Chemistry (0620) Study Notes: Relative Masses of Atoms and Molecules (Stoichiometry Section 3.2)

Hello future chemist! This chapter is your foundation for understanding how much substance you actually have and how much product you can make. Atoms are incredibly tiny, so we can't weigh them individually. Instead, we use a system of comparison called Relative Mass. Think of it like a universal scale where everything is measured against a standard reference point.

Mastering these definitions and calculations is crucial because they lead directly into all future stoichiometry topics (like the mole concept, even though we won't cover that here!).

1. The Universal Standard: Carbon-12

Since atoms are too small to weigh on a normal scale, scientists needed a fixed, stable reference atom to compare everything else to. They chose a specific isotope of carbon.

The Relative Mass Scale

The standard reference unit for measuring the mass of atoms is the mass of one atom of the Carbon-12 isotope (\({}^{12}\text{C}\)).

  • The mass of one Carbon-12 atom is defined as exactly 12 units.
  • The unit used for comparison is 1/12th of the mass of a single Carbon-12 atom.

Analogy: Imagine $\text{Carbon-12}$ is a chemistry currency worth 12 units. If an Oxygen atom weighs 16 of those units, its relative mass is 16. We are comparing weights, not measuring absolute grams.

Key Takeaway (The Most Important Rule):

Relative masses have NO units! They are ratios (comparisons). We write them simply as a number, like 1.0 or 35.5.

2. Relative Atomic Mass (\(A_r\))

The Relative Atomic Mass (\(A_r\)) is the key value you find on the Periodic Table.

Definition of \(A_r\) (Core Syllabus Requirement 3.2.1)

The relative atomic mass (\(A_r\)) of an element is the average mass of the isotopes of that element compared to 1/12th of the mass of an atom of $\text{Carbon-12}$ (\({}^{12}\text{C}\)).

  • Average Mass: We use the average mass because most elements have isotopes (atoms of the same element with different numbers of neutrons).
  • The $\text{A}_r$ takes into account the different masses of these isotopes and how common each one is (its abundance).

Example: Chlorine atoms exist mostly as $\text{Cl-35}$ and $\text{Cl-37}$. Since there is a lot more $\text{Cl-35}$, the average mass (the $\text{A}_r$) is $\mathbf{35.5}$, which is closer to 35.

Quick Review: Where do I find \(A_r\)?

You find the \(A_r\) for every element on the Periodic Table (it's usually the larger number, or the mass number/nucleon number).

3. Calculating Relative Atomic Mass from Isotopes (Extended Content)

For Extended students, you must be able to calculate the $\text{A}_r$ if you are given the masses and abundances of the isotopes (Syllabus 2.3.4).

Step-by-Step Calculation

If an element has two isotopes, Isotope A and Isotope B, the formula is:

$$A_r = \frac{(\text{Mass A} \times \% \text{Abundance A}) + (\text{Mass B} \times \% \text{Abundance B})}{100}$$

Example: Neon (\text{Ne})

Neon has two major isotopes:

  • \({}^{20}\text{Ne}\): Mass = 20.0, Abundance = 90%
  • \({}^{22}\text{Ne}\): Mass = 22.0, Abundance = 10%

Step 1: Multiply mass by percentage for each isotope.
\(\text{Isotope 1:} 20.0 \times 90 = 1800\)
\(\text{Isotope 2:} 22.0 \times 10 = 220\)

Step 2: Add the results together.
\(\text{Total mass contribution:} 1800 + 220 = 2020\)

Step 3: Divide by 100 to find the average.
\(\text{Average Relative Atomic Mass (}A_r\text{):} \frac{2020}{100} = 20.2\)

The Relative Atomic Mass of Neon is 20.2.

Common Mistake Alert!

Don't forget to divide by 100 at the end! Students often do the multiplication and addition but forget the final step that turns the 'total contribution' back into the 'average mass'.

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

When atoms join together to form molecules (covalent compounds) or ions join together (ionic compounds), we need to find the total relative mass of the resulting substance.

Definition of \(M_r\) (Core Syllabus Requirement 3.2.2)

The Relative Molecular Mass (\(M_r\)) (used for simple covalent molecules) or Relative Formula Mass (\(M_r\)) (used for ionic compounds) is simply the sum of the Relative Atomic Masses (\(A_r\)) of all the atoms shown in the formula.

We use the symbol $\mathbf{M_r}$ for both types of calculation in IGCSE Chemistry.

Calculating \(M_r\) Step-by-Step

To calculate $\text{M}_r$, you must know the chemical formula and the $\text{A}_r$ values (usually provided in the exam or found on the Periodic Table).

We will use the following standard \(A_r\) values for examples: H=1.0, C=12.0, O=16.0, Na=23.0, Cl=35.5.

