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CIE iGCSE Chemistry Relative masses of atoms and molecules Study Notes

Relative masses of atoms and molecules- CIE iGCSE Chemistry Notes - New Syllabus

Relative masses of atoms and molecules Notes for iGCSE

Core Syllabus

  • Describe relative atomic mass, Ar , as the average mass of the isotopes of an element compared to 1/12th of the mass of an atom of 12C
  •  Define relative molecular mass, Mr , as the sum of the relative atomic masses. Relative formula mass, Mr , will be used for ionic compounds
  •  Calculate reacting masses in simple proportions. Calculations will not involve the mole concept

iGCSE Chemistry Notes – All Topics

Relative Atomic Mass and Relative Molecular Mass

Relative Atomic Mass (\( A_r \))

The relative atomic mass of an element, denoted by \( A_r \), is the average mass of all the naturally occurring isotopes of that element, compared to 1/12th the mass of a carbon-12 atom.

Important Points:

  • It has no units (it is a ratio).
  • It is a weighted average based on the abundance of each isotope.
  • The standard used is the carbon-12 isotope, which is exactly 12.

Why is it an average?

Most elements have more than one isotope. Each isotope has a different mass, and the average depends on how abundant each isotope is.

Formula to calculate \( A_r \):

\( A_r = \frac{\text{(mass of isotope 1 × % abundance) + (mass of isotope 2 × % abundance) + …}}{100} \)

Example:
Chlorine has two isotopes:

    • Cl-35 (about 75%)
    • Cl-37 (about 25%)

Calculate \( A_r \) of chlorine:

\( A_r = \frac{(35 × 75) + (37 × 25)}{100} = \frac{2625 + 925}{100} = 35.5 \)

So, the relative atomic mass of chlorine is 35.5

Reminder:

Values of \( A_r \) are found on the Periodic Table, usually as decimal numbers (e.g. H = 1.0, C = 12.0, Cl = 35.5, Mg = 24.3)

Relative Molecular Mass and Relative Formula Mass (\( M_r \))

The relative molecular mass, \( M_r \), of a molecule is the sum of the relative atomic masses (\( A_r \)) of all atoms in the molecule.

The relative formula mass, also written as \( M_r \), is used for ionic compounds and is calculated in the same way — by adding the \( A_r \) values of the ions present in the formula unit.

Key points:

  • \( M_r \) has no units.
  • It is calculated using the atomic masses from the Periodic Table.
  • Molecular compounds use “relative molecular mass”, while ionic compounds use “relative formula mass”.

Examples:

1. Molecular compound: Water \( \text{H}_2\text{O} \)

\( M_r = (2 × A_r \text{ of H}) + (1 × A_r \text{ of O}) = (2 × 1.0) + (1 × 16.0) = 18.0 \)

2. Ionic compound: Sodium chloride \( \text{NaCl} \)

\( M_r = A_r \text{ of Na} + A_r \text{ of Cl} = 23.0 + 35.5 = 58.5 \)

3. Covalent molecule: Carbon dioxide \( \text{CO}_2 \)

\( M_r = A_r \text{ of C} + (2 × A_r \text{ of O}) = 12.0 + (2 × 16.0) = 44.0 \)

Example

A compound contains chlorine, hydrogen, and oxygen. It is called chloric acid, with the formula \( \text{HClO}_3 \).

Given:

  • Chlorine has two isotopes: Cl-35 (75%) and Cl-37 (25%)
  • Relative atomic masses: H = 1, O = 16

Calculate the following:

  1. The relative atomic mass (\( A_r \)) of chlorine
  2. The relative molecular mass (\( M_r \)) of chloric acid \( \text{HClO}_3 \)
  3. The formula mass of an ionic compound: sodium carbonate \( \text{Na}_2\text{CO}_3 \)
▶️Answer/Explanation

1. Relative Atomic Mass of Chlorine:
\( A_r = \frac{(75 × 35) + (25 × 37)}{100} = \frac{2625 + 925}{100} = 35.5 \)

2. Relative Molecular Mass of \( \text{HClO}_3 \):
\( M_r = (1 × 1) + 35.5 + (3 × 16) = 1 + 35.5 + 48 = 84.5 \)

3. Relative Formula Mass of \( \text{Na}_2\text{CO}_3 \):
\( M_r = (2 × 23) + 12 + (3 × 16) = 46 + 12 + 48 = 106 \)

Calculating Reacting Masses in Simple Proportions

Calculating Reacting Masses in Simple Proportions

Calculation of the mass of a product or reactant using a balanced chemical equation and the given mass of another substance can be done using the formula mass (Mr) without involving the mole concept.

Steps:

  1. Write a balanced chemical equation.
  2. Use relative formula masses (\( M_r \)) to find the mass ratio from the equation.
  3. Set up a proportion using the known and unknown masses.

Key Formula (Ratio Method):

If you know the mass of substance A and need the mass of substance B:

Mass of B = \( \frac{\text{Mass of A} × M_r\text{(B)} × \text{coeff. of B}}{M_r\text{(A)} × \text{coeff. of A}} \)

Example

Hydrogen reacts with oxygen to form water:
\( 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \)

If 4 g of hydrogen reacts, calculate the mass of water produced.

▶️Answer/Explanation

Step 1: Find \( M_r \) values:

\( \text{H}_2 = 2 \) (2 × 1)

\( \text{H}_2\text{O} = 18 \) (2 + 16)

Step 2: According to the equation:
2 g of H₂ → 18 g of H₂O
So, 4 g of H₂ → \( \frac{18 × 4}{2} = 36 \) g of H₂O

Answer: 36 g of water is formed.

Example

Zinc reacts with hydrochloric acid:
\( \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \)

Calculate the mass of zinc chloride formed when 6.5 g of zinc reacts completely.
(Relative atomic masses: Zn = 65, Cl = 35.5)

▶️Show Answer

\( M_r(\text{Zn}) = 65 \)
\( M_r(\text{ZnCl}_2) = 65 + (2 × 35.5) = 136 \)

65 g Zn → 136 g ZnCl₂
So, 6.5 g Zn → \( \frac{136 × 6.5}{65} = 13.6 \) g ZnCl₂

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