol = 1 mole.
    • Second, find the number of molecules: 1 mole of H₂O contains 6.022 × 10²³ molecules.
    • Finally, find H atoms: Each molecule has 2 H atoms. Total H atoms = (6.022 × 10²³) × 2 = 044 × 10²³ atoms.

You’ve brilliantly connected the mole to mass and particle count. Now, we’ll explore the third and final connection: the relationship between the mole and the volume of a gas. This involves a very famous law and another magic number for you to learn.

Step 8: Avogadro’s Law and the Molar Volume of Gases

The Italian scientist Amedeo Avogadro, the same person for whom we named Avogadro’s Number, proposed a revolutionary idea about gases.

Avogadro’s Law states that:

Equal volumes of all gases, under the same conditions of temperature and pressure, contain an equal number of molecules.

Analogy: The Identical Boxes 📦 Imagine you have three identical, empty boxes. The fact they are identical means they have the same volume. Let’s say you keep them in the same room, so they have the same temperature and pressure.

Avogadro’s Law says that if you fill one box completely with light hydrogen gas, one with heavier oxygen gas, and one with very heavy carbon dioxide gas, all three boxes will hold the exact same number of molecules.

This is a unique property of gases. It doesn’t work for solids or liquids!

The Molar Volume of a Gas

Avogadro’s law leads to a very useful conclusion. If equal volumes have an equal number of molecules, then a fixed number of molecules must always occupy the same volume.

What is our favourite fixed number of molecules? One mole (6.022 × 10²³).

This means that one mole of ANY gas will occupy the same volume, as long as we keep the temperature and pressure the same. To make it easy for scientists to compare results, they defined a set of standard conditions.

STP = Standard Temperature and Pressure

  • Standard Temperature: 0° Celsius (273 Kelvin)
  • Standard Pressure: 1 atmosphere (atm)

At these specific conditions (STP), scientists measured the volume of one mole of many gases and found a magic number.

At STP, one mole of any gas occupies a volume of 22.4 litres.

This volume is called the Molar Volume. (Note: 1 litre is the same as 1 cubic decimetre, or dm³).

  • 1 mole of H₂ (mass = 2g) occupies 22.4 L at STP.
  • 1 mole of O₂ (mass = 32g) occupies 22.4 L at STP.
  • 1 mole of CO₂ (mass = 44g) occupies 22.4 L at STP.

Even though their masses are very different, they all take up the same amount of space at STP!

Practice Questions (Step 8):

  1. What is the volume occupied by 1 mole of helium gas at STP?
  2. What does Avogadro’s Law state about the number of molecules in equal volumes of different gases?
  3. What do the letters S, T, and P stand for, and what are their standard values?
  4. You have two identical balloons at the same temperature and pressure. One is filled with air, and the other is filled with hydrogen gas. Which balloon contains more gas molecules?
  5. What special name is given to the volume of 22.4 litres that one mole of a gas occupies at STP?

Answers to Practice Questions (Step 8):

  1. The volume occupied by 1 mole of any gas at STP is 4 litres.
  2. Avogadro’s Law states that equal volumes of all gases, under the same conditions of temperature and pressure, contain an equal number of molecules.
  3. STP stands for Standard Temperature and Pressure. The standard values are 0° Celsius (or 273 Kelvin) and 1 atmosphere.
  4. According to Avogadro’s Law, since the balloons have identical volumes and are at the same temperature and pressure, they contain the same number of molecules.
  5. The special name given to this volume is the Molar Volume.

Excellent! You’ve perfectly understood the concept of Molar Volume. Now, just as we did with mass and particles, let’s use our new magic number (22.4 L/mol) as a practical conversion factor. This step is all about converting between the volume of a gas at STP and the number of moles it contains.

Step 9: Mole-Volume Conversions for Gases

This is the third and final type of direct mole conversion. Remember, this only works for gases and only at STP.

The Molar Volume (22.4 L/mol) is our new bridge, connecting the land of ‘Volume at STP’ to the land of ‘Moles’.

The Two Key Formulas

  1. To find the number of moles (n) when you know the volume of a gas at STP: Number of Moles = Given Volume (in L) / 22.4 L/mol
  2. To find the volume of a gas at STP when you know the number of moles: Volume (in L) = Number of Moles × 22.4 L/mol

Example 1: Converting Volume to Moles

Problem: A balloon contains 5.6 litres of helium gas (He) at STP. How many moles of helium are in the balloon?

Step A: Use the formula.

  • Number of Moles (n) = Volume / 22.4
  • n = 5.6 L / 22.4 L/mol = 0.25 moles. (Tip: It’s useful to remember that 5.6 is 1/4 of 22.4, and 11.2 is 1/2 of 22.4. These numbers appear often in questions!)

Example 2: Converting Moles to Volume

Problem: What is the volume occupied by 2 moles of hydrogen gas (H₂) at STP?

Step A: Use the formula.

  • Volume = Number of Moles × 22.4
  • Volume = 2 mol × 22.4 L/mol = 44.8 litres.

Important Note on Units: The Molar Volume is in litres (L). A litre is the same as a cubic decimetre (dm³). If a question gives you a volume in millilitres (ml) or cubic centimetres (cm³), you must convert it to litres first by dividing by 1000.

  • For example, 500 cm³ = 0.5 L.

Practice Questions (Step 9):

  1. What is the volume occupied by 0.5 moles of carbon dioxide (CO₂) at STP?
  2. How many moles of gas are present in a 44.8 dm³ container of nitrogen (N₂) at STP?
  3. A small cylinder contains 2.24 L of oxygen (O₂) gas at STP. How many moles of oxygen are in the cylinder?
  4. Challenge: What is the volume occupied by 8 grams of methane gas (CH₄) at STP? [C=12, H=1] (Hint: First find the number of moles).
  5. Challenge: What is the mass of 5.6 litres of ammonia gas (NH₃) at STP? [N=14, H=1]

Answers to Practice Questions (Step 9):

  1. Volume of 0.5 moles of CO₂: Volume = Moles × 22.4 = 0.5 mol × 22.4 L/mol = 2 litres.
  2. Moles in 44.8 dm³ of N₂: Moles = Volume / 22.4 = 44.8 L / 22.4 L/mol = 2 moles.
  3. Moles in 2.24 L of O₂: Moles = Volume / 22.4 = 2.24 L / 22.4 L/mol = 1 moles.
  4. Challenge (Mass to Volume):
    • Step 1 (Mass → Moles): Molar Mass of CH₄ = 16 g/mol. Moles = 8 g / 16 g/mol = 0.5 moles.
    • Step 2 (Moles → Volume): Volume = 0.5 mol × 22.4 L/mol = 2 litres.
  5. Challenge (Volume to Mass):
    • Step 1 (Volume → Moles): Moles = 5.6 L / 22.4 L/mol = 0.25 moles.
    • Step 2 (Moles → Mass): Molar Mass of NH₃ = 17 g/mol. Mass = 0.25 mol × 17 g/mol = 25 grams.

