The Mistake Bank
Chapter 1: Some Basic Concepts of Chemistry
Don't lose marks to silly errors. Learn from the "Graveyard of Marks".
The "Limiting Reagent" Trap
StoichiometryScenario: You are given 5 moles of \(A\) and 6 moles of \(B\) for the reaction:
$$ 2A + 3B \rightarrow A_2B_3 $$
Which is the Limiting Reagent (LR)?
Student compares moles directly:
$$ 5 \text{ moles } A < 6 \text{ moles } B $$
Student assumes A is the LR simply because the number is smaller.
Divide Moles by Coefficient!
- For A: \( \frac{5}{2} = \mathbf{2.5} \)
- For B: \( \frac{6}{3} = \mathbf{2.0} \)
Since \( 2.0 < 2.5 \), B is the actual Limiting Reagent.
Molarity vs. Molality
ConcentrationScenario: Calculate the molality of a solution containing 10g \(NaOH\) in 500mL of solution (density = 1.2 g/mL).
Student divides moles of solute by the Volume of Solution (in Liters).
$$ M = \frac{n}{V_{solution}} $$
Result: Calculates Molarity instead of Molality.
$$ m = \frac{\text{Moles}}{\text{Mass of Solvent (kg)}} $$
You must subtract solute mass from solution mass.
\( \text{Mass}_{soln} = 500 \times 1.2 = 600g \)
\( \text{Mass}_{solvent} = 600 - 10 = 590g \)
The Vapour Density Blunder
Molecular MassScenario: A question gives the Vapour Density (VD) of a gas as 22. Find the mass of 0.5 moles.
Student uses 22 as the Molar Mass directly.
$$ \text{Mass} = 0.5 \times 22 = 11g $$
(Wrong Answer)
Remember the relationship:
$$ \text{Molar Mass} = 2 \times \text{VD} $$
First, convert VD: \( 2 \times 22 = 44 \text{ g/mol} \) (likely \( CO_2 \)).
Then calculate mass: \( 0.5 \times 44 = \mathbf{22g} \).
The "Water at STP" Trick
Molar VolumeScenario: Find the volume occupied by 1 mole of \(H_2O\) at STP conditions.
Student blindly applies the gas law constant:
$$ V = 1 \text{ mole} \times 22.4 \text{ L} = 22.4 \text{ L} $$
This is the most common negative marking trap!
22.4 L is ONLY for Gases!
Water is a LIQUID at STP.
$$ \text{Mass} = 18g $$
$$ \text{Density} \approx 1 g/mL $$
$$ \text{Volume} = \frac{18g}{1 g/mL} = 18 \text{ mL} $$
Significant Figures in Addition
PrecisionScenario: Calculate \( 12.11 + 18.0 + 1.012 \).
Student adds them up to get \( 31.122 \).
Then counts total significant figures (often 4) and rounds to \( 31.12 \).
This rule applies to multiplication, not addition.
Look at Decimal Places (DP), not Sig Figs!
- 12.11 (2 DP)
- 18.0 (1 DP) ← Least Precise
- 1.012 (3 DP)
Sum = 31.122. Round to 1 decimal place.
Correct Answer: 31.1
Average Atomic Mass
IsotopesScenario: Chlorine exists as \( ^{35}Cl \) (75%) and \( ^{37}Cl \) (25%). Find the atomic mass.
Student takes a simple average of the mass numbers:
$$ \frac{35 + 37}{2} = 36 \text{ u} $$
This ignores natural abundance.
Use Weighted Average!
$$ \text{Avg Mass} = \frac{(35 \times 75) + (37 \times 25)}{100} $$
$$ = \frac{2625 + 925}{100} = \mathbf{35.5 \text{ u}} $$
Mixing Solutions
DilutionScenario: 1 L of 2M HCl is mixed with 2 L of 0.5M HCl. What is the resulting Molarity?
Student adds the molarities directly:
$$ 2M + 0.5M = 2.5M $$
Concentration is an intensive property; it cannot be added directly.
Use the Mixing Formula:
$$ M_{mix} = \frac{M_1V_1 + M_2V_2}{V_1 + V_2} $$
$$ = \frac{(2 \times 1) + (0.5 \times 2)}{1 + 2} $$
$$ = \frac{2 + 1}{3} = \frac{3}{3} = \mathbf{1M} $$
Confess Your Sins!
"The only real mistake is the one from which we learn nothing."
Did one of these catch you? Or do you have a different horror story from your last exam?
Scroll down to the comments section below and tell us:
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