9.2.8 Relative Atomic Mass

This is the eighth lecture from Chapter 2: ‘Atomic Structure’ of the new book for Class 9 Chemistry (Punjab Board – PCTB). It discusses the relative atomic mass and its calculation from isotopic abundance. The lecture includes a multiple-choice quiz, short-answer questions, and detailed long-answer notes.

MCQs Based Quiz

1 / 12

What standard is used for measuring relative atomic masses?

Oxygen-16 atom

Hydrogen-1 atom

Carbon-12 atom

Nitrogen-14 atom

2 / 12

What fraction of carbon-12's mass equals 1 atomic mass unit (amu)?

1/10th

1/12th

1/14th

1/16th

3 / 12

The relative atomic mass of hydrogen-1 is approximately:

1.000 amu

1.007 amu

1.500 amu

2.000 amu

4 / 12

Why was a standardized atomic mass scale needed?

To count electrons accurately

To measure radioactivity

To identify radioactive isotopes

To compare masses of different atoms

5 / 12

If an element has a relative atomic mass of 32 amu, its mass is compared to:

1/12th of hydrogen-1

1/12th of carbon-12

1/16th of oxygen-16

1/14th of nitrogen-14

6 / 12

The atomic mass unit (amu) is defined in terms of:

Kilograms

Grams

Pounds

Moles

7 / 12

What practical benefit does the relative atomic mass scale provide in chemistry?

Predicting the reactivity of atoms

Determining the ratio of masses of reactants

Measuring temperature of reaction

Determining electron configuration

8 / 12

What information is needed to calculate relative atomic mass of an element?

Only the number of protons

Only the number of electrons

Relative isotopic masses and their abundances

9 / 12

If an element has two isotopes with masses 10 (20% abundance) and 11 (80% abundance), its relative atomic mass is:

5.32

0.99

1.21

10.8

10 / 12

The unit mass on relative atomic mass scale is ___ of the mass of C-12.

1/12th

12th

1/6th

6th

11 / 12

The mass of one protium atom on relative atomic mass scale is ___.

1.007

2.007

3.007

4.007

12 / 12

1 amu is equal to:

$1.67377 × 10^{-26} {kg}$

$1.67377 × 10^{-27} {kg}$

$1.67377 × 10^{-28} {kg}$

$1.67377 × 10^{-29} {kg}$

Your score is

The average score is 50%

Short Questions

Q1. Why is relative atomic mass necessary in chemical reactions?

This is because without knowing the relative atomic mass of the reactants, we would not be able to mix reactants in proper ratios of their masses to carry out chemical reactions.

Q2. How would you define relative atomic mass of an element?

Relative atomic mass of an element is the average of all of its atoms compared to the mass of C-12 atom taken as 12. It is expressed in atomic mass unit (amu).

(Correction: Relative atomic mass has no units.)

Q3. How would you define atomic mass unit?

Atomic mass unit (amu) is define as 1/12^th the mass of an atom of carbon-12 taken as 12.

$1 \text{ amu} = 1.67377 \times 10^{-27} \text{ kg}$

Q5. How would you compare the masses of the atoms of Mg and Cl?

We compare the masses of these atoms using their relative atomic masses. The relative atomic mass of magnesium (Mg) is 24.305, while that of chlorine (Cl) is 35.446. By comparing these values, we find that a chlorine atom is almost 1.46 times heavier than a magnesium atom.

Q6. What is relative isotopic mass?

The mass number of a specific isotope of an element is called relative isotopic mass.

(Correction: Relative isotopic mass is mass of that isotope not its mass number.)

Q7. What is meant by isotopic abundance?

The percentage of a particular isotope in a natural sample is called isotopic abundance.

Q8. How can we calculate relative atomic mass of an element?

To calculate the relative atomic mass of an element we would need the relative isotopic masses (m) and isotopic abundance (p).

$Relative Atomic Mass = {\Large \frac{m_1p_1\; +\; m_2p_2\; +\; m_3p_3\; +\; \cdots}{100}}$

Descriptive Question

Q1. Write an explanatory note on relative atomic mass and its calculation using isotopic abundance.