You have now mastered the three bridges that connect the mole to the measurable world: mass, particles, and volume. For this next step, we’ll combine them into one powerful visual tool that will be your guide for solving almost any problem in this chapter.

Step 10: The Unified Mole Map

Think of the Mole as a central railway station. If you want to get from the city of ‘Grams’ to the city of ‘Litres’, you can’t take a direct train. You must first travel to the central ‘Mole’ station and then take a connecting train to your final destination.

This idea can be drawn as a simple map that is incredibly useful for solving problems.

How to Use the Mole Map

The map shows that to convert between any of the three outer boxes (Mass, Particles, Volume), you must always go through the Mole hub in the center. Let’s solve a problem using the map.

Problem: What is the volume occupied by 3.011 × 10²³ molecules of hydrogen gas (H₂) at STP?

  1. Identify your Start and End points:
    • Starting Point: Particles (3.011 × 10²³ molecules).
    • Destination: Volume (in Litres).
  2. Follow the Path on the Map: The path is Particles → Moles → Volume.
  3. Perform the Calculations:
    • First journey (Particles → Moles): The map says to “÷ Avogadro’s Number”.
      • Moles = (3.011 × 10²³) / (6.022 × 10²³) = 5 moles.
    • Second journey (Moles → Volume): The map says to “× 22.4 L”.
      • Volume = 0.5 mol × 22.4 L/mol = 2 litres.

The map turns a complex problem into a simple, two-step journey!

Practice Questions (Step 10):

  1. What is the mass of 44.8 litres of Sulphur Dioxide gas (SO₂) at STP? [S=32, O=16]
  2. How many molecules are present in 7 grams of Nitrogen gas (N₂)? [N=14, Nₐ=6×10²³]
  3. What is the volume at STP that would be occupied by 1.2 × 10²⁴ molecules of methane gas (CH₄)? [Nₐ=6×10²³]
  4. Using the mole map, calculate the mass of a single atom of Helium (He). [He=4, Nₐ=6×10²³] (Hint: the number of particles is 1).
  5. Which has a greater mass: 11.2 L of Hydrogen (H₂) at STP or 11.2 L of Nitrogen (N₂) at STP? Justify your answer with calculations.

Answers to Practice Questions (Step 10):

  1. Mass of 44.8 L of SO₂:
    • Path: Volume → Moles → Mass.
    • Moles = 44.8 L / 22.4 L/mol = 2 moles.
    • Molar Mass of SO₂ = 32 + (2×16) = 64 g/mol.
    • Mass = 2 mol × 64 g/mol = 128 grams.
  2. Molecules in 7g of N₂:
    • Path: Mass → Moles → Particles.
    • Molar Mass of N₂ = 28 g/mol. Moles = 7 g / 28 g/mol = 0.25 moles.
    • Molecules = 0.25 mol × (6 × 10²³) = 5 × 10²³ molecules.
  3. Volume of 1.2 × 10²⁴ molecules of CH₄:
    • Path: Particles → Moles → Volume.
    • Moles = (1.2 × 10²⁴) / (6 × 10²³) = 2 moles.
    • Volume = 2 mol × 22.4 L/mol = 8 litres.
  4. Mass of a single He atom:
    • Path: Particles → Moles → Mass. (Number of particles = 1).
    • Moles = 1 / (6 × 10²³) moles.
    • Molar Mass of He = 4 g/mol.
    • Mass = [1 / (6 × 10²³)] × 4 = 67 × 10⁻²⁴ grams (approx).
  5. Which has greater mass?:
    • At STP, equal volumes of gases have an equal number of moles. So, 11.2 L of H₂ is 0.5 moles, and 11.2 L of N₂ is also 0.5 moles.
    • Mass of H₂ = 0.5 mol × 2 g/mol = 1 gram.
    • Mass of N₂ = 0.5 mol × 28 g/mol = 14 grams.
    • 2 L of Nitrogen (N₂) has a greater mass.

You have now built the complete Mole Map, your master tool for conversions! Now, we’ll use these skills to become chemical detectives. Imagine a lab gives you the composition of an unknown substance. Your job is to figure out its chemical formula. The first step is to find the Empirical Formula.

Step 11: Empirical Formula – Finding the Simplest Ratio

What is an Empirical Formula? It is the simplest whole-number ratio of atoms of each element present in a compound.

Analogy: A Recipe’s Simplest Form The actual, true formula for the sugar glucose is C₆H₁₂O₆. The ratio of atoms is 6 (Carbon) : 12 (Hydrogen) : 6 (Oxygen). If you were simplifying this recipe, you could divide all numbers by 6 to get the simplest ratio: 1 : 2 : 1. So, the Empirical Formula for glucose is CH₂O.

The true formula (C₆H₁₂O₆) is called the Molecular Formula, which we’ll cover in the next step. For now, let’s learn the 4-step recipe to find the Empirical Formula from percentage composition data.

The 4-Step Method

Problem: An unknown compound is found to contain 40% Carbon, 6.67% Hydrogen, and 53.33% Oxygen by mass. Find its empirical formula. [C=12, H=1, O=16]

We’ll use a table to keep our work organized.

Element Step 1: Mass in 100g Step 2: Moles (Mass ÷ Molar Mass) Step 3: Divide by Smallest (3.33) Step 4: Simplest Ratio
Carbon (C) 40 g 40 g / 12 g/mol = 3.33 3.33 / 3.33 = 1 1
Hydrogen(H) 6.67 g 6.67 g / 1 g/mol = 6.67 6.67 / 3.33 ≈ 2 2
Oxygen (O) 53.33 g 53.33 g / 16 g/mol ≈ 3.33 3.33 / 3.33 = 1 1

Conclusion: The simplest ratio of atoms C : H : O is 1 : 2 : 1. Therefore, the Empirical Formula is CH₂O.

A special note on Step 4: If Step 3 gives you decimals like 1.5, 2.5, or 1.33, you must multiply ALL the ratios by a small whole number (like 2 or 3) to make them all whole numbers. For example, if you get a ratio of 1 : 1.5, you would multiply both by 2 to get the final whole-number ratio of 2 : 3.

