Relative masses of atoms and molecules
We define the mass of one atom of ^{12}\text{C} as exactly 12 units. Therefore, one-twelfth (1/12) of the mass of a ^{12}\text{C} atom is defined as 1 atomic mass unit.
This standard allows us to express the masses of all other atoms relative to this fixed value.
Definition: The relative atomic mass, A_r, is the weighted average mass of an atom of an element compared to 1/12th of the mass of an atom of carbon-12.
Key Concept: Weighted Average
Most elements exist as a mixture of isotopes (atoms with the same number of protons but different numbers of neutrons). Because isotopes have different masses, we cannot just pick one. Instead, we calculate an average that accounts for how common each isotope is.
- If an element has only one stable isotope (e.g., Fluorine), its A_r equals the mass number of that isotope.
- If an element has multiple isotopes, the A_r is closer to the mass number of the most abundant isotope.
Step 1: Multiply each isotope's mass number by its percentage abundance.
(24 \times 80) + (25 \times 10) + (26 \times 10)
Step 2: Add these values together.
1920 + 250 + 260 = 2430
Step 3: Divide by the total percentage (which is always 100).
A_r = \frac{2430}{100} = 24.3
Answer: The relative atomic mass of magnesium is 24.3.
The Correction: You must calculate the weighted average. Even if one isotope is dominant, the presence of heavier or lighter isotopes shifts the average slightly. Always perform the calculation unless the question states the element has only one isotope.
Context: When asked to define A_r in Paper 3 or 4.
Why Examiners Accept This: The definition is precise and relies on the specific standard. Vague terms like 'average mass' are insufficient without the reference point.
Correct Phrasing: "The weighted average mass of an atom of an element compared to 1/12th of the mass of an atom of carbon-12."
Example Usage:
- Incorrect: "The average mass of an atom."
- Correct: "The average mass of an atom compared to 1/12th of the mass of a carbon-12 atom."
Ionic Compounds: For ionic compounds (which form giant lattices rather than discrete molecules), we use the term Relative Formula Mass, but the symbol is still M_r. The calculation method is identical: sum the A_r values of all atoms in the formula unit.
Formula:
M_r = \sum A_r(\text{all atoms})
Step 1: Identify the number of each atom. The brackets ( )_2 mean everything inside is multiplied by 2.
- Ca: 1 atom
- N: 1 \times 2 = 2 atoms
- O: 3 \times 2 = 6 atoms
Step 2: Multiply each count by its A_r.
- Ca: 1 \times 40 = 40
- N: 2 \times 14 = 28
- O: 6 \times 16 = 96
Step 3: Sum the results.
M_r = 40 + 28 + 96 = 164
Answer: The relative formula mass is 164.
The Correction: Always expand the formula first. Write out the total count of each element before multiplying by A_r. For \text{Ca(NO}_3)_2, explicitly write CaN_2O_6 in your working.
Why Examiners Accept This: The examiner looks for the ratio of the total mass of that specific element to the total mass of the compound.
Correct Phrasing/Method:
\text{Percentage by mass} = \frac{\text{Total mass of element in formula}}{M_r \text{ of compound}} \times 100
Example Usage:
For \text{Fe}_2\text{O}_3 (A_r: Fe=56, O=16):
- Total mass of Fe = 2 \times 56 = 112
- M_r of \text{Fe}_2\text{O}_3 = (2 \times 56) + (3 \times 16) = 160
- Calculation: \frac{112}{160} \times 100 = 70%
Note: A common error is using the mass of one iron atom (56) instead of the total mass in the formula (112).
The Principle: The coefficients in a balanced equation tell us the ratio of moles. By multiplying these coefficients by the respective M_r values, we get the mass ratio.
\text{Mass Ratio} = \text{Coefficient} \times M_r
This mass ratio remains constant regardless of the scale of the reaction.
Step 1: Calculate the theoretical mass ratio using coefficients and M_r.
- Reactant Mg: Coefficient 2 \rightarrow Mass = 2 \times 24 = 48
- Product MgO: Coefficient 2 \rightarrow Mass = 2 \times (24+16) = 80
Step 2: Set up the proportion.
We know that 48 g of Mg produces 80 g of MgO.
Step 3: Calculate for the given mass (6.0 g).
\text{Mass of MgO} = \frac{80}{48} \times 6.0
\text{Mass of MgO} = 10.0 \text{ g}
Answer: 10.0 g of MgO is produced.
The Correction: Always use the full formula mass (M_r) of the species as it appears in the balanced equation, multiplied by its coefficient. Do not use A_r for diatomic molecules like \text{O}_2 or \text{Cl}_2.
Why Examiners Accept This: Theoretical calculations assume perfect conditions. Real reactions are rarely 100% efficient. The examiner expects you to adjust the theoretical mass.
Correct Phrasing/Method:
\text{Actual Mass} = \text{Theoretical Mass} \times \frac{\text{Percentage Yield}}{100}
Example Usage:
If the theoretical mass calculated in the previous example was 10.0 g, and the yield is 75%:
\text{Actual Mass} = 10.0 \times 0.75 = 7.5 \text{ g}
Note: Do not apply the percentage to the reactant mass first. Calculate the theoretical product mass first, then apply the yield.