Topic 1.4 Notes – Composition of Mixtures

1. Pure Substances vs Mixtures

Everything here starts with one question: Are we dealing with one type of particle, or more than one?

Pure Substance

A pure substance contains only one type of particle:

• Element → one type of atom (e.g., $\ce{Cu}$)
• Molecular compound → identical molecules (e.g., $\ce{CO2}$)
• Ionic compound → identical formula units (e.g., $\ce{NaCl}$)

Key idea:

• Fixed composition
• Every sample has the same ratio of elements.
• That means a constant percent by mass of each element.

This is the Law of Definite Proportions in action. If it’s truly a compound, the ratio does not change.

Mixture

A mixture contains two or more different types of particles physically combined.

• The substances are not chemically bonded to each other.
• Composition is variable.
• Percent by mass depends on how much of each substance you mixed together.

Two samples of the same mixture can have totally different mass percentages. That’s your red flag that it’s not a pure compound.

2. Types of Mixtures

The next question becomes: does it look uniform?

Homogeneous Mixtures

A homogeneous mixture (solution):

• Uniform throughout
• One visible phase
• Particles evenly distributed at the molecular level

Examples:

• Saltwater
• Air

You cannot see different parts. If you scoop from the top or bottom, composition is the same.

Heterogeneous Mixtures

A heterogeneous mixture:

• Not uniform
• Multiple visible phases
• Different parts are physically distinct

Examples:

• Sand + water
• Oil + water

Here’s a quick side-by-side comparison:

This distinction shows up in lab-based FRQs where you must choose a separation method.

3. Quantitative Composition by Mass

Now we connect particles to numbers.

Percent by Mass

$$\% \text{ by mass} = \frac{\text{mass of component}}{\text{total mass}} \times 100$$

In a Pure Compound

Percent composition comes directly from the formula.

Example: Suppose a compound has formula $\ce{MgCl2}$.

1. Find molar masses
   • Mg = 24.31 g/mol
   • Cl = 35.45 g/mol × 2 = 70.90 g/mol
2. Total molar mass = 95.21 g/mol
3. Percent Mg:

   $$\frac{24.31}{95.21} \times 100 \approx 25.5\%$$

Every pure sample of $\ce{MgCl2}$ will be 25.5% Mg by mass.

In a Mixture

If you mix 10 g salt and 90 g sand:

• Total mass = 100 g
• Salt = 10%

Change the masses and the percent changes. There’s no fixed ratio.

On tests, they often give experimental percent data and ask whether it matches a known compound. Small differences are experimental error. Big differences mean mixture or impurity.

4. Elemental Analysis and Determining Composition

Elemental analysis gives mass percent of each element in a sample.

From that, you can determine ratios of atoms.

Finding an Empirical Formula

If a compound is 40.0% C, 6.7% H, 53.3% O:

1. Assume 100 g → 40.0 g C, 6.7 g H, 53.3 g O
2. Convert to moles
   • C: $40.0/12.01 \approx 3.33$
   • H: $6.7/1.01 \approx 6.63$
   • O: $53.3/16.00 \approx 3.33$
3. Divide by smallest (3.33)
   • C: 1
   • H: 2
   • O: 1

Empirical formula = $\ce{CH2O}$

That mole ratio only works if it’s a pure compound.

Determining Purity

If experimental percent composition does not consistently match the theoretical percent for a known formula, the sample likely contains impurities or is a mixture.

If two samples of a “compound” have different mass ratios of elements, it cannot be a pure compound.

That logic shows up in reasoning-heavy multiple choice questions.

5. Separating Mixtures Based on Physical Properties

Mixtures can be separated because components differ in physical properties.

Filtration

• Separates insoluble solid from liquid.
• Works for heterogeneous mixtures.
• Based on particle size and phase difference.

Distillation

• Separates liquids using boiling point differences.
• Lower boiling point vaporizes first.
• Used for homogeneous liquid mixtures.

Thin-Layer Chromatography

TLC separates substances based on polarity and attraction to two phases.

• Stationary phase → usually polar silica
• Mobile phase → solvent moving up the plate

Key idea:

• Polar substances stick more to the polar plate and travel less.
• Nonpolar substances travel farther in a nonpolar solvent.

Retention factor:

$$Rf = \frac{\text{distance traveled by substance}}{\text{distance traveled by solvent front}}$$

• Always between 0 and 1.
• Used to compare substances.

On exams, they love giving two plates with different solvent heights. You must compare Rf values, not just raw distance.