Practice Questions (Step 11):

  1. What is the empirical formula of a compound whose molecular formula is C₄H₁₀ (butane)?
  2. A compound is made of 75% Carbon and 25% Hydrogen by mass. Find its empirical formula. [C=12, H=1]
  3. Find the empirical formula of a compound that contains 80% Copper and 20% Sulphur. [Cu=64, S=32]
  4. An iron ore is found to contain 70% Iron and 30% Oxygen by mass. What is its empirical formula? [Fe=56, O=16]
  5. What is the simplest whole-number ratio of atoms in a compound called?

Answers to Practice Questions (Step 11):

  1. Empirical Formula of C₄H₁₀: The ratio of atoms is 4:10. Dividing both by the greatest common divisor (2) gives the simplest ratio of 2:5. The empirical formula is C₂H₅.
  2. 75% C and 25% H:
    • Moles of C = 75 / 12 = 6.25. Moles of H = 25 / 1 = 25.
    • Dividing by the smallest (6.25): C = 1, H = 4.
    • The empirical formula is CH₄.
  3. 80% Cu and 20% S:
    • Moles of Cu = 80 / 64 = 1.25. Moles of S = 20 / 32 = 0.625.
    • Dividing by the smallest (0.625): Cu = 2, S = 1.
    • The empirical formula is Cu₂S.
  4. 70% Fe and 30% O:
    • Moles of Fe = 70 / 56 = 1.25. Moles of O = 30 / 16 = 1.875.
    • Dividing by the smallest (1.25): Fe = 1, O = 1.5.
    • Since we cannot have 1.5 atoms, we multiply both by 2 to get whole numbers. The ratio becomes 2:3.
    • The empirical formula is Fe₂O₃.
  5. The simplest whole-number ratio of atoms in a compound is called the Empirical Formula.

Perfect! You can now find the simplest formula. But what if the simplest formula isn’t the true formula? We know the simplest ratio for glucose is CH₂O, but the real molecule is C₆H₁₂O₆. How do we find the true formula? We need one more clue.

Step 12: Molecular Formula – Finding the True Formula

The Molecular Formula (MF) shows the actual number of atoms of each element in one molecule of a substance. It represents the true, complete molecule.

The Molecular Formula is always a simple whole-number multiple of the Empirical Formula.

Molecular Formula = (Empirical Formula)n where n is a whole number like 1, 2, 3, etc.

To find this number n, we need to compare the mass of the true molecule (the Molecular Mass) with the mass of our simple-ratio formula (the Empirical Formula Mass).

n = Molecular Mass / Empirical Formula Mass

The Complete Method

Problem: A compound made of carbon and hydrogen is found to be 92.3% Carbon. Its Relative Molecular Mass is 78 amu. Find its molecular formula. [C=12, H=1]

Step 1: Find the Empirical Formula (EF).

  • The compound is 92.3% C, so the rest must be Hydrogen: 100 – 92.3 = 7.7% H.
  • C: Moles = 92.3 / 12 ≈ 7.7.
  • H: Moles = 7.7 / 1 = 7.7.
  • Ratio: Divide by the smallest (7.7): C = 1, H = 1.
  • The Empirical Formula is CH.

Step 2: Calculate the Empirical Formula Mass.

  • Mass of CH = 12 + 1 = 13 amu.

Step 3: Find the value of ‘n’.

  • We are given that the true Molecular Mass is 78 amu.
  • n = Molecular Mass / Empirical Formula Mass
  • n = 78 / 13 = 6.

Step 4: Find the Molecular Formula (MF).

  • MF = (EF)n
  • MF = (CH)₆
  • MF = C₆H₆ The true formula of the compound is C₆H₆ (which is benzene).

Practice Questions (Step 12):

  1. The empirical formula of a compound is C₂H₅ and its molecular mass is 58 amu. What is its molecular formula? [C=12, H=1]
  2. A compound contains 40% Carbon, 6.67% Hydrogen, and 53.33% Oxygen. Its molecular mass is 60 amu. Find its molecular formula. [C=12, H=1, O=16]
  3. The simplest formula of a compound is P₂O₅. If its molecular mass is 284 amu, what is its molecular formula? [P=31, O=16]
  4. If a compound’s empirical formula is NO₂ and its molecular mass is 92 amu, find its molecular formula. [N=14, O=16]
  5. What two pieces of information are essential to determine a compound’s molecular formula if you only know its percentage composition?

Answers to Practice Questions (Step 12):

  1. Compound with EF C₂H₅:
    • Empirical Formula Mass = (2 × 12) + (5 × 1) = 29 amu.
    • n = Molecular Mass / EF Mass = 58 / 29 = 2.
    • Molecular Formula = (C₂H₅)₂ = C₄H₁₀.
  2. Compound with 40% C, 6.67% H, 53.33% O:
    • From Step 11, we know the Empirical Formula is CH₂O.
    • EF Mass = 12 + (2×1) + 16 = 30 amu.
    • n = Molecular Mass / EF Mass = 60 / 30 = 2.
    • Molecular Formula = (CH₂O)₂ = C₂H₄O₂ (Acetic Acid).
  3. Compound with EF P₂O₅:
    • EF Mass = (2 × 31) + (5 × 16) = 62 + 80 = 142 amu.
    • n = Molecular Mass / EF Mass = 284 / 142 = 2.
    • Molecular Formula = (P₂O₅)₂ = P₄O₁₀.
  4. Compound with EF NO₂:
    • EF Mass = 14 + (2 × 16) = 46 amu.
    • n = Molecular Mass / EF Mass = 92 / 46 = 2.
    • Molecular Formula = (NO₂)₂ = N₂O₄.
  5. To find the molecular formula from percentage composition, you need:
    1. The percentage composition (to find the empirical formula).
    2. The molecular mass (to find the multiplier n).

You’ve mastered finding the true formula of a compound. But what if you’re not given the Molecular Mass directly? In the 19th century, chemists found a clever way to determine a molecule’s mass by working with gases. This property is called Vapour Density.

Step 13: Vapour Density – A Clever Shortcut

Before modern machines, it was very hard to find the mass of a single molecule. However, it was much easier to weigh gases. Scientists found they could determine a molecule’s mass by comparing its gas form to the lightest gas of all: hydrogen.

Vapour Density (V.D.) is a ratio. It tells you how many times a gas is heavier than an equal volume of hydrogen gas (measured under the same conditions). V.D. = (Mass of ‘V’ litres of a gas) / (Mass of ‘V’ litres of hydrogen)

The Magic Formula

Through Avogadro’s Law, this definition simplifies to one of the most useful shortcuts in chemistry:

Relative Molecular Mass (RMM) = 2 × Vapour Density

This simple formula is a powerful tool. If you can measure the Vapour Density of an unknown gas, you can immediately find its Molecular Mass.

