Percent Composition by Mass

A chemical formula like H₂O₂ is a powerful blueprint. It tells us the precise ratio of atoms in a molecule: two hydrogen atoms for every two oxygen atoms. But what if we wanted a different kind of blueprint, one based on mass? For example, in a 100 g sample of hydrogen peroxide, how many grams are hydrogen, and how many are oxygen?

This “recipe by mass” is known as the percent composition by mass (often written as “% by mass” or “mass %”). It tells us what percentage of a compound’s total mass comes from each of its constituent elements.

To calculate this, we use the concept of mass fraction (w). The mass fraction of an element E within a compound X is the ratio of the total mass of that element, m(E), to the total mass of the compound, m(X):

\[ w(\text{E}) = \frac{m(\text{E})}{m(\text{X})} \]

Because mass fraction is a ratio, we can also calculate it directly using the dimensionless standard atomic weights (A_r°) from the periodic table. The equation becomes:

\[ w(\text{E}) = \frac{N \times A_{\text{r}}^{\circ}(\text{E})}{M_{\text{r}}(\text{X})} \]

where

N is the number of atoms of element E in the formula of X.
A_r°(E) is the standard atomic weight of element E.
M_r(X) is the relative formula (or molecular) mass of the compound X, found by summing the standard atomic weights of all atoms in the formula.

A Note on Symbols: A_r° vs. M_r

You will use two different but related symbols when working with the dimensionless values from the periodic table. It is crucial to use the correct symbol for the correct entity.

A_r°(E) represents the Standard Atomic Weight.
- Use this for: A single element (E).
- Source: This is the value you look up directly on the periodic table for one element (e.g., A_r°(C) = 12.01).
M_r(X) represents the Relative Formula Mass (or Relative Molecular Mass).
- Use this for: A compound or molecule (X).
- Source: This is a calculated value. You find it by summing the standard atomic weights of all the atoms in the chemical formula (e.g., M_r(H₂O) = 2 A_r°(H) + A_r°(O)).

The Takeaway: Use A_r° for a single element, and M_r for a whole compound.

The percent composition by mass (or mass percent) is simply the mass fraction expressed as a percentage:

\[ w(\mathrm{E})~\% = w(\text{E}) \times 100~\% \]

A Note on Symbols: Mass Fraction vs. Mass Percent

The formal IUPAC symbol for the mass fraction of a substance is a lowercase w. Mass fraction is a dimensionless ratio of the mass of a constituent to the total mass of the mixture.

However, in practice, we almost always express this value as a percentage. To make this distinction clear, this textbook will use two related symbols:

w(X): Represents the mass fraction of element X (e.g., 0.711).
w(X) %: Represents the mass percent of element X (e.g., 71.1 %), which is simply the mass fraction multiplied by 100%.

In some other texts, particularly in engineering, you may see the lowercase Greek letter omega (ω) used for mass fraction to avoid confusion with the symbol for work (w). This book will adhere to the formal IUPAC symbol, w.

Calculating Percent Composition from a Formula

The most common task is to determine a compound’s percent composition directly from its chemical formula. This is done by using the standard atomic weights from the periodic table. The strategy involves three steps:

Calculate the Relative Formula Mass (M_r): Find the total relative mass of the compound (the denominator) by summing the standard atomic weights of all atoms in the formula.
Calculate the Total Relative Mass of Each Element: For each element, find its total relative mass contribution (the numerator) by multiplying its standard atomic weight by its subscript in the formula.
Calculate the Percentage: For each element, divide its relative mass contribution by the total relative formula mass and multiply by 100 %.

Example: Mass % of Elements in a Compound

Determine the percent composition by mass of hydrogen and oxygen in hydrogen peroxide (H₂O₂).

Solution

\[ \begin{align*} w(\mathrm{H})~\% &= w(\mathrm{H}) \times 100~\% \\[1.5ex] &= \left ( \dfrac{2~A_{\mathrm{r}}^{\circ}(\mathrm{H})}{M_{\mathrm{r}}(\mathrm{H_2O_2})} \right ) \times 100~\% \\[1.5ex] &= \left ( \dfrac{2 \times 1.0\bar{1}}{34.0\bar{2}} \right ) \times 100~\% \\[1.5ex] &= 5.9\bar{3}76~\% \\[1.5ex] &= 5.94~\% \\[3.0ex] \end{align*} \]

