What is the mass percent of carbon in carbon dioxide? (i) 0.034% (ii) 27.27% (iii) 3.4% (iv) 28.7%

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Q. What is the mass percent of carbon in carbon dioxide?

(i) 0.034%

(ii) 27.27%

(iii) 3.4%

(iv) 28.7%

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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 mass percent of carbon in carbon dioxide is found by dividing the atomic mass of carbon by the molecular mass of CO₂ and multiplying by 100. The result is 27.27%, which corresponds to option (ii).

Concept and Intuition

Mass percent (or weight percent) tells you what fraction of a compound’s total mass comes from a particular element. For carbon dioxide (CO₂), we want to know: “If I have one molecule of CO₂, how much of its mass is due to carbon, and how much is due to oxygen?”

The key idea is simple: the mass percent of an element in a compound is just (mass of that element in one formula unit) divided by (total mass of the formula unit), multiplied by 100. This works because the ratio of masses is the same whether you consider one molecule, one mole, or any amount — the percentage is fixed by the chemical formula.

For CO₂, the formula tells us there is one carbon atom and two oxygen atoms. So we need the atomic masses of carbon and oxygen from the periodic table. Standard values (to one decimal place) are:

Carbon (C): 12.0 u (atomic mass units)
Oxygen (O): 16.0 u

Watch out

A common mistake is to forget that CO₂ has two oxygen atoms, not one. Using just 16.0 u for oxygen instead of 32.0 u would give a wildly wrong percentage (42.9% instead of 27.3%). Always count the atoms in the formula carefully.

Step-by-Step Calculation

Find the molecular mass of CO₂.

The molecule contains 1 carbon atom and 2 oxygen atoms.

Mass of carbon = $1 \times 12.0 = 12.0$ u

Mass of oxygen = $2 \times 16.0 = 32.0$ u

Total molecular mass = $12.0 + 32.0 = 44.0$ u
Identify the mass contributed by carbon.

That’s just the mass of the single carbon atom: $12.0$ u.
Set up the mass percent formula.

\text{Mass percent of C} = \frac{\text{mass of C}}{\text{mass of CO}_2} \times 100\%

Plug in the numbers.

\frac{12.0}{44.0} \times 100\% = \frac{12}{44} \times 100\%

Simplify the fraction.

$\frac{12}{44} = \frac{3}{11}$ (divide numerator and denominator by 4).

So the calculation becomes $\frac{3}{11} \times 100\%$ .
Compute the decimal. …

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Common Mistakes on "Mass Percent of Carbon in CO₂"

This question tests your understanding of mass percent composition — a fundamental concept in stoichiometry. Let's break down the common errors.

✗ Mistake 1: Using atomic numbers instead of atomic masses

What students do wrong:

They take the atomic number of carbon (6) and oxygen (8) instead of the atomic masses (≈12 g/mol and ≈16 g/mol).

Why it happens:

Students confuse the periodic table columns — atomic number (Z) is at the top, atomic mass is at the bottom.

How to avoid:

Always check the mass number (bottom value) on the periodic table. For carbon, use 12 g/mol; for oxygen, use 16 g/mol.

✗ Mistake 2: Forgetting there are two oxygen atoms

What students do wrong:

They calculate:

$\text{Mass of CO}_2 = 12 + 16 = 28 \text{ g/mol}$

Then:

$\text{Mass percent of C} = \frac{12}{28} \times 100 \approx 42.86\%$

(which isn't even an option — a red flag missed!)

Why it happens:

Rushing and not writing the molecular formula carefully.

How to avoid:

Always write the full molecular formula first:

$\text{CO}_2$ means 1 C atom + 2 O atoms.

So:

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

✗ Mistake 3: Calculating mass percent of oxygen instead of carbon

What students do wrong:

They compute:

$\frac{32}{44} \times 100 \approx 72.73\%$

and then pick 27.27% (option (ii)) by mistake — or worse, pick 72.73% if it were listed.

Why it happens:

Misreading the question — "carbon" vs "oxygen."

How to avoid:

Circle the element asked in the question. Underline "carbon" before starting.

✗ Mistake 4: Reporting the decimal as a percentage incorrectly

What students do wrong:

They compute:

$\frac{12}{44} \approx 0.2727$

and write 0.2727% instead of 27.27%.

Why it happens:

Forgetting to multiply by 100.

How to avoid:

Mass percent formula: …

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Q.What is the mass percent of carbon in carbon dioxide?

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: …

Concept and Intuition

Step-by-Step Calculation

Concept: Mass Percent Composition

Method: Formula-Based Mass Percent Calculation

Steps

Common Mistakes on "Mass Percent of Carbon in CO₂"

✗ Mistake 1: Using atomic numbers instead of atomic masses

✗ Mistake 2: Forgetting there are two oxygen atoms

✗ Mistake 3: Calculating mass percent of oxygen instead of carbon

✗ Mistake 4: Reporting the decimal as a percentage incorrectly

Q.What is the mass percent of carbon in carbon dioxide?

Concept and Intuition

Step-by-Step Calculation

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