Calculate the average atomic mass of hydrogen using the following data: Isotope $^1H$ — % Natural abundance: 99.985, Molar mass: 1 Isotope $^2H$ — % Natural abundance: 0.015, Molar mass: 2

Calculate the average atomic mass of hydrogen usin – Class 11 Chemistry

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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 average atomic mass of hydrogen is a weighted mean of its isotopes' masses, using their natural abundances as weights. The result is 1.00015 u.

Why a weighted average?

An element's atomic mass on the periodic table isn't the mass of a single atom — it's the average mass of all naturally occurring atoms of that element. Hydrogen exists as two stable isotopes: protium ( $^1H$ , mass ≈ 1 u) and deuterium ( $^2H$ , mass ≈ 2 u). Since protium is vastly more common (99.985% of all hydrogen atoms), the average should be very close to 1, but slightly higher because of the tiny fraction of heavier deuterium atoms.

The formula is straightforward:

\text{Average atomic mass} = \frac{\sum (\text{isotope mass} \times \text{\% abundance})}{100}

The division by 100 converts percentage to a decimal fraction.

Step-by-step calculation

Identify the data
- $^1H$ : mass = 1 u, abundance = 99.985%
- $^2H$ : mass = 2 u, abundance = 0.015%
Multiply each isotope's mass by its percentage abundance
- For $^1H$ : $1 \times 99.985 = 99.985$
- For $^2H$ : $2 \times 0.015 = 0.030$
Add these products

99.985 + 0.030 = 100.015

Divide by 100 (because abundances are in percent) $\frac{100.015}{100} = 1.00015$ …

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Common Mistakes & How to Avoid Them

Mistake 1: Using the wrong formula (averaging the masses directly)

The error:

Students often take a simple average:

\frac{1 + 2}{2} = 1.5

This is wrong because it ignores the fact that $^1H$ is far more abundant.

Why it's wrong:

Average atomic mass is a weighted average, not an arithmetic mean. The weight is the natural abundance (as a fraction or percentage).

How to avoid:

Always use the weighted average formula:

\text{Average atomic mass} = \frac{\sum (\text{isotopic mass} \times \text{\% abundance})}{100}

For this problem:

\frac{(1 \times 99.985) + (2 \times 0.015)}{100}

Mistake 2: Forgetting to convert percentage to decimal

The error:

Using abundances as decimals (0.99985 and 0.00015) but then not dividing by 100 — or vice versa.

Why it's wrong:

If you use percentages directly, you must divide by 100. If you use decimals, you don't divide again. Mixing these gives a wrong result.

How to avoid:

Pick one method and stick to it:

Method A (percentages):

\frac{(1 \times 99.985) + (2 \times 0.015)}{100}

Method B (decimals):

(1 \times 0.99985) + (2 \times 0.00015)

Both give the same answer: 1.00015 u

Mistake 3: Rounding too early

The error:

Rounding 99.985 to 100 or 0.015 to 0 before calculating.

Why it's wrong:

This destroys the precision. The contribution from $^2H$ (0.00015 u) is small but real — rounding it to zero gives exactly 1 u, which is incorrect.

How to avoid:

Carry all digits through the calculation. Round only at the final step to the required significant figures.

Mistake 4: Confusing "molar mass" with "mass number"

The error:

Using mass numbers (1 and 2) directly without realising they are approximate values of the actual isotopic masses.

Why it's wrong:

For hydrogen, the mass of $^1H$ is actually 1.007825 u, not exactly 1 u. The problem gives "molar mass: 1" as a simplification, but in real exam problems, using exact masses changes the answer slightly.

How to avoid:

If the problem gives exact masses (like 1.007825), use those. …

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

AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
$3\times10^{22}$
›Reveal solutionSolution
Converting each quantity to grams shows 0.5 mol of ozone (24 g) has the highest mass among the four options.

Concept and Intuition

This tests careful unit conversion between grams, moles, molecules, and STP volumes into a common basis (mass in grams) so they can be directly compared.

Step-by-Step Solution

(A) 0.5 g-atom of oxygen = 0.5 mol O atoms $\times$ 16 g/mol = 8 g.

(B) 0.5 mol of ozone (O₃, molar mass 48 g/mol) = $0.5\times48=24$ g.

(C) $3\times10^{22}$ molecules of N₂: moles $=\dfrac{3\times10^{22}}{6.022\times10^{23}}\approx0.0498$ mol $\times28\ \text{g/mol}\approx1.39$ g.

(D) 5.6 L CO₂ at STP: moles $=\dfrac{5.6}{22.4}=0.25$ mol $\times44\ \text{g/mol}=11$ g. …
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AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
$^{54}Fe$
›Reveal solutionSolution
Atomic mass is the abundance-weighted average of isotopic masses; for Fe here it works out to $55.85$ .

