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Mole Calculations

CSCA Mole Calculations study guide organized around the publicly available CSCA syllabus. Practice Chemistry questions on aicsca.com.

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Core Concepts: Mole Calculations

This section is the "heart" of chemical calculations. Whether the question asks for mass, volume, or particle count, the path to the solution almost always requires **finding the Amount of Substance ($n$) first**. Remember the golden rule: **"Convert everything to Moles first."**

#### 1. Core Concepts: The Bridge between Micro and Macro

* **Amount of Substance ($n$)**: Measures the number of microscopic particles (atoms, molecules, ions). Unit: **Mole (mol)**.

* **Avogadro Constant ($N_A$)**: The number of particles in $1 \text{ mol}$. Value $\approx \mathbf{6.02 \times 10^{23} \text{ mol}^{-1}}$.

* *Analogy*: Just as "1 Dozen" = 12, "1 Mole" = $6.02 \times 10^{23}$.

* **Molar Mass ($M$)**: The mass of $1 \text{ mol}$ of a substance. Numerically equal to the relative atomic/molecular mass. Unit: **g/mol**.

* *Example*: Relative molecular mass of $H_2O$ is 18, so $M(H_2O) = 18 \text{ g/mol}$.

#### 2. The Four Key Formulas (The Mole Network)

Place Amount of Substance ($n$) in the center and connect it to four physical quantities. Mastering these conversions is key to CSCA calculation questions.

1. **Mass $\leftrightarrow$ Moles**:

$$n = \frac{m}{M}$$

($m$: Mass, in **g**)

2. **Particle Count $\leftrightarrow$ Moles**:

$$n = \frac{N}{N_A}$$

($N$: Number of particles)

3. **Gas Volume $\leftrightarrow$ Moles (STP Only)**:

$$n = \frac{V}{V_m}$$

* **STP Definition**: $0^\circ\text{C} (273.15 \text{ K})$ and $101.3 \text{ kPa} (1 \text{ atm})$.

* **Value**: $V_m \approx 22.4 \text{ L/mol}$.

* *Trap*: If the substance is a **Liquid** (like water) or **Solid**, or under **Non-Standard Conditions**, **NEVER use 22.4**!

4. **Solution Concentration $\leftrightarrow$ Moles**:

$$n = c \cdot V_{\text{solution}}$$

($c$: Molarity in mol/L; $V$: Volume in **L**)

#### 3. The "Three-Step" Strategy

**Example**: At STP, $11.2 \text{ L}$ of $CO_2$ is passed into excess limewater. What is the mass of the precipitate formed?

* **Step 1: Known to Moles**

Given $V(CO_2) = 11.2 \text{ L}$ (STP).

$$n(CO_2) = \frac{11.2 \text{ L}}{22.4 \text{ L/mol}} = 0.5 \text{ mol}$$

* **Step 2: Use Stoichiometry (The Ratio)**

Reaction: $CO_2 + Ca(OH)_2 \rightarrow CaCO_3\downarrow + H_2O$

Ratio $n(CO_2) : n(CaCO_3) = 1 : 1$.

Therefore, $n(CaCO_3) = 0.5 \text{ mol}$.

* **Step 3: Moles to Target**

Target is mass $m$. $M(CaCO_3) = 100 \text{ g/mol}$.

$$m = n \cdot M = 0.5 \text{ mol} \times 100 \text{ g/mol} = 50 \text{ g}$$

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#### Common Traps

1. **Unit Trap**: In $n=m/M$, $m$ must be in **grams**. In $n=c\cdot V$, $V$ must be in **Liters** (divide mL by 1000).

2. **Condition Trap**: If the problem states "Room Temperature", gas molar volume is usually NOT 22.4 L/mol (often 24 or 24.5). Check the given data or use $PV=nRT$.

3. **State Trap**: $H_2O$ at STP ($0^\circ\text{C}$) is liquid/ice, **NOT a gas**. You cannot use 22.4 L/mol for water.