Calculate the amount of carbon dioxide that could be produced when (i) 1 mole of carbon is burnt in air. (ii)

Exercises · 1.4

Q. Calculate the amount of carbon dioxide that could be produced when

(i) 1 mole of carbon is burnt in air.

(ii) 1 mole of carbon is burnt in 16 g of dioxygen.

(iii) 2 moles of carbon are burnt in 16 g of dioxygen.

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Concept understanding — Molecular Mass Calculation

What is Molecular Mass? The Intuition

Imagine you're at a market buying apples. You don't weigh each apple individually — you put a dozen on a scale. The total weight tells you something about the apples, but it also depends on how many apples you have.

Atoms and molecules are unimaginably tiny. A single water molecule ( $H_2O$ ) weighs about $3 \times 10^{-23}$ grams. That number is useless for practical chemistry. So instead of working with individual molecules, chemists count them in huge fixed numbers — specifically, $6.022 \times 10^{23}$ molecules, which is called one mole.

Molecular mass is simply the mass of one mole of a substance, expressed in grams per mole (g/mol). It answers the question: "If I have $6.022 \times 10^{23}$ molecules of this compound, how much would they weigh on a lab balance?"

Note

The number $6.022 \times 10^{23}$ is Avogadro's constant. It's chosen so that the mass of one mole of carbon-12 atoms is exactly 12 grams — matching the atomic mass unit scale perfectly.

The Precise Definition

Molecular mass (also called molar mass) is the mass of one mole of a molecular substance. It is numerically equal to the sum of the atomic masses of all atoms in the molecule, expressed in g/mol.

For example:

Water ( $H_2O$ ): 2 hydrogen atoms + 1 oxygen atom
Atomic mass of H = 1.008 g/mol
Atomic mass of O = 16.00 g/mol
Molecular mass of $H_2O$ = $2(1.008) + 16.00 = 18.016$ g/mol

\text{Molecular mass} = \sum (\text{number of atoms of each element} \times \text{atomic mass of that element})

How to Calculate It — Step by Step

Let's take glucose, $C_6H_{12}O_6$ , as a worked example.

Step 1: Identify each element and its count

Carbon (C): 6 atoms
Hydrogen (H): 12 atoms
Oxygen (O): 6 atoms

Step 2: Look up atomic masses (from the periodic table)

C: 12.01 g/mol
H: 1.008 g/mol
O: 16.00 g/mol

Step 3: Multiply and add

\text{Molecular mass} = 6(12.01) + 12(1.008) + 6(16.00)

= 72.06 + 12.096 + 96.00

= 180.156 \text{ g/mol}

Tip

Always keep at least 2 decimal places from the periodic table. For exam problems, they usually give you atomic masses — use exactly what's provided.

Why This Matters

Molecular mass is the bridge between the microscopic world (atoms and molecules) and the macroscopic world (grams you can weigh). Once you know the molecular mass, you can:

Convert grams to moles: $\text{moles} = \frac{\text{mass in grams}}{\text{molecular mass}}$
Convert moles to grams: $\text{mass} = \text{moles} \times \text{molecular mass}$
Determine the number of molecules: $\text{molecules} = \text{moles} \times 6.022 \times 10^{23}$

Watch out

Do not confuse molecular mass with atomic mass. Atomic mass refers to a single element (like oxygen = 16.00 g/mol). Molecular mass refers to a compound (like $CO_2$ = 44.01 g/mol). Also, for ionic compounds like NaCl, we use formula mass (same calculation, but the substance isn't molecular).

Common Exam Pitfalls

Forgetting to multiply by the subscript. In $H_2SO_4$ , there are 2 hydrogens, not 1. …

Why this formula?

Stoichiometry & Mole Calculation: The "Why" Behind the Formula

Let's build this from the ground up — not as a list of formulas to memorise, but as a logical chain of reasoning.

1. The Core Question: What is a Mole?

A mole is simply a counting unit, like a dozen (12) or a gross (144). But instead of 12, a mole contains $6.022 \times 10^{23}$ particles (Avogadro's number, $N_A$ ).