Example: The Vapour Density of carbon dioxide is measured to be 22.

  • What is its Molecular Mass?
  • RMM = 2 × V.D.
  • RMM = 2 × 22 = 44 amu. This matches our calculation for CO₂ (12 + 32 = 44). It works perfectly!

Using V.D. to Find a Molecular Formula

Now we can solve problems where the Molecular Mass isn’t given directly.

Problem: A gas has a Vapour Density of 15. It is found to contain 80% Carbon and 20% Hydrogen. Find its Molecular Formula.

Step A: Find the Molecular Mass from Vapour Density.

  • RMM = 2 × V.D. = 2 × 15 = 30 amu.

Step B: Find the Empirical Formula.

  • C: Moles = 80 / 12 = 6.67.
  • H: Moles = 20 / 1 = 20.
  • Ratio: Divide by smallest (6.67): C = 1, H = 3.
  • The Empirical Formula is CH₃.

Step C: Find the Molecular Formula.

  • Empirical Formula Mass = 12 + 3 = 15 amu.
  • n = Molecular Mass / EF Mass = 30 / 15 = 2.
  • Molecular Formula = (CH₃)₂ = C₂H₆.
  • The unknown gas is Ethane.

Practice Questions (Step 13):

  1. The Vapour Density of a gas is 32. What is its Relative Molecular Mass?
  2. Calculate the Vapour Density of ammonia gas (NH₃). [N=14, H=1]
  3. An unknown gas has a Vapour Density of 2. What is the mass of one mole of this gas?
  4. The empirical formula of a compound is CH₂. Its Vapour Density is 21. Find its molecular formula. [C=12, H=1]
  5. What does it mean if a gas has a Vapour Density of 1?

Answers to Practice Questions (Step 13):

  1. RMM from V.D.:
    • RMM = 2 × V.D. = 2 × 32 = 64 amu.
  2. D. of Ammonia (NH₃):
    • RMM of NH₃ = 14 + (3×1) = 17 amu.
    • D. = RMM / 2 = 17 / 2 = 8.5.
  3. Mass of one mole:
    • RMM = 2 × V.D. = 2 × 2 = 4 amu.
    • The mass of one mole (Molar Mass) is numerically equal to the RMM. So, the mass is 4 grams.
  4. Molecular Formula from V.D.:
    • RMM = 2 × V.D. = 2 × 21 = 42 amu.
    • Empirical Formula Mass of CH₂ = 12 + 2 = 14 amu.
    • n = RMM / EF Mass = 42 / 14 = 3.
    • Molecular Formula = (CH₂)₃ = C₃H₆.
  5. A Vapour Density of 1 means the gas is exactly as dense as hydrogen gas. Since RMM = 2 × 1 = 2 amu, the gas must be hydrogen (H₂) itself.

You are now a pro at figuring out the true formula of a compound. Let’s switch gears slightly. Instead of looking at single substances, we’ll now start looking at chemical reactions. We’ll begin with a special law that applies only when gases react with each other.

Step 14: Gay-Lussac’s Law of Combining Volumes

In the early 1800s, the French chemist Joseph Louis Gay-Lussac noticed something remarkable about how gases react. His discovery is now a scientific law.

Gay-Lussac’s Law states that:

When gases react, the volumes in which they do so bear a simple whole-number ratio to one another, and to the volumes of the products (if gaseous), provided the temperature and pressure remain constant.

Analogy: A Simple Recipe This sounds complicated, but it’s not. Think of a simple recipe: “To make a fruit smoothie, combine 1 part banana, 2 parts strawberries, to make 3 parts of smoothie.” The ratio of volumes is 1:2:3. You would never find a recipe that says “combine 1.37 parts banana with 2.89 parts strawberries.” Gay-Lussac found that gases always react in such simple, whole-number ratios.

The Link to Balanced Equations

Here’s the most useful part of this law: The “simple whole-number ratio” of volumes is exactly the same as the ratio of the coefficients (the big numbers in front) in the balanced chemical equation!

Example: Making Ammonia N₂(g) + 3H₂(g) → 2NH₃(g)

  • The coefficients are: 1 for N₂, 3 for H₂, and 2 for NH₃.
  • The ratio of the coefficients is 1 : 3 : 2.
  • Therefore, the ratio of the reacting volumes is also 1 : 3 : 2.

This means:

  • 1 litre of Nitrogen reacts with 3 litres of Hydrogen to produce 2 litres of Ammonia.
  • 10 mL of Nitrogen reacts with 30 mL of Hydrogen to produce 20 mL of Ammonia.

A Solved Problem

Problem: If 100 cm³ of ammonia (NH₃) is catalytically oxidized, what volume of oxygen (O₂) is required, and what volume of nitric oxide (NO) is formed? Equation: 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

  1. Find the Ratio from the Equation: Look at the coefficients of the gases we care about: NH₃, O₂, and NO. The ratio is 4 : 5 : 4.
  2. Apply the Ratio:
    • For Oxygen (O₂): The ratio of O₂ to NH₃ is 5:4. Volume of O₂ = (5/4) × Volume of NH₃ = (5/4) × 100 cm³ = 125 cm³.
    • For Nitric Oxide (NO): The ratio of NO to NH₃ is 4:4, or 1:1. Volume of NO = (4/4) × Volume of NH₃ = 1 × 100 cm³ = 100 cm³.

Important: This law only works for the volumes of gases. It does not work for solids or liquids.

Practice Questions (Step 14):

  1. State Gay-Lussac’s Law of Combining Volumes in your own words.
  2. For the reaction 2CO(g) + O₂(g) → 2CO₂(g), what is the ratio of the volumes of the reactants and the product?
  3. If 40 litres of carbon monoxide (CO) are burned in the reaction above, what volume of oxygen is needed and what volume of carbon dioxide is produced?
  4. In the reaction H₂(g) + Cl₂(g) → 2HCl(g), what volume of chlorine is required to produce 50 mL of hydrogen chloride gas?
  5. Does Gay-Lussac’s Law apply to the reaction CaCO₃(s) → CaO(s) + CO₂(g)? Why or why not?