Example 1: Water (\(\text{H}_2\text{O}\)) - A covalent molecule

  • Formula contains: 2 Hydrogen atoms (H) and 1 Oxygen atom (O).
  • Calculation:
    \(M_r = (2 \times A_r(\text{H})) + (1 \times A_r(\text{O}))\)
    \(M_r = (2 \times 1.0) + (1 \times 16.0)\)
    \(M_r = 2.0 + 16.0 = 18.0\)

Example 2: Sodium Chloride (\(\text{NaCl}\)) - An ionic compound (Relative Formula Mass)

  • Formula contains: 1 Sodium ion ($\text{Na}^+$) and 1 Chloride ion ($\text{Cl}^-$).
  • Calculation:
    \(M_r = (1 \times A_r(\text{Na})) + (1 \times A_r(\text{Cl}))\)
    \(M_r = (1 \times 23.0) + (1 \times 35.5)\)
    \(M_r = 23.0 + 35.5 = 58.5\)

Example 3: Magnesium Nitrate (\(\text{Mg}(\text{NO}_3)_2\)) - Complex ionic compound

  • Formula contains: 1 Magnesium ($\text{Mg}$), 2 Nitrogen ($\text{N}$) and \(2 \times 3 = 6\) Oxygen ($\text{O}$) atoms. (Assume $A_r$: $\text{Mg}=24.0$, $\text{N}=14.0$, $\text{O}=16.0$)
  • Calculation:
    $\text{M}_r = (1 \times 24.0) + (2 \times 14.0) + (6 \times 16.0)$
    $\text{M}_r = 24.0 + 28.0 + 96.0 = 148.0$
Key Takeaway for \(M_r\)

Always count every single atom in the formula, especially when subscripts or brackets are involved! The $M_r$ is a simple addition problem using the $A_r$ values.

5. Calculations of Reacting Masses in Simple Proportions (Core Content)

This section uses the relative masses you just calculated to work out how much of a reactant you need, or how much product you will get. Crucially, the IGCSE Core syllabus requires these calculations to be done using simple proportions without involving the mole concept (Syllabus 3.2.3).

This means we use the $\text{M}_r$ values to set up a ratio based on the balanced chemical equation.

Step-by-Step Reacting Mass Calculation

Scenario: Hydrogen gas reacts with Oxygen gas to form water.

$$\mathbf{2\text{H}_2} + \mathbf{\text{O}_2} \rightarrow \mathbf{2\text{H}_2\text{O}}$$

Question: If 8.0 g of Hydrogen gas ($\text{H}_2$) reacts completely, what mass of Water ($\text{H}_2\text{O}$) is produced?

(Use \(A_r\): H=1.0, O=16.0)

Step 1: Calculate the total relative mass for the substances involved in the ratio.

  • \(\text{Total } M_r \text{ for } \mathbf{2\text{H}_2}\text{:}\)
    \(\text{Mass of one } \text{H}_2 \text{ molecule} = 2 \times 1.0 = 2.0\)
    \(\text{Mass of two } \text{H}_2 \text{ molecules} = 2 \times 2.0 = \mathbf{4.0}\)
  • \(\text{Total } M_r \text{ for } \mathbf{2\text{H}_2\text{O}}\text{:}\)
    \(\text{Mass of one } \text{H}_2\text{O} \text{ molecule} = (2 \times 1.0) + 16.0 = 18.0\)
    \(\text{Mass of two } \text{H}_2\text{O} \text{ molecules} = 2 \times 18.0 = \mathbf{36.0}\)

Step 2: Set up the ratio based on the balanced equation.

The equation tells us that 4.0 mass units of $\text{H}_2$ produce 36.0 mass units of $\text{H}_2\text{O}$.

$$\frac{\text{Mass of } \text{H}_2}{\text{Mass of } \text{H}_2\text{O}} = \frac{4.0}{36.0}$$

Step 3: Use simple proportion to find the unknown mass.

We start with 8.0 g of $\text{H}_2$. Let \(x\) be the mass of water produced.

$$\frac{8.0 \text{ g } (\text{H}_2)}{x \text{ g } (\text{H}_2\text{O})} = \frac{4.0}{36.0}$$

To solve for \(x\), notice that the mass of $\text{H}_2$ (8.0 g) is exactly double the theoretical mass (4.0 units). Therefore, the mass of $\text{H}_2\text{O}$ must also be double the theoretical mass:

$$x = 36.0 \times 2 = \mathbf{72.0 \text{ g}}$$

72.0 g of water are produced.

Did you know? (Connection to real life)

Stoichiometry is used in all industrial chemical processes, from making fertilizer to producing plastics. Companies need these calculations to ensure they use the minimum amount of expensive raw materials and maximize their product yield, saving vast amounts of money!

Summary of Reacting Masses

The relative masses obtained from the balanced equation give you the fundamental mass ratio. You can use this ratio to scale up (or scale down) for any given mass in the question.

Tip for Struggling Students: Always start by writing the balanced equation and then listing the $\text{M}_r$ below the formulas involved. This clearly shows the ratio you must use!