A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carb

Exercises · 1.34

Q. A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate

(i) empirical formula,

(ii) molar mass of the gas, and

(iii) molecular formula.

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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 problem gives combustion masses and STP density to find the empirical formula, molar mass, and molecular formula. The empirical formula is CH, the molar mass is 26.0 g/mol, and the molecular formula is C₂H₂.

Why This Approach Works

We have two independent pieces of data. The combustion data tells us the ratio of carbon to hydrogen atoms in the compound — that gives the empirical formula. The STP density data tells us the actual mass of one mole of the gas — that gives the molar mass. Comparing the molar mass to the empirical formula mass reveals how many empirical units fit into one molecule, yielding the molecular formula.

The key insight: combustion converts all carbon to CO₂ and all hydrogen to H₂O. By measuring the masses of these products, we can work backwards to find the masses of C and H in the original sample. The ratio of these masses gives the simplest whole-number atom ratio.

Step-by-Step Solution

1. Find masses of carbon and hydrogen from combustion data

All carbon in the sample ends up in CO₂. The molar mass of CO₂ is $12.01 + 2 \times 16.00 = 44.01 \text{ g/mol}$ .

Mass of carbon in 3.38 g CO₂:

\text{Mass of C} = \frac{12.01}{44.01} \times 3.38 = 0.922 \text{ g}

All hydrogen ends up in H₂O. Molar mass of H₂O is $2 \times 1.008 + 16.00 = 18.016 \text{ g/mol}$ .

Mass of hydrogen in 0.690 g H₂O:

\text{Mass of H} = \frac{2 \times 1.008}{18.016} \times 0.690 = 0.0772 \text{ g}

Note

We use $2 \times 1.008$ because each water molecule contains two hydrogen atoms.

2. Find the mole ratio of C to H

Moles of C:

\frac{0.922}{12.01} = 0.0768 \text{ mol}

Moles of H:

\frac{0.0772}{1.008} = 0.0766 \text{ mol}

The ratio is essentially $0.0768 : 0.0766 \approx 1 : 1$ .

So the empirical formula is CH.

Watch out

Don't round too aggressively. Here the numbers are nearly identical, so 1:1 is clear. If the ratio were, say, 1:1.33, you'd multiply by 3 to get whole numbers.

3. Calculate the molar mass from STP data

At STP (0°C, 1 atm), one mole of any ideal gas occupies 22.4 L. …

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✗ Mistake 1: Forgetting that carbon in CO₂ comes only from the sample

What students do wrong:

They take the mass of CO₂ (3.38 g) and treat it as the mass of carbon in the sample.

Why it’s wrong:

CO₂ contains both carbon and oxygen. Only the carbon part comes from the fuel.

How to avoid:

Always convert mass of CO₂ → moles of CO₂ → moles of C → mass of C.

Moles of CO₂ = $\frac{3.38}{44.0} = 0.0768$ mol
Moles of C = same as moles of CO₂ = 0.0768 mol
Mass of C = $0.0768 \times 12.0 = 0.922$ g

Similarly for water:

Moles of H₂O = $\frac{0.690}{18.0} = 0.0383$ mol
Moles of H = $2 \times 0.0383 = 0.0766$ mol
Mass of H = $0.0766 \times 1.0 = 0.0766$ g

✓ Key rule:

Mass of element in compound = (moles of that element) × (atomic mass)

✗ Mistake 2: Not checking that the sample mass equals C + H mass

What students do wrong:

They calculate C and H masses but don’t verify if oxygen is present.

Why it’s wrong:

The problem says the gas contains only carbon and hydrogen. If C + H mass doesn’t match the sample mass, you’ve made an error.

How to avoid:

Add the masses:

$0.922 + 0.0766 = 0.9986$ g (≈ 1.00 g sample).

Since it matches, no oxygen — correct.

✓ Always check:

If sum < sample mass → oxygen present. If sum = sample mass → only C & H.

✗ Mistake 3: Using mass instead of moles for empirical formula ratio

What students do wrong:

They write the ratio of masses (0.922 : 0.0766) and try to simplify.

Why it’s wrong:

Empirical formula is based on mole ratio, not mass ratio.

How to avoid:

Convert masses to moles first:

Moles of C = 0.0768
Moles of H = 0.0766

Ratio C : H = 0.0768 : 0.0766 ≈ 1 : 1

So empirical formula = CH

✓ Always:

Mass → moles → simplest whole-number ratio

✗ Mistake 4: Using wrong molar volume at STP

What students do wrong:

They use 22.4 L/mol correctly but then miscalculate molar mass.

Why it’s wrong:

Sometimes they forget that STP means 22.4 L per mole for any gas.

How to avoid:

Given: 10.0 L weighs 11.6 g at STP.

Moles of gas = $\frac{10.0}{22.4} = 0.446$ mol …

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Q.A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate

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: …

Why This Approach Works

Step-by-Step Solution

1. Find masses of carbon and hydrogen from combustion data

2. Find the mole ratio of C to H

3. Calculate the molar mass from STP data

Method: Combustion Analysis + Molar Mass from STP Density

Step 1: Find moles of C and H from combustion products

Step 2: Find simplest whole-number ratio (empirical formula)

Step 3: Find molar mass from STP data

✗ Mistake 1: Forgetting that carbon in CO₂ comes only from the sample

✗ Mistake 2: Not checking that the sample mass equals C + H mass

✗ Mistake 3: Using mass instead of moles for empirical formula ratio

✗ Mistake 4: Using wrong molar volume at STP

Q.A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate

Why This Approach Works

Step-by-Step Solution

1. Find masses of carbon and hydrogen from combustion data

2. Find the mole ratio of C to H

3. Calculate the molar mass from STP data

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