Relative Atomic Mass:

The relative atomic mass scale was adopted in 1961. It is based on the carbon-12 (C-12) isotope as a standard. In this scale:

The mass of one C-12 atom is set as 12.
The unit of mass is defined as 1/12th the mass of a C-12 atom.

So, the relative atomic mass of an element is the average mass of all of its atoms, compared to the mass of a C-12 atom taken as 12. Its unit is atomic mass unit (amu).

$1 \text{ amu} = 1.67377 \times 10^{-27} \text{ kg}$

(Correction: Relative atomic mass has no units.)

The mass of ¹H on this sale is 1.007 amu and the mass of ³²S is 31.972 amu.

The relative atomic mass scale was developed to:

To compare the masses of different atoms.
To measure the exact mass ratio of reactants involved in a chemical reaction.

Calculation of Relative Atomic Mass from Isotopic Abundance:

We calculate the relative atomic mass of an element using relative isotopic masses and isotopic abundances.

The mass number of a specific isotope of an element is called relative isotopic mass (m).

(Correction: Relative isotopic mass is mass of that isotope not its mass number.)

The percentage of a particular isotope in a natural sample is called isotopic abundance (p).

$Relative Atomic Mass = {\Large \frac{m_1p_1\; +\; m_2p_2\; +\; m_3p_3\; +\; \cdots}{100}}$

Q2. The element Krypton (Kr) has five isotopes. Their relative isotopic masses and isotopic abundances are given in the table below. Calculate the relative atomic mass of Krypton.

Relative Isotopic Mass

Isotopic Abundance

2.0%

12.0%

57.0%

17.0%

$\begin{aligned} \text{Relative atomic mass of krypton} &= \frac{80 \times 2.0 + 82 \times 12.0 + 83 \times 12.0 + 84 \times 57.0 + 86 \times 17.0}{100} \\ &= 83.9 \end{aligned}$

Q3. Calculate the relative atomic mass of light isotope of chlorine when its relative atomic mass is taken as 35.45.

Chlorine element has two isotopes: ³⁵Cl and ³⁷Cl. We are asked to find the relative atomic (isotopic) mass of ³⁵Cl.

$\begin{aligned}
\text{Isotopic abundance of Cl-37} &= 24.23\% \\
\text{Isotopic abundance of Cl-35} &= 75.77\% \\
\text{Relative atomic mass of chlorine element} &= 35.45 \\[10pt]
\text{Calculation:} \\
35.45 &= \frac{m_1 \times 75.77 + 37 \times 24.23}{100} \\
3545 &= m_1 \times 75.77 + 37 \times 24.23 \\
3545 &= 75.77m_1 + 896.51 \\
3545 – 896.51 &= 75.77m_1 \\
2648.49 &= 75.77m_1 \\
m_1 &= \frac{2648.49}{75.77} \\
m_1 &= \boxed{34.96}
\end{aligned}$

The relative atomic (isotopic) mass of Cl-35 is 34.96.

Q3. Calculate the relative atomic mass of light isotope of chlorine when its relative atomic mass is taken as 35.45.

$\begin{aligned}
\text{Given isotopic abundances:} \\
\text{Isotopic abundance of }^{204}\text{Pb} &= 2.0\% \\
\text{Isotopic abundance of }^{206}\text{Pb} &= 24.0\% \\
\text{Isotopic abundance of }^{207}\text{Pb} &= 22.0\% \\
\text{Isotopic abundance of }^{208}\text{Pb} &= 52.0\% \\[10pt]
\text{Calculation:} \\
\text{Relative atomic mass of Pb} &= \frac{(204 \times 2.0) + (206 \times 24.0) + (207 \times 22.0) + (208 \times 52.0)}{100} \\
&= \frac{408 + 4944 + 4554 + 10816}{100} \\&= \frac{20722}{100} \\
&= \boxed{207.22} \quad \text{(Relative atomic mass of lead)}
\end{aligned}$