Why this number?

It was chosen so that 1 mole of any substance has a mass in grams equal to its atomic/molecular mass in amu.

Example: 1 atom of carbon-12 has mass 12 amu.
1 mole of carbon-12 has mass 12 grams.

This is the bridge between the microscopic (atoms/molecules) and the macroscopic (grams we can weigh).

2. The Fundamental Relationship

The key formula is:

n = \frac{m}{M}

Where:

$n$ = number of moles
$m$ = mass of substance (in grams)
$M$ = molar mass (in g/mol)

Why does this work?

Think of it as a conversion factor:

If 1 mole of a substance weighs $M$ grams, then $m$ grams contains $\frac{m}{M}$ moles.

Derivation logic:

Molar mass $M$ tells you: "1 mol = $M$ g"
So the conversion factor is $\frac{1 \text{ mol}}{M \text{ g}}$
Multiply mass $m$ by this factor: $m \times \frac{1}{M} = \frac{m}{M}$ moles

3. Connecting to Number of Particles

n = \frac{N}{N_A}

Where:

$N$ = number of particles (atoms, molecules, ions)
$N_A = 6.022 \times 10^{23}$ particles/mol

Why?

1 mole = $N_A$ particles
So $N$ particles = $\frac{N}{N_A}$ moles

Combine both formulas:

\frac{m}{M} = \frac{N}{N_A}

This single equation ties mass, molar mass, number of particles, and Avogadro's number together.

4. The Gas Volume Connection (for gases at STP)

For gases only:

n = \frac{V}{22.4 \text{ L/mol}}

Why 22.4 L?

From the ideal gas law: $PV = nRT$

At STP (Standard Temperature and Pressure: 0°C, 1 atm):

$P = 1$ atm
$T = 273.15$ K
$R = 0.0821$ L·atm/(mol·K)

For $n = 1$ mole:

V = \frac{nRT}{P} = \frac{1 \times 0.0821 \times 273.15}{1} \approx 22.4 \text{ L}

So 1 mole of any ideal gas occupies 22.4 L at STP. This is a consequence of the gas laws, not a definition.

5. The Stoichiometry Chain: From One Substance to Another

In a balanced chemical equation like:

aA + bB \rightarrow cC + dD

The coefficients tell you the mole ratio:

\frac{\text{moles of A}}{\text{moles of B}} = \frac{a}{b}

Why this works: …

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The key idea is to use the balanced combustion reaction $\mathrm{C + O_2 \rightarrow CO_2}$ and identify the limiting reagent in each case. The amounts of $\mathrm{CO_2}$ produced are: (i) 44 g,

(ii) 22 g,

(iii) 22 g.

This problem is a classic exercise in stoichiometry — the art of relating quantities of reactants and products using a balanced chemical equation. The combustion of carbon in oxygen is beautifully simple: one atom of carbon combines with one molecule of dioxygen to form one molecule of carbon dioxide. The balanced equation is:

\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)}

From this, the molar ratios are crystal clear: 1 mole of C reacts with 1 mole of $\mathrm{O_2}$ to produce 1 mole of $\mathrm{CO_2}$ . The molar mass of $\mathrm{CO_2}$ is $12 + 2 \times 16 = 44\ \mathrm{g/mol}$ .

Now, the three parts differ only in the amounts of reactants given. The trick is to check whether both reactants are present in the exact 1:1 mole ratio, or if one of them runs out first — that’s the limiting reagent. The product amount is always determined by the limiting reagent.

Let’s work through each part.

1 mole of carbon burnt in air. Air contains plenty of oxygen (about 21% by volume), so oxygen is in vast excess. The carbon is the limiting reagent. Since 1 mole of C gives 1 mole of $\mathrm{CO_2}$ , the mass produced is:

1\ \mathrm{mol} \times 44\ \mathrm{g/mol} = 44\ \mathrm{g}

1 mole of carbon burnt in 16 g of dioxygen. First, find how many moles of $\mathrm{O_2}$ are in 16 g. Molar mass of $\mathrm{O_2}$ is $32\ \mathrm{g/mol}$ , so:

\text{Moles of } \mathrm{O_2} = \frac{16}{32} = 0.5\ \mathrm{mol}

The balanced equation requires 1 mole of $\mathrm{O_2}$ for 1 mole of C. Here, we have only 0.5 mole of $\mathrm{O_2}$ but 1 mole of C. Oxygen is the limiting reagent.