Answers to Practice Questions (Step 14):

  1. Gay-Lussac’s Law: When gases react, the volumes they consume and produce are in simple, whole-number ratios, as long as the temperature and pressure are kept the same.
  2. Ratio for CO combustion: The coefficients are 2, 1, and 2. So, the volume ratio of CO : O₂ : CO₂ is 2 : 1 : 2.
  3. Applying the ratio:
    • Oxygen needed: The ratio of O₂ to CO is 1:2. Volume of O₂ = (1/2) × 40 L = 20 litres.
    • CO₂ produced: The ratio of CO₂ to CO is 2:2 or 1:1. Volume of CO₂ = 40 L.
  4. HCl production: The ratio of Cl₂ to HCl is 1:2. This means you need half the volume of chlorine to produce a certain volume of HCl. Volume of Cl₂ = (1/2) × 50 mL = 25 mL.
  5. Applicability of the Law: No, the law does not fully apply. Gay-Lussac’s Law is specifically for gases. Since Calcium Carbonate (CaCO₃) and Calcium Oxide (CaO) are solids (s), you cannot use their coefficients in a volume ratio.

You’ve mastered the laws that govern reacting gases. Now, let’s learn a much more powerful and universal method that works for solids, liquids, and gases. This is the heart of all chemical calculations, a topic called Stoichiometry.

Step 15: Stoichiometry I – The Chemical Recipe (Mole Ratios)

Think of a balanced chemical equation as a perfect recipe for making molecules. For example, the “recipe” for making a simple cheese sandwich could be written as: 2 Bread Slices + 1 Cheese Slice → 1 Sandwich

This tells you the exact ratio needed. The same is true in chemistry. Let’s look at the recipe for making ammonia: N₂(g) + 3H₂(g) → 2NH₃(g)

This balanced equation is a quantitative recipe that tells us:

1 mole of Nitrogen reacts with 3 moles of Hydrogen to produce 2 moles of Ammonia.

The coefficients (the numbers in front) in a balanced equation give us the mole ratio for the reaction. This is the single most important concept in stoichiometry.

Using the Mole Ratio

We can use this ratio as a conversion factor to figure out how much of one substance reacts with or produces another.

Problem: Using the ammonia reaction N₂(g) + 3H₂(g) → 2NH₃(g), if we want to produce 10 moles of ammonia (NH₃), how many moles of hydrogen (H₂) do we need?

  1. Identify the Mole Ratio from the balanced equation. We are comparing H₂ and NH₃. The ratio is 3 moles of H₂ : 2 moles of NH₃.
  2. Set up the Conversion. We want to find moles of H₂, so we write the ratio with H₂ on top: (3 moles H₂ / 2 moles NH₃).
  3. Moles of H₂ = (moles of NH₃ we want) × (the ratio) Moles of H₂ = 10 moles NH₃ × (3 moles H₂ / 2 moles NH₃) Moles of H₂ = (10 × 3) / 2 = 15 moles of H₂.

So, to produce 10 moles of ammonia, you need 15 moles of hydrogen.

Practice Questions (Step 15):

  1. For the reaction CH₄ + 2O₂ → CO₂ + 2H₂O, what do the coefficients tell you about the relationship between moles of methane (CH₄) and moles of oxygen (O₂)?
  2. In the reaction 2H₂S + 3O₂ → 2SO₂ + 2H₂O, what is the mole ratio of oxygen (O₂) to sulphur dioxide (SO₂)?
  3. Using the reaction above, if you start with 6 moles of H₂S, how many moles of O₂ are required for a complete reaction?
  4. The equation for rusting iron is 4Fe + 3O₂ → 2Fe₂O₃. If 1 mole of iron rusts completely, how many moles of iron(III) oxide (rust) are formed?
  5. In the decomposition of hydrogen peroxide, 2H₂O₂ → 2H₂O + O₂, how many moles of H₂O₂ are needed to produce 5 moles of oxygen (O₂)?

Answers to Practice Questions (Step 15):

  1. The coefficients in the reaction CH₄ + 2O₂ → CO₂ + 2H₂O tell you that 1 mole of methane reacts with 2 moles of oxygen.
  2. In the reaction 2H₂S + 3O₂ → 2SO₂ + 2H₂O, the mole ratio of oxygen (O₂) to sulphur dioxide (SO₂) is 3 : 2.
  3. The ratio of H₂S to O₂ is 2:3. To find the moles of O₂ needed, you calculate: 6 moles H₂S × (3 moles O₂ / 2 moles H₂S) = 9 moles of O₂.
  4. The ratio of Fe to Fe₂O₃ is 4:2, which simplifies to 2:1. This means you produce half as many moles of rust as the moles of iron you start with. So, 1 mole of Fe will produce 5 moles of Fe₂O₃.
  5. The ratio of H₂O₂ to O₂ is 2:1. To find the moles of H₂O₂ needed, you calculate: 5 moles O₂ × (2 moles H₂O₂ / 1 mole O₂) = 10 moles of H₂O₂.

Perfect! You’ve mastered the heart of chemical recipes—the mole ratio. But in a real lab, we don’t measure moles; we measure mass in grams. So, a real-world question is always, “If I use X grams of a reactant, how many grams of product will I get?” This next step teaches you exactly how to answer that.

Step 16: Stoichiometry II – Mass-Mass Relationships

This is the most common and practical type of chemical calculation. We will use the mole ratio you just learned, but we’ll add our mole-mass conversion skills from the Mole Map.

The Three-Step Path

You cannot compare the mass of one substance to the mass of another directly. You must always go through the central ‘Mole’ station. The path is always the same:

  1. Grams of A → Moles of A: Convert the mass of your starting substance (A) into moles.
  2. Moles of A → Moles of B: Use the mole ratio from the balanced equation to find how many moles of your target substance (B) will be produced or are needed.
  3. Moles of B → Grams of B: Convert the moles of your target substance (B) back into mass.

Flowchart: Mass (A) → Moles (A) → Moles (B) → Mass (B)

A Complete Solved Example

Problem: How many grams of iron (Fe) can be produced from 320 grams of iron(III) oxide (Fe₂O₃) in the following reaction? Fe₂O₃ + 3CO → 2Fe + 3CO₂ [Atomic Masses: Fe=56, O=16]

  1. Grams of Fe₂O₃ → Moles of Fe₂O₃
    • First, find the Molar Mass of Fe₂O₃ = (2 × 56) + (3 × 16) = 112 + 48 = 160 g/mol.
    • Moles = Mass / Molar Mass = 320 g / 160 g/mol = 2 moles of Fe₂O₃.
  2. Moles of Fe₂O₃ → Moles of Fe
    • Look at the balanced equation. The mole ratio of Fe₂O₃ to Fe is 1 : 2.
    • Moles of Fe = 2 moles Fe₂O₃ × (2 moles Fe / 1 mole Fe₂O₃) = 4 moles of Fe.
  3. Moles of Fe → Grams of Fe
    • The Molar Mass (atomic mass) of Fe is 56 g/mol.
    • Mass = Moles × Molar Mass = 4 mol × 56 g/mol = 224 grams of Fe.

