Percentage Composition by Mass for HSC Chemistry

A 20 g sample contains 5 g of salt. Another 100 g sample contains 15 g of salt. Which one is “saltier”?

The second sample contains more salt overall, but that isn’t enough information. What matters is the fraction of each sample that is salt.

For the first sample:

\[
\frac{5}{20}\times 100 = 25\%
\]

For the second:

\[
\frac{15}{100}\times 100 = 15\%
\]

So the smaller sample is actually saltier. Percentage composition by mass lets us make this comparison fairly. It tells us how much of the total mass comes from one component.

The calculation itself is usually easy. The harder part in HSC Chemistry is deciding what belongs on top of the fraction, and what belongs on the bottom.

01The basic idea: part of the mass divided by total mass

Imagine a 200 g bag of trail mix containing 50 g of almonds.

The almonds make up:

\[
\frac{50}{200}\times100 = 25\%
\]

Chemistry uses exactly the same idea.

\[
\text{percentage by mass}=
\frac{\text{mass of component}}{\text{total mass}}
\times100
\]

The component is whatever the question asks about.

The total mass is the mass of the entire substance or mixture being considered.

That gives us the most useful decision rule in this topic:

Numerator = the thing you want. Denominator = the whole thing.

Before going further, predict this:

A 40 g mixture contains 8 g of copper. Should the denominator be 8 g or 40 g?

It must be 40 g, because 40 g is the mass of the whole mixture.

\[
\frac{8}{40}\times100=20\%
\]

So copper makes up 20% of the mixture by mass.

02There are two versions of the same calculation

HSC questions commonly ask percentage composition in two different situations.

The underlying idea is identical. Only the way you obtain the masses changes.

03Percentage composition of an element in a pure compound

Suppose you are asked:

What percentage of water’s mass comes from oxygen?

You usually aren’t given the mass of an actual water sample. Instead, you use the chemical formula, \(\ce{H2O}\), and relative atomic masses.

Using approximately:

• H = \(1.008\)
• O = \(16.00\)

One mole of water contains:

\[
2(1.008)+16.00=18.016\text{ g mol}^{-1}
\]

Of that \(18.016\text{ g}\), oxygen contributes \(16.00\text{ g}\).

Therefore:

\[
\%\text{ O}=
\frac{16.00}{18.016}\times100
=88.81\%
\]

This works because a chemical formula tells us the fixed ratio of atoms in a pure compound.

It doesn’t matter whether you have 18 g of water or 18 tonnes of water. In pure \(\ce{H2O}\), the proportion of the mass contributed by oxygen is always the same.

Quick check

What percentage of \(\ce{NaCl}\) must therefore be chlorine?

Because sodium chloride contains only sodium and chlorine:

\[
100\%-39.34\%=60.66\%
\]

So:

\[
\boxed{60.66\%\text{ Cl by mass}}
\]

For a pure compound, the percentages of all elements should add to approximately 100%. Small differences can appear because of rounding.

04Watch the subscripts carefully

The most common mistake in compound questions is forgetting that a subscript changes how many atoms contribute to the molar mass.

Consider \(\ce{CaCl2}\).

It contains:

• one Ca atom
• two Cl atoms

So the chlorine contribution is not \(35.45\). It is:

\[
2(35.45)=70.90
\]

The subscript is basically chemistry saying, “Yes, you need two of these.”

05Percentage composition in a mixture

Now suppose the substances are physically mixed rather than chemically bonded.

For example, a sample contains:

• 12 g of sodium chloride
• 48 g of water

The total mass of the mixture is:

\[
12+48=60\text{ g}
\]

The percentage by mass of sodium chloride is therefore:

\[
\frac{12}{60}\times100=20\%
\]

So the mixture is:

\[
\boxed{20\%\text{ NaCl by mass}}
\]

In this case, you do not need atomic masses. The masses of the substances themselves are already given.

This distinction is worth remembering:

• For a compound, the chemical formula tells you how the mass is divided between elements.
• For a mixture, the measured masses tell you how the total mass is divided between components.

If you need to refresh the distinction between different types of mixtures, see tutorgum’s guides to homogeneous mixtures and heterogeneous mixtures.

06The denominator trap

Suppose a mixture contains 10 g of ethanol and 90 g of water.

A student calculates:

\[
\frac{10}{90}\times100=11.1\%
\]

That looks reasonable, but it is wrong.

Why?

Because 90 g is the mass of the water, not the mass of the whole mixture.

The total mass is:

\[
10+90=100\text{ g}
\]

Therefore:

\[
\%\text{ ethanol}
=
\frac{10}{100}\times100
=
10\%
\]

The student’s calculation actually compared ethanol mass with water mass. That is a mass ratio, not percentage composition by mass.

Use this check:

If the denominator does not represent the whole sample, you are probably not calculating a percentage composition.

