Calculate the atomic mass (average) of chlorine using the following data: | | % Natural Abundance | Molar Mass | |---|---|---| | $^{35}Cl$ | 75.77 | 34.9689 | | $^{37}Cl$ | 24.23 | 36.9659 |

Calculate the atomic mass (average) of chlorine us – 1st PUC Chemistry

Exercises · 1.9

Calculate the atomic mass (average) of chlorine using the following data:

	% Natural Abundance	Molar Mass
$^{35}Cl$	75.77	34.9689
$^{37}Cl$	24.23	36.9659

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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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Given data

	% Natural Abundance	Molar Mass
$^{35}Cl$	75.77	34.9689
$^{37}Cl$	24.23	36.9659

Concept: Average atomic mass from isotopic abundance

Natural chlorine is a mixture of two isotopes. The average atomic mass is the weighted mean of the individual isotopic masses, where the weights are their fractional abundances.

Step 1: Convert percentages to fractions.

f_{35} = \frac{75.77}{100} = 0.7577, \quad f_{37} = \frac{24.23}{100} = 0.2423

Step 2: Multiply each isotopic mass by its fraction and sum. …

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The average atomic mass of an element is the weighted mean of its isotopes' masses, where the weights are their natural abundances. For chlorine, this gives 35.45 u.

Why weighted averages matter in atomic mass

When you pick up a sample of chlorine from nature, you're not getting just one isotope—you're getting a mixture. About three-quarters of the atoms are $^{35}\text{Cl}$ and one-quarter are $^{37}\text{Cl}$ . The atomic mass on the periodic table reflects this reality: it's not the mass of any single isotope, but rather the average mass you'd measure if you weighed a large collection of randomly selected chlorine atoms.

The calculation is a weighted average because the isotopes don't contribute equally. The more abundant isotope pulls the average closer to its own mass.

\text{Average Atomic Mass} = \sum (\text{fractional abundance})_i \times (\text{molar mass})_i

Step-by-step calculation

1. Convert percentages to fractions

Natural abundance is given as a percentage, but we need it as a decimal fraction for the calculation:

$^{35}\text{Cl}$ : $\frac{75.77}{100} = 0.7577$
$^{37}\text{Cl}$ : $\frac{24.23}{100} = 0.2423$

2. Multiply each isotope's mass by its fractional abundance

This gives the contribution of each isotope to the overall average:

Contribution from $^{35}\text{Cl}$ : $0.7577 \times 34.9689 = 26.4959$ u
Contribution from $^{37}\text{Cl}$ : $0.2423 \times 36.9659 = 8.9568$ u

3. Sum the contributions …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KCET 2024Set B-21 markMCQ
$_2$
›Reveal solutionSolution

Every carbon atom burns to $CO_2$ , so convert the $CO_2$ mass to carbon mass with the factor $12/44$ , then express it as a percentage of the sample.

Step 1 — The principle (Liebig's combustion method)

In quantitative combustion analysis the organic compound is burnt completely in excess oxygen. All its carbon is converted to $CO_2$ , which is absorbed and weighed. So the carbon in the $CO_2$ is the carbon that was in the compound — a simple mass-conservation argument.

Step 2 — The carbon fraction of $CO_2$
$M(CO_2) = 12 + 2(16) = 44\ \text{g mol}^{-1}, \qquad M(C) = 12\ \text{g mol}^{-1}$
So every 44 g of $CO_2$ contains 12 g of carbon:
$\text{fraction of C in } CO_2 = \frac{12}{44}$
Step 3 — Mass of carbon in the sample
$m_C = \frac{12}{44}\times m_{CO_2} = \frac{12}{44}\times 0.22 = \frac{2.64}{44} = 0.06\ \text{g}$
Step 4 — Percentage of carbon
$\%\,C = \frac{\text{mass of carbon}}{\text{mass of compound}}\times 100 = \frac{0.06}{0.48}\times 100$ $= \frac{6}{48}\times 100 = 0.125 \times 100 = 12.5\%$
Step 5 — Sanity check

The standard working formula for this experiment is …
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COMEDK 2023Set 2023-E1 markMCQ
$5.8 \mathrm{~g}$
›Reveal solutionSolution

Molar mass: M = mass / n = 5.8 / 0.145 = 40 g/mol

Concept: ideal gas equation, PV = nRT. Same V and same P for both gases, so nT = PV/R = constant:

n_gas x T_gas = n_H2 x T_H2

Hydrogen:

n_H2 = 0.368 / 2 = 0.184 mol, T_H2 = 17 C = 290 K

Unknown gas:

T = 95 C = 368 K, mass = 5.8 g …
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KCET 2021Set B-21 markMCQ
$300$
›Reveal solutionSolution

Apply the unit-cell density formula $\rho = ZM/(a^3N_A)$ with $Z=2$ for a body-centred cubic lattice and solve for $M$ .

Step 1 — The concept.

A crystal's macroscopic density is just the mass of one unit cell divided by its volume. A unit cell of edge $a$ contains $Z$ formula units, each of mass $M/N_A$ :
$\rho = \frac{Z\,(M/N_A)}{a^3} = \frac{Z M}{a^3 N_A}.$
Step 2 — Fix $Z$ for BCC.

BCC has $8$ corner atoms shared by $8$ cells each ( $8\times\tfrac18 = 1$ ) plus $1$ atom fully inside at the body centre. So
$Z = 1 + 1 = 2.$
Step 3 — Convert the edge length to cm (so it matches $\mathrm{g\,cm^{-3}}$ ):
$a = 300\,\text{pm} = 300\times10^{-12}\,\mathrm{m} = 3\times10^{-8}\,\mathrm{cm},$ $a^3 = (3\times10^{-8})^3 = 27\times10^{-24} = 2.7\times10^{-23}\,\mathrm{cm^3}.$
Step 4 — Solve for $M$ . …
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COMEDK 2021Set 2021-B1 markMCQ
$Q. In a given sample of air the ratio between the masses of O2 gas and N2 gas is 6 : 7. What would be the ratio of their moles? (A) Ratio of the molecules of O2:N2 = 2 : 3 (B) Ratio of the molecules of O2:N2 = 1 : 4 (C) Ratio of the molecules of O2:N2 = 2 : 5 (D) Ratio of the molecules of O2:N2 = 3 : 4$
›Reveal solutionSolution

Dividing the given masses by molar masses gives a mole (and molecule) ratio of 3 : 4.

Moles from mass:
$n_{O_2} = \frac{6}{32} = 0.1875, \qquad n_{N_2} = \frac{7}{28} = 0.25$
Ratio:

$\frac{n_{O_2}}{n_{N_2}} = \frac{0.1875}{0.25} = \frac{3}{4}$ …
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←Previous question1.8 — Determine the molecular formula of an oxide of iron, in which the mass…→Next question1.10 — In three moles of ethane (C₂H₆), calculate the following: (i) Number o…

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 weighted averages matter in atomic mass

Step-by-step calculation

Method: Weighted Average Method

Steps

Why this works …

✗ Mistake 1: Using the percentage as a decimal incorrectly

✗ Mistake 2: Using rounded mass numbers instead of given molar masses

✗ Mistake 3: Forgetting to add both contributions

✗ Mistake 4: Misreading the table (swapping columns)

Step 1 — The principle (Liebig's combustion method)

Step 2 — The carbon fraction of $CO_2$

Step 3 — Mass of carbon in the sample

Step 4 — Percentage of carbon

Step 5 — Sanity check

Why weighted averages matter in atomic mass

Step-by-step calculation

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