Conclusion: 320 grams of iron(III) oxide can produce 224 grams of pure iron.

Practice Questions (Step 16):

  1. In the reaction 2H₂ + O₂ → 2H₂O, what mass of water is produced when 8 grams of hydrogen (H₂) is completely reacted? [H=1, O=16]
  2. CaCO₃ → CaO + CO₂. If 50 grams of calcium carbonate (CaCO₃) are heated, what mass of carbon dioxide (CO₂) is produced? [Ca=40, C=12, O=16]
  3. What is the first thing you must always have before you can solve a stoichiometry problem?
  4. In the reaction C + O₂ → CO₂, how many grams of carbon (C) are needed to produce 11 grams of carbon dioxide (CO₂)? [C=12, O=16]
  5. According to the equation 2KClO₃ → 2KCl + 3O₂, what mass of potassium chlorate (KClO₃) is needed to produce 48 grams of oxygen (O₂)? [K=39, Cl=35.5, O=16]

Answers to Practice Questions (Step 16):

  1. Mass of water from 8g of hydrogen:
    • Moles of H₂ = 8 g / 2 g/mol = 4 moles.
    • From the 2:2 (or 1:1) ratio, 4 moles of H₂ produce 4 moles of H₂O.
    • Mass of H₂O = 4 mol × 18 g/mol = 72 grams.
  2. Mass of CO₂ from 50g of CaCO₃:
    • Moles of CaCO₃ = 50 g / 100 g/mol = 0.5 moles.
    • From the 1:1 ratio, 0.5 moles of CaCO₃ produce 0.5 moles of CO₂.
    • Mass of CO₂ = 0.5 mol × 44 g/mol = 22 grams.
  3. The first thing you must always have is a balanced chemical equation.
  4. Grams of chlorine to produce 234g of NaCl:
    • Moles of NaCl = 234 g / 58.5 g/mol = 4 moles.
    • From the 2:1 ratio (NaCl to Cl₂), you need 2 moles of Cl₂ to produce 4 moles of NaCl.
    • Mass of Cl₂ = 2 mol × 71 g/mol = 142 grams.
  5. Mass of oxygen to burn 8g of methane:
    • Moles of CH₄ = 8 g / 16 g/mol = 0.5 moles.
    • From the 1:2 ratio (CH₄ to O₂), you need 1 mole of O₂.
    • Mass of O₂ = 1 mol × 32 g/mol = 32 grams.

You’re an expert at mass-based calculations now! Let’s bring gases into our reactions. What if a question asks, “If I use X grams of this solid, what volume of gas is produced?” This step combines stoichiometry with our knowledge of Molar Volume and Gay-Lussac’s Law.

Step 17: Stoichiometry III – Calculations Involving Volume

There are two main types of problems involving volume in reactions. The method you choose depends on what information you’re given.

  1. Mass-Volume Stoichiometry

This is for problems that connect the mass of one substance to the volume of another (which must be a gas at STP). We use the same 3-step path, but our final step uses the Molar Volume (22.4 L) instead of Molar Mass.

The Path: Mass (A) → Moles (A) → Moles (B) → Volume (B) at STP

Example: What volume of carbon dioxide (CO₂) is produced at STP when 60 grams of carbon (C) is burned? C(s) + O₂(g) → CO₂(g) [Atomic Masses: C=12, O=16]

  1. Grams of C → Moles of C:
    • Moles = Mass / Molar Mass = 60 g / 12 g/mol = 5 moles of C.
  2. Moles of C → Moles of CO₂:
    • From the equation, the mole ratio of C to CO₂ is 1 : 1.
    • So, 5 moles of C will produce 5 moles of CO₂.
  3. Moles of CO₂ → Volume of CO₂:
    • Volume = Moles × Molar Volume = 5 mol × 22.4 L/mol = 112 litres of CO₂.
  1. Volume-Volume Stoichiometry (The Shortcut!)

This is for problems where you are given the volume of a gas and asked to find the volume of another gas in the same reaction. Thanks to Gay-Lussac’s Law, we have a great shortcut!

The Shortcut: The ratio of coefficients in the balanced equation is the same as the ratio of volumes. You can skip the mole step completely!

Example: What volume of oxygen is required to completely burn 500 cm³ of ammonia (NH₃)? 4NH₃(g) + 3O₂(g) → 2N₂(g) + 6H₂O(g)

  1. Find the Volume Ratio:
    • We are comparing NH₃ and O₂. From the equation, the ratio of the coefficients is 4 : 3.
  2. Apply the Ratio directly to the Volumes:
    • Volume of O₂ = Volume of NH₃ × (Ratio)
    • Volume of O₂ = 500 cm³ NH₃ × (3 volumes O₂ / 4 volumes NH₃)
    • Volume of O₂ = (500 × 3) / 4 = 375 cm³.

Remember: This shortcut only works when dealing with gas to gas volumes, measured at the same temperature and pressure.

Practice Questions (Step 17):

  1. In the reaction Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g), what volume of hydrogen gas (H₂) is produced at STP when 6.5 grams of zinc (Zn) reacts? [Zn=65]
  2. From the equation N₂(g) + 3H₂(g) → 2NH₃(g), what volume of hydrogen is needed to react completely with 20 litres of nitrogen?
  3. What mass of water (H₂O) must be electrolyzed (2H₂O(l) → 2H₂(g) + O₂(g)) to produce 5.6 dm³ of oxygen gas at STP? [H=1, O=16]
  4. What is the shortcut for solving volume-volume stoichiometry problems, and what scientific law is it based on?
  5. What volume of carbon dioxide is formed in the complete combustion of 40 ml of ethyne (2C₂H₂(g) + 5O₂(g) → 4CO₂(g) + 2H₂O(g))?

Answers to Practice Questions (Step 17):

  1. Volume of H₂ from Zn:
    • Path: Mass Zn → Moles Zn → Moles H₂ → Volume H₂.
    • Moles of Zn = 6.5 g / 65 g/mol = 0.1 moles.
    • The mole ratio of Zn to H₂ is 1:1, so 0.1 moles of H₂ are produced.
    • Volume of H₂ = 0.1 mol × 22.4 L/mol = 24 litres.
  2. Volume of O₂ from C₃H₈:
    • This is a volume-volume problem, so we use the shortcut. The ratio of C₃H₈ to O₂ is 1:5.
    • Volume of O₂ = 10 L C₃H₈ × (5 L O₂ / 1 L C₃H₈) = 50 litres.
  3. Mass of H₂O to produce O₂:
    • Path: Volume O₂ → Moles O₂ → Moles H₂O → Mass H₂O.
    • Moles of O₂ = 5.6 L / 22.4 L/mol = 0.25 moles.
    • The mole ratio of O₂ to H₂O is 1:2. Moles of H₂O needed = 0.25 × 2 = 0.5 moles.
    • Mass of H₂O = 0.5 mol × 18 g/mol = 9 grams.
  4. The shortcut is to use the ratio of coefficients from the balanced equation directly as the ratio of volumes. It is based on Gay-Lussac’s Law of Combining Volumes.
  5. Volume of CO₂ from C₂H₂:
    • This is a volume-volume problem. The ratio of C₂H₂ to CO₂ is 2:4, or 1:2.
    • Volume of CO₂ = 40 mL C₂H₂ × (4 volumes CO₂ / 2 volumes C₂H₂) = 40 mL × 2 = 80 mL.