Concept and Intuition

The atomic mass listed on the periodic table is not any single isotope's mass — it is the average of all naturally occurring isotopes, weighted by how abundant each one is.

Step-by-Step Solution

Multiply each isotope's mass by its fractional abundance: $^{54}Fe: 0.10\times54=5.4$ ; $^{56}Fe: 0.85\times56=47.6$ ; $^{57}Fe: 0.05\times57=2.85$ .

Sum: $5.4+47.6+2.85=55.85$ . …
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AP EAPCET 2021Set ap-2021-10-05-FN1 markMCQ
$NO_2$
›Reveal solutionSolution
Using vapour density to get the average molar mass of the $NO_2$ / $N_2O_4$ mixture and a mole-fraction balance gives $0.437$ mol of $NO_2$ in $100$ g of mixture.

Concept and Intuition

Vapour density relates to molar mass by $M=2\times VD$ ; for a mixture of two related gases, the observed (average) molar mass is a mole-fraction-weighted average of the pure components' molar masses.

Step-by-Step Solution

Average molar mass of the mixture: $M_{avg}=2\times38.3=76.6\ g/mol$ .

Let $x$ = mole fraction of $NO_2$ ( $M=46$ ), so $(1-x)$ = mole fraction of $N_2O_4$ ( $M=92$ ): $46x+92(1-x)=76.6$ .

Solve: $92-46x=76.6\Rightarrow46x=15.4\Rightarrow x=0.3348$ .

Total moles of mixture in 100 g: $n=\dfrac{100}{76.6}=1.305\ mol$ . …
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AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
$6.643 \times 10^{-23}\ g$
›Reveal solutionSolution
Tests converting single-atom mass to molar mass via Avogadro's number, then finding moles in a bulk sample. Answer: 1250 moles.

Concept and Intuition

The mass of a single atom, multiplied by Avogadro's number ( $6.022\times10^{23}$ , the number of atoms in one mole), gives the molar mass of the element. Once we know the molar mass, converting a bulk mass into moles is a straightforward division.

Step-by-Step Solution

Molar mass $M = (\text{mass of one atom}) \times N_A = 6.643\times10^{-23}\ g \times 6.022\times10^{23}\ mol^{-1}$ .

$M \approx 6.643 \times 6.022 \approx 40.01\ g/mol$ (this is calcium, atomic mass 40). …
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AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
$Fe$
›Reveal solutionSolution
Equivalent weight of an element in a compound is its atomic mass divided by its valence (oxidation number) in that compound; for Fe(III) in $Fe_2O_3$ , this gives $56/3 \approx 18.6$ .

Concept and Intuition

Equivalent weight expresses how much mass of an element corresponds to a single 'unit of combining power' (one unit of charge/valence). It is defined as $\text{Equivalent weight} = \dfrac{\text{Atomic mass}}{\text{Valence}}$ . The valence to use is the oxidation state the element actually has in the given compound.

Step-by-Step Solution

In $Fe_2O_3$ , oxygen is $-2$ ; overall neutral, so $2\times(\text{Fe oxidation state}) + 3\times(-2) = 0 \Rightarrow$ Fe oxidation state $=+3$ .

Equivalent weight of Fe $= \dfrac{\text{Atomic mass of Fe}}{\text{valence}} = \dfrac{56}{3}$ . …
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AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
$3.011 \times 10^{22}$
›Reveal solutionSolution
Converting the given number of atoms to moles (via Avogadro's number) and dividing the given mass by that mole count gives an atomic mass of $23\ amu$ — consistent with sodium.

Concept and Intuition

The mole concept links a countable number of atoms to a measurable mass via Avogadro's number ( $6.022\times10^{23}$ per mole) and molar mass (mass per mole). Given both the atom count and the corresponding mass, we can directly compute the molar (atomic) mass.

Step-by-Step Solution

Convert atom count to moles:

$n = \frac{3.011\times10^{22}}{6.022\times10^{23}} = 0.05\ mol$

Atomic mass = mass per mole:

$M = \frac{\text{given mass}}{n} = \frac{1.15\ g}{0.05\ mol} = 23\ g/mol$

So the atomic mass is $23\ amu$ .

Common Mistakes …
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Q.Calculate the average atomic mass of hydrogen using the following data: Isotope 1H^1H1H — % Natural abundance: 99.985, Molar mass: 1 Isotope 2H^2H2H — % Natural abundance: 0.015, Molar mass: 2

Why a weighted average?

Step-by-step calculation

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Q.Calculate the average atomic mass of hydrogen using the following data: Isotope $^1H$ — % Natural abundance: 99.985, Molar mass: 1
Isotope $^2H$ — % Natural abundance: 0.015, Molar mass: 2

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