From 0.5 mole of $\mathrm{O_2}$ , we get 0.5 mole of $\mathrm{CO_2}$ . Mass:

$0.5\ \mathrm{mol} \times 44\ \mathrm{g/mol} = 22\ \mathrm{g}$ …

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Here are the common mistakes students make when solving molecular mass and limiting reagent problems like this one, along with how to avoid each.

1. Forgetting the Balanced Chemical Equation

Mistake: Jumping straight to arithmetic without writing the reaction.

Why it’s wrong: You can’t determine the mole ratio or the limiting reagent without the balanced equation.

How to avoid: Always write the balanced equation first:

\text{C} + \text{O}_2 \rightarrow \text{CO}_2

This tells you: 1 mole C + 1 mole O $_2$ → 1 mole CO $_2$ .

2. Confusing Mass with Moles (Part ii & iii)

Mistake: Treating 16 g of O $_2$ as 1 mole of O $_2$ .

Why it’s wrong: 1 mole of O $_2$ has a mass of 32 g, not 16 g. So 16 g is only 0.5 mole of O $_2$ .

How to avoid: Always convert mass to moles using:

\text{Moles} = \frac{\text{Given mass (g)}}{\text{Molar mass (g/mol)}}

For O $_2$ : molar mass = $2 \times 16 = 32$ g/mol.

So 16 g O $_2$ = $\frac{16}{32} = 0.5$ moles.

3. Ignoring the Limiting Reagent (Part iii)

Mistake: Assuming both reactants are fully consumed.

Why it’s wrong: In part (iii), you have 2 moles C but only 0.5 moles O $_2$ . Oxygen runs out first — it is the limiting reagent.

How to avoid:

Compare the mole ratio from the equation (1:1).
The reactant with fewer moles (after accounting for the ratio) limits the product.
Here: 0.5 mole O $_2$ can only produce 0.5 mole CO $_2$ , regardless of excess carbon.

4. Using Wrong Molar Mass for CO $_2$

Mistake: Using 44 g/mol for CO $_2$ but then miscalculating the mass.

Why it’s wrong: Even if you get the moles right, a slip in molar mass ruins the final answer.

How to avoid: Memorise and double-check:

\text{Molar mass of CO}_2 = 12 + (2 \times 16) = 44 \text{ g/mol}

Then mass = moles $\times$ 44.

5. Not Stating the Final Answer in Grams

Mistake: Leaving the answer in moles when the question asks for “amount” (mass). …

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Q.Calculate the amount of carbon dioxide that could be produced when

Concept understanding — Molecular Mass Calculation

What is Molecular Mass? The Intuition

The Precise Definition

How to Calculate It — Step by Step

Why This Matters

Common Exam Pitfalls

Why this formula?

Stoichiometry & Mole Calculation: The "Why" Behind the Formula

1. The Core Question: What is a Mole?

2. The Fundamental Relationship

Why does this work?

3. Connecting to Number of Particles

Why?

4. The Gas Volume Connection (for gases at STP)

Why 22.4 L?

5. The Stoichiometry Chain: From One Substance to Another

Why this works: …

Method: Limiting Reagent & Stoichiometric Calculation

Step 1 — Write the balanced equation

Step 2 — Solve each case

(i) 1 mole of carbon burnt in air

(ii) 1 mole of carbon burnt in 16 g of dioxygen

(iii) 2 moles of carbon burnt in 16 g of dioxygen

1. Forgetting the Balanced Chemical Equation

2. Confusing Mass with Moles (Part ii & iii)

3. Ignoring the Limiting Reagent (Part iii)

4. Using Wrong Molar Mass for CO $_2$

5. Not Stating the Final Answer in Grams

Q.Calculate the amount of carbon dioxide that could be produced when

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