Incredible work! You’ve handled every type of calculation we’ve seen so far. But all our “recipes” have assumed we have the perfect amount of every ingredient. In a real lab, that’s rarely true. One ingredient always runs out first. This next step teaches you how to deal with that.

Step 18: The Limiting Reagent Concept

The Sandwich Analogy  Let’s go back to our sandwich recipe: 2 Bread Slices + 1 Cheese Slice → 1 Sandwich. Now, imagine you look in your fridge and you have 10 bread slices but only 3 cheese slices.

  • How many sandwiches can you possibly make? Only 3.
  • Which ingredient ran out first and limited how many sandwiches you could make? The cheese.
  • Which ingredient was left over? The bread. (You used 6 slices, so 4 are left over).

In chemistry:

  • The Limiting Reagent is the reactant that gets used up completely first and determines the maximum amount of product that can be formed (like the cheese).
  • The Excess Reagent is the reactant that is left over after the reaction stops (like the bread).

The Chemical Method

We can’t just compare grams to find the limiting reagent; we must compare moles.

Method:

  1. Calculate the number of moles of each reactant you have.
  2. For each reactant, calculate how much product it could make if it reacted completely.
  3. The reactant that produces the LEAST amount of product is the Limiting Reagent. This least amount is the actual amount of product that will be formed.

A Solved Example: If 56 grams of Nitrogen (N₂) are mixed with 12 grams of Hydrogen (H₂) to produce ammonia, which is the limiting reagent and what is the maximum mass of ammonia (NH₃) that can be formed? N₂(g) + 3H₂(g) → 2NH₃(g) [N=14, H=1]

  1. Calculate moles of each reactant:
    • Moles of N₂ = 56 g / 28 g/mol = 2 moles N₂.
    • Moles of H₂ = 12 g / 2 g/mol = 6 moles H₂.
  2. Calculate potential product from each:
    • From N₂: The N₂ to NH₃ ratio is 1:2.
      • Potential NH₃ = 2 moles N₂ × (2 moles NH₃ / 1 mole N₂) = 4 moles NH₃.
    • From H₂: The H₂ to NH₃ ratio is 3:2.
      • Potential NH₃ = 6 moles H₂ × (2 moles NH₃ / 3 moles H₂) = 4 moles NH₃.
  1. Identify the Limiting Reagent:
    • In this case, both reactants produce the same amount of product (4 moles). This means the reactants were mixed in the perfect stoichiometric ratio! Neither is limiting.
    • Let’s change the problem slightly to see the effect. What if we only had 3 grams of Hydrogen?
      • Moles of H₂ = 3 g / 2 g/mol = 5 moles H₂.
      • Potential NH₃ from H₂ = 1.5 moles H₂ × (2 moles NH₃ / 3 moles H₂) = 1 mole NH₃.
    • Now, N₂ can make 4 moles of NH₃, but H₂ can only make 1 mole. Since 1 is the smaller amount, Hydrogen (H₂) is the limiting reagent.
  2. Calculate final mass of product:
    • The reaction will only produce the smaller amount, which is 1 mole of NH₃.
    • Mass of NH₃ = 1 mol × 17 g/mol = 17 grams.

Practice Questions (Step 18):

  1. In the sandwich recipe (2 Bread + 1 Cheese → 1 Sandwich), if you have 8 bread slices and 5 cheese slices, which is the limiting reagent?
  2. For the reaction 2H₂ + O₂ → 2H₂O, if you mix 4 moles of H₂ with 3 moles of O₂, which is the limiting reagent?
  3. In the reaction C + O₂ → CO₂, if 36 grams of Carbon are burned with 64 grams of Oxygen, which is the limiting reagent? [C=12, O=16]
  4. Using the reaction from Q3, what is the maximum mass of carbon dioxide that can be formed?
  5. What is the name for the reactant that is left over after a chemical reaction is complete?

Answers to Practice Questions (Step 18):

  1. Sandwich Analogy: 8 bread slices can make 8 / 2 = 4 sandwiches. 5 cheese slices can make 5 / 1 = 5 sandwiches. Since the bread produces the smaller number of sandwiches, the bread is the limiting reagent.
  2. H₂ vs O₂:
    • From 4 moles of H₂, you can make 4 × (2/2) = 4 moles of H₂O.
    • From 3 moles of O₂, you can make 3 × (2/1) = 6 moles of H₂O.
    • Since H₂ produces the lesser amount, Hydrogen (H₂) is the limiting reagent and Oxygen (O₂) is the excess reagent.
  3. C vs O₂:
    • Moles of C = 36 g / 12 g/mol = 3 moles.
    • Moles of O₂ = 64 g / 32 g/mol = 2 moles.
    • From C, you can make 3 moles of CO₂ (1:1 ratio).
    • From O₂, you can make 2 moles of CO₂ (1:1 ratio).
    • Since O₂ produces the lesser amount, Oxygen (O₂) is the limiting reagent.
  4. Maximum Mass of CO₂: The amount of product is determined by the limiting reagent (O₂). The reaction can produce a maximum of 2 moles of CO₂.
    • Mass of CO₂ = 2 mol × 44 g/mol = 88 grams.
  5. The reactant that is not completely used up is called the Excess Reagent.

You’ve done an incredible job with one of the most challenging concepts in stoichiometry! For our next-to-last step, we’ll add two more layers of real-world chemistry to our problems: Percentage Purity and Percentage Yield.

Step 19: Purity and Yield – Chemistry in the Real World

In all our problems so far, we’ve made two big assumptions:

  1. The chemicals we start with are 100% pure.
  2. The reactions work perfectly and produce the maximum possible amount of product.

In reality, neither of these is true!

  1. Percentage Purity

Most chemicals, especially those mined from the earth like ores, contain impurities. Only the pure part of the substance will actually react.