\[ \begin{align*} w(\mathrm{O})~\% &= w(\mathrm{O}) \times 100~\% \\[1.5ex] &= \left ( \dfrac{2~A_{\mathrm{r}}^{\circ}(\mathrm{O})}{M_{\mathrm{r}}(\mathrm{H_2O_2})} \right ) \times 100~\% \\[1.5ex] &= \left ( \dfrac{2 \times 16.0\bar{0}}{34.0\bar{2}} \right ) \times 100~\% \\[1.5ex] &= 94.0\bar{6}2~\%\\[1.5ex] &= 94.06~\% \\[1.5ex] \textrm{or}\\[1.5ex] w(\mathrm{O})~\% &= 100~\% - w(\mathrm{H})\% \\[0.5ex] &= 100~\% - 5.9\bar{3}76~\% \\[0.5ex] &= 94.0\bar{6}2~\% \\[0.5ex] &= 94.06~\% \end{align*} \]

Using Percent Composition as an Experimental Tool

The previous example showed how to calculate a compound’s theoretical percent composition from its formula. We can also flip this process around. Because every pure compound has a fixed percent composition (the Law of Definite Proportions), we can use experimentally measured masses to identify an unknown compound or verify a proposed formula.

This is one of the foundational techniques in analytical chemistry. Let’s see how it works.

Example: Verifying a Formula with Percent Composition

A 352.0 g sample of a compound containing only chromium and oxygen is heated until it completely decomposes, yielding 240.84 g of pure chromium and 111.16 g of oxygen gas. Based on this experimental data, is the formula of the original compound Cr₂O₃?

Strategy

To solve this, we will:

Calculate the experimental percent composition of chromium from the given masses.
Calculate the theoretical percent composition of chromium for the formula Cr₂O₃.
Compare the two percentages. If they match, the formula is verified.

Solution

Step 1: Determine the % by mass of Cr and O in the 352 g sample.

\[ \begin{align*} w(\mathrm{Cr})~\% &= w(\mathrm{Cr}) \times 100~\% \\[1.5ex] &= \dfrac{m(\textrm{Cr})}{m(\mathrm{Cr_2O_3})} \times 100~\% \\[1.5ex] &= \dfrac{240.8\bar{4}~\mathrm{g}}{352.\bar{0}~\mathrm{g}} \times 100~\% \\[1.5ex] &= 68.4\bar{2}0~\% \end{align*} \]

\[ \begin{align*} w(\mathrm{O})~\% &= w(\mathrm{O}) \times 100~\% \\[1.5ex] &= \dfrac{m(\textrm{O})}{m(\mathrm{Cr_2O_3})} \times 100~\% \\[1.5ex] &= \dfrac{111.1\bar{6}~\mathrm{g}}{352.\bar{0}~\mathrm{g}} \times 100~\% \\[1.5ex] &= 31.5\bar{7}9~\% \end{align*} \]

Step 2: Determine the % by mass of Cr and O in Cr₂O₃.

\[ \begin{align*} w(\mathrm{Cr})~\% &= w(\mathrm{Cr}) \times 100~\% \\[1.5ex] &= \dfrac{2~A_{\textrm{r}}^{\circ}(\textrm{Cr})} {M_{\textrm{r}}(\mathrm{Cr_2O_3})} \times 100~\% \\[1.5ex] &= \dfrac{\left ( 2 \times 52.0\bar{0} \right )} {152.0\bar{0}} \times 100~\% \\[1.5ex] &= 68.4\bar{2}10~\% \end{align*} \]

\[ \begin{align*} w(\mathrm{O})~\% &= w(\mathrm{O}) \times 100~\% \\[1.5ex] &= \dfrac{3~A_{\textrm{r}}^{\circ}(\textrm{O})} {M_{\textrm{r}}(\mathrm{Cr_2O_3})} \times 100~\% \\[1.5ex] &= \dfrac{\left ( 3 \times 16.0\bar{0} \right )} {152.0\bar{0}} \times 100~\% \\[1.5ex] &= 31.5\bar{7}89~\% \end{align*} \]

Step 3: Compare the % by mass in the sample to the compound

\[ \begin{align*} w(\mathrm{Cr})~\% ~ \textrm{in sample} = 68.42 \% \\[1.5ex] w(\mathrm{Cr})~\% ~ \textrm{in compound} = 68.42 \% \\[3ex] w(\mathrm{O})~\% ~ \textrm{in sample} = 31.58 \% \\[1.5ex] w(\mathrm{O})~\% ~ \textrm{in compound} = 31.58 \% \end{align*} \]

Because these values reasonably match, the compound is Cr₂O₃.