07Don’t confuse mass percentage with percentage of atoms

Consider \(\ce{CO2}\).

There is one carbon atom and two oxygen atoms. You might be tempted to say oxygen makes up:

\[
\frac{2}{3}\times100=66.7\%
\]

But that is the percentage of atoms, not percentage by mass.

Carbon and oxygen atoms don’t have the same mass.

Using C = 12.01 and O = 16.00:

\[
M(\ce{CO2})
=
12.01+2(16.00)
=
44.01\text{ g mol}^{-1}
\]

Oxygen contributes:

\[
32.00\text{ g mol}^{-1}
\]

Therefore:

\[
\%\text{ O}
=
\frac{32.00}{44.01}\times100
=
72.71\%
\]

So although oxygen accounts for two out of every three atoms in \(\ce{CO2}\), it accounts for about 72.71% of the mass.

Check your understanding

Which calculation gives the percentage by mass of hydrogen in \(\ce{NH3}\)?

A. \(\frac{3}{4}\times100\)

B. \(\frac{3(1.008)}{14.01+3(1.008)}\times100\)

C. \(\frac{1.008}{14.01}\times100\)

Answer: B.

The three hydrogen atoms contribute \(3(1.008)\) to the molar mass, and the denominator must be the molar mass of the entire compound.

\[
\%\text{ H}
=
\frac{3.024}{17.034}\times100
\approx17.75\%
\]

Option A counts atoms rather than mass. Option C compares one hydrogen atom with one nitrogen atom rather than comparing hydrogen’s total mass contribution with the whole compound.

08What if the question gives percentages and asks for a mass?

The same relationship can be rearranged.

Suppose a 250 g mixture is 12% sodium chloride by mass. How much sodium chloride does it contain?

Start from:

\[
\text{percentage by mass}
=
\frac{\text{mass of component}}{\text{total mass}}\times100
\]

Convert 12% to a decimal fraction:

\[
12\%=0.12
\]

Then:

\[
m_{\text{component}}
=
0.12\times250\text{ g}
=
30\text{ g}
\]

So the mixture contains:

\[
\boxed{30\text{ g NaCl}}
\]

You can think of this in ordinary language:

12% of 250 g = 30 g.

Practice question

A mineral sample is 35% calcium carbonate by mass. What mass of calcium carbonate is present in a 480 g sample?

Step 1: Convert the percentage to a decimal

\[
35\%=0.35
\]

Step 2: Multiply by the total mass

\[
m(\ce{CaCO3})
=
0.35\times480
=
168\text{ g}
\]

09A reliable HSC method

When you see a percentage composition question, don’t immediately start punching numbers into the calculator.

First decide which situation you have.

If it is a pure compound

1. Write the chemical formula.
2. Count the atoms of each required element.
3. Calculate the molar mass of the entire compound.
4. Calculate the mass contribution from the required element.
5. Use:

\[
\%\text{ element}
=
\frac{\text{element’s mass contribution}}{\text{molar mass of compound}}
\times100
\]

If it is a mixture

1. Identify the component asked for.
2. Find its mass.
3. Find the total mass of the mixture.
4. Use:

\[
\%\text{ component}
=
\frac{\text{mass of component}}{\text{total mass of mixture}}
\times100
\]

The mathematics is almost always the easy part. Correctly identifying the part and the whole is what prevents most mistakes.

10One final mixed problem

A 75.0 g mixture contains 18.0 g of pure calcium carbonate, \(\ce{CaCO3}\). What percentage of the entire mixture’s mass is oxygen?

There are two layers here.

First, only part of the mixture is calcium carbonate. Second, only part of the calcium carbonate’s mass is oxygen.

Predict the strategy before reading on.

You could calculate the percentage of oxygen in \(\ce{CaCO3}\), use that to find the mass of oxygen in 18.0 g of \(\ce{CaCO3}\), and then compare that oxygen mass with the full 75.0 g mixture.

That is exactly the right approach.

Step 1: Calculate the molar mass of calcium carbonate

Using Ca = 40.08, C = 12.01, and O = 16.00:

\[
M(\ce{CaCO3})
=
40.08+12.01+3(16.00)
=
100.09\text{ g mol}^{-1}
\]

Step 2: Calculate the fraction of calcium carbonate that is oxygen

Oxygen contributes:

\[
3(16.00)=48.00\text{ g mol}^{-1}
\]

So the mass fraction of oxygen is:

\[
\frac{48.00}{100.09}=0.4796
\]

Step 3: Find the mass of oxygen in 18.0 g of calcium carbonate

\[
m(\ce{O})
=
18.0\times0.4796
=
8.63\text{ g}
\]

Step 4: Compare that oxygen mass with the entire mixture

\[
\%\text{ O in mixture}
=
\frac{8.63}{75.0}\times100
=
11.5\%
\]