Analogy: Peanuts in Shells 🥜 If you buy a 1 kg bag of peanuts with their shells, you can’t use the whole 1 kg in a recipe. The shells are an impurity. If the peanuts themselves weigh 700g, the sample is 70% pure. Only the 700g of pure peanuts will react with your hunger!

How to Calculate: The very first step in a problem with an impure reactant is to find the mass of the pure substance. Mass of Pure Reactant = Total Mass of Sample × (Purity % / 100)

Example: What volume of hydrogen gas (H₂) is produced at STP when 100 grams of zinc ore, which is 80% pure zinc, reacts with acid? Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g) [Zn=65]

  1. Find the mass of PURE Zinc:
    • Mass of pure Zn = 100 g × (80 / 100) = 80 grams of Zn.
  2. Use this pure mass in your stoichiometry calculation:
    • Moles of Zn = 80 g / 65 g/mol ≈ 1.23 moles.
    • The mole ratio of Zn to H₂ is 1:1, so moles of H₂ = 1.23 moles.
    • Volume of H₂ = 1.23 mol × 22.4 L/mol = 55 Litres.
  1. Percentage Yield

Chemical reactions are often inefficient. Some product can be lost during the experiment.

  • Theoretical Yield: The maximum amount of product you calculate on paper. This is what we’ve been calculating in all our problems.
  • Actual Yield: The amount of product you actually get in a real lab experiment.

Analogy: The Baker’s Loss 🎂 A recipe might say it produces a 1 kg cake (Theoretical Yield). But after baking, a few crumbs are lost and some batter sticks to the bowl. You might only end up with 950g of cake (Actual Yield).

The Percentage Yield tells you how successful the reaction was. % Yield = (Actual Yield / Theoretical Yield) × 100

Example: A chemist calculated that a reaction should produce 40 grams of a product (theoretical yield). After performing the experiment, they only collected 35 grams (actual yield). What was the percentage yield?

  • % Yield = (35 g / 40 g) × 100 = 87.5%.

Practice Questions (Step 19):

  1. A 250-gram rock of limestone is 92% pure calcium carbonate (CaCO₃). What is the mass of pure CaCO₃ in the rock?
  2. A student calculates that a reaction should theoretically produce 15 grams of a salt. They perform the reaction and obtain 12 grams. What is the percentage yield?
  3. Which term describes the amount of impurity in a reactant sample: Purity or Yield?
  4. How many grams of pure iron (Fe) can be extracted from 500 kg of an iron ore that is 60% pure Fe₂O₃? The reaction is Fe₂O₃ + 3CO → 2Fe + 3CO₂. [Fe=56, O=16]
  5. The theoretical yield of a reaction is 80 grams. If the reaction has an 85% yield, what is the actual mass of product that will be obtained?

Answers to Practice Questions (Step 19):

  1. Mass of pure CaCO₃:
    • Mass = Total Mass × (Purity % / 100) = 250 g × (92 / 100) = 230 grams.
  2. Percentage Yield:
    • % Yield = (Actual Yield / Theoretical Yield) × 100 = (12 g / 15 g) × 100 = 80%.
  3. Describing Impurity: The term that describes the amount of impurity in a reactant sample is Purity.
  4. Iron from Ore:
    • Step 1 (Pure Mass): 500 kg = 500,000 g. Mass of pure Fe₂O₃ = 500,000 g × (60/100) = 300,000 g.
    • Step 2 (Stoichiometry): Moles of Fe₂O₃ = 300,000 g / 160 g/mol = 1875 moles. The mole ratio of Fe₂O₃ to Fe is 1:2, so moles of Fe = 1875 × 2 = 3750 moles.
    • Mass of Fe = 3750 mol × 56 g/mol = 210,000 g or 210 kg.
  5. Actual Yield:
    • Actual Yield = Theoretical Yield × (% Yield / 100) = 80 g × (85 / 100) = 68 grams.

You have now mastered every individual concept, from the simplest definition to the practical realities of purity and yield. For our final step, we will put everything together to solve a single, comprehensive problem that will test all the skills you’ve learned. This is the final challenge.

Step 20: Putting It All Together – The Final Challenge

This problem will require you to use concepts from across our entire 20-step journey. Think of it as the final exam, designed to prove your mastery.

The Master Problem:

A chemist’s goal is to produce chlorine gas (Cl₂). The plan is to use an impure sample of manganese dioxide (MnO₂) ore. The ore sample has a total mass of 200 grams and is 87% pure MnO₂. It reacts with excess hydrochloric acid according to the following equation:

MnO₂(s) + 4HCl(aq) → MnCl₂(aq) + 2H₂O(l) + Cl₂(g)

The reaction is known to be inefficient and has a 90% yield. [Atomic Masses: Mn=55, O=16, Cl=35.5, H=1]

Calculate: (a) The mass of pure MnO₂ available to react. (b) The number of moles of pure MnO₂. (c) The theoretical volume of chlorine gas (Cl₂) that can be produced at STP. (d) The actual volume of chlorine gas (Cl₂) collected at STP, considering the yield. (e) The mass of the actual volume of chlorine gas collected.

Solution Breakdown:

(a) Find the mass of pure MnO₂: (Using our Purity skill from Step 19)

  • Pure Mass = Total Mass × (Purity % / 100) = 200 g × (87 / 100) = 174 grams.

(b) Find the moles of pure MnO₂: (Using our Mole-Mass skill from Step 6)

  • Molar Mass of MnO₂ = 55 + (2 × 16) = 87 g/mol.
  • Moles = Mass / Molar Mass = 174 g / 87 g/mol = 2 moles.

(c) Find the theoretical volume of Cl₂: (Using our Stoichiometry skills from Step 17)

  • From the equation, the mole ratio of MnO₂ to Cl₂ is 1 : 1.
  • So, 2 moles of MnO₂ will theoretically produce 2 moles of Cl₂.
  • Theoretical Volume = Moles × Molar Volume = 2 mol × 22.4 L/mol = 44.8 litres.

(d) Find the actual volume of Cl₂: (Using our Yield skill from Step 19)

  • Actual Yield = Theoretical Yield × (% Yield / 100)
  • Actual Volume = 44.8 L × (90 / 100) = 40.32 litres.

(e) Find the mass of the actual volume of Cl₂: (Using our Mole Map skills from Step 10)

  • Path: Volume → Moles → Mass.
  • Actual Moles of Cl₂ = Actual Volume / 22.4 = 40.32 L / 22.4 L/mol = 1.8 moles.
  • Molar Mass of Cl₂ = 2 × 35.5 = 71 g/mol.
  • Actual Mass = 1.8 mol × 71 g/mol = 127.8 grams.

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