Q.Calculate the mole fraction of ethylene glycol ($C_2H_6O_2$) in a solution containing 20% of $C_2H_6O_2$ by mass.

Concept: Mass Percentage → Mole Fraction

Mass percentage gives the mass of solute per 100 g of solution. Convert masses to moles, then use the mole fraction formula.

Step 1 – Masses from percentage

In 100 g of solution:

Mass of $C_2H_6O_2$ = 20 g

Mass of water = 80 g

Step 2 – Moles of each component

Molar mass of $C_2H_6O_2$ = $2(12) + 6(1) + 2(16) = 62$ g/mol

Moles of ethylene glycol = $\frac{20}{62} = 0.3226$ mol

Molar mass of water = 18 g/mol

Moles of water = $\frac{80}{18} = 4.444$ mol

Step 3 – Mole fraction

Step 4 – Mole fraction of water

Since the mole fractions must sum to 1:

Why mass percentage works as a starting point

When a problem says "20% by mass," it means that in every 100 grams of solution, 20 grams are the solute (ethylene glycol) and the remaining 80 grams are the solvent (water). This is the most direct way to get actual masses without any extra information. The mole fraction asks for the ratio of moles of one component to the total moles of all components — so we need to convert these masses into moles using molar masses.

Step-by-step calculation

1. Find the molar masses

Ethylene glycol is $C_2H_6O_2$.

Carbon: $2 \times 12 = 24$

Hydrogen: $6 \times 1 = 6$

Oxygen: $2 \times 16 = 32$

Molar mass of glycol = $24 + 6 + 32 = 62 \text{ g/mol}$.

Water is $H_2O$: $2 \times 1 + 16 = 18 \text{ g/mol}$.

2. Determine the masses in 100 g of solution

Mass of glycol = 20 g

Mass of water = 80 g

3. Convert masses to moles

Moles of glycol:

$\displaystyle \frac{20}{62} = 0.3226 \text{ mol}$ (approximately)

Moles of water:

$\displaystyle \frac{80}{18} = 4.4444 \text{ mol}$ (approximately)

4. Calculate total moles

Total moles = $0.3226 + 4.4444 = 4.7670 \text{ mol}$

5. Find the mole fraction of glycol

Mole fraction of glycol = $\displaystyle \frac{\text{moles of glycol}}{\text{total moles}} = \frac{0.3226}{4.7670} = 0.0677$

Rounding to three significant figures gives 0.068.

6. Find the mole fraction of water

Because the mole fractions of all components of a solution add up to 1, we can find the mole fraction of water in the same way:

Mole fraction of water = $\displaystyle \frac{\text{moles of water}}{\text{total moles}} = \frac{4.4444}{4.7670} = 0.932$

As a quick check, it can also be obtained directly from the glycol value:

$x_{\text{water}} = 1 - x_{\text{glycol}} = 1 - 0.068 = 0.932$, confirming that the two mole fractions sum to 1.

Method: Mass-to-Mole Conversion via Mass Percentage

This is a mass percentage → mole fraction problem. The key insight: mass percentage gives you a ratio by mass, and mole fraction requires a ratio by moles — so you must convert mass to moles using molar masses.

Steps

Step 1: Assume a convenient sample mass

Since the solution is 20% ethylene glycol by mass, take 100 g of solution.

• Mass of $C_2H_6O_2$ = $20\%$ of 100 g = 20 g
• Mass of water (solvent) = $100 - 20 = \textbf{80 g}$

Step 2: Calculate moles of each component

Molar mass of $C_2H_6O_2$:

$2(12) + 6(1) + 2(16) = 24 + 6 + 32 = 62 \text{ g/mol}$

Moles of ethylene glycol:

Molar mass of water ($H_2O$): $18 \text{ g/mol}$

Moles of water:

Step 3: Apply mole fraction formula

Mole fraction of ethylene glycol:

Substitute:

Step 4: Compute final result

Final Answer:

Why this works

• Mass percentage gives a fixed ratio by mass, so any sample size yields the same mole fraction.
• Choosing 100 g avoids decimals in the mass values and simplifies calculation.
• The conversion from mass to moles is the critical bridge between the two types of concentration units.

Here are the most common mistakes students make when solving this exact problem, along with the concept-first reasoning to avoid each.

Mistake 1: Confusing “20% by mass” with “20 g in 100 mL”

The error:

Students assume 20% by mass means 20 g of solute in 100 mL of solution. This is wrong — mass percentage is mass of solute per 100 g of solution, not per 100 mL.

How to avoid:

Always read “X% by mass” as:

X g of solute in 100 g of solution.

So here:

• Mass of ethylene glycol = 20 g
• Mass of water = 100 g – 20 g = 80 g

Mistake 2: Using the wrong molar mass

The error:

Students miscalculate the molar mass of $C_2H_6O_2$ (ethylene glycol). Common slip-ups:

• Forgetting the two oxygen atoms
• Using atomic masses incorrectly (e.g., C = 12, H = 1, O = 16 — correct, but adding wrong)

How to avoid:

Write the formula clearly and sum step-by-step:

Molar mass of water = $18 \text{ g/mol}$.

Mistake 3: Swapping solute and solvent in mole fraction formula

The error:

Mole fraction of solute =

Students sometimes put solvent moles in the numerator.

How to avoid:

Remember: mole fraction is always “part over whole” — the part you want divided by total moles of all components.

Mistake 4: Forgetting to convert mass to moles

The error:

Plugging masses directly into the mole fraction formula.

How to avoid:

Always convert mass → moles first using:

For this problem:

Mistake 5: Rounding too early

The error:

Rounding intermediate values (e.g., $20/62 \approx 0.32$) leads to an inaccurate final answer.

How to avoid:

Keep at least 4 decimal places during calculation. Round only at the final step.

✓ Correct final answer (for reference)

Final answer: $\boxed{0.068}$ (rounded to 3 decimal places)

Quick checklist to avoid all mistakes