Empirical and Molecular Formulas

While a compound’s percent composition is a useful descriptor, its most powerful application is in determining a compound’s chemical formula from experimental data. This process allows chemists to identify unknown substances.

When we determine a formula from percent composition, we are finding the simplest whole-number ratio of atoms in the compound. This is known as the empirical formula.

However, the empirical formula may not be the same as the actual molecular formula, which gives the true number of atoms in a single molecule. Consider these three very different compounds:

All three share the same empirical formula (CH₂O) because they have the same ratio of C:H:O atoms. The molecular formula is always a whole-number multiple of the empirical formula.

Determining the Empirical Formula from Percent Composition

The general strategy for finding a compound’s empirical formula from its percent composition by mass is a classic four-step process:

Assume an exact 100 g Sample: This is a convenient trick that allows us to directly convert the given percentages into masses in grams.
Convert Mass to Moles: Use the molar mass of each element to convert the mass of each element into moles.
Find the Simplest Mole Ratio: Divide the mole amount of each element by the smallest mole amount calculated in the previous step. This will yield a ratio of numbers that are very close to whole integers.
Write the Empirical Formula: Use the whole numbers from the mole ratio as the subscripts for each element in the empirical formula.

Example: Determining an Empirical Formula

An unknown compound is found to contain 27.29 % carbon and 72.71 % oxygen by mass. What is its empirical formula?

Solution

Step 1: Assume a 100 g Sample and Find Masses

If we assume we have exactly 100 g of the compound, the percentages directly become masses:

m(C) = 27.29 g
m(O) = 72.71 g

Step 2: Convert Mass of Each Element to Moles

Now, we use the molar masses of carbon (12.01 g mol⁻¹) and oxygen (16.00 g mol⁻¹) to find the number of moles of each.

\[ \begin{align*} n(\mathrm{C}) &= m(\mathrm{C}) ~ M(\mathrm{C})^{-1} \\[1.5ex] &= 27.2\bar{9}~\mathrm{g} \left ( \frac{1~\mathrm{mol}}{12.0\bar{1}~\mathrm{g}} \right ) \\[1.5ex] &= 2.27\bar{2}27~\mathrm{mol} \end{align*} \]

\[ \begin{align*} n(\mathrm{O}) &= m(\mathrm{O}) ~ M(\mathrm{O})^{-1} \\[1.5ex] &= 72.7\bar{1}~\mathrm{g} \left ( \frac{1~\mathrm{mol}}{16.0\bar{0}~\mathrm{g}} \right ) \\[1.5ex] &= 4.54\bar{4}37~\mathrm{mol} \end{align*} \]

Step 3: Find the Simplest Mole Ratio

At this point, we know the formula is approximately C_2.272O_4.545. To get whole numbers, we divide both mole amounts by the smallest value (2.272 mol):

For Carbon: 2.272 mol / 2.272 mol = 1.00
For Oxygen: 4.545 mol / 2.272 mol = 2.00

The mole ratio of carbon to oxygen is 1 to 2.

Step 4: Write the Empirical Formula

Using the whole-number ratio, the empirical formula of the compound is

CO₂

Determining the Molecular Formula

To find the actual molecular formula from the empirical formula, we need one more piece of information: the molar mass of the compound.

The relationship is straightforward:

\[ \text{Molecular Formula} = (\text{Empirical Formula})_n \]

Where n is a whole number found by:

\[ n = \frac{\text{Molar Mass of Compound}}{\text{Empirical Formula Mass}} \]

For our example above, the empirical formula is CO₂ and the empirical formula mass is 12.01 g mol⁻¹ + (2 × 16.00 g mol⁻¹) = 44.01 g mol⁻¹. If an experiment determined that the molar mass of the unknown compound was also ~44 g mol⁻¹, then n = 44.01 / 44 ≈ 1. Therefore,

\[ \begin{align*} \text{Molecular Formula} &= (\text{Empirical Formula})_n \\[1.5ex] &= (\mathrm{CO_2})_1 \\[1.5ex] &= (\mathrm{C_1O_{2\times 1}})\\[1.5ex] &= \mathrm{CO_2} \end{align*} \]

In this case, the molecular formula is the same as the empirical formula: CO₂.

Extra Practice

Practice

What is the empirical formula for the following molecules?

C₂H₆
C₆H₁₄
C₆H₁₂O₆

Solution

CH₃
C₃H₇
CH₂O

Practice

A compound used in the synthesis of polymers is found to be composed of 48.64 % carbon, 8.16 % hydrogen, and 43.20 % oxygen by mass. Determine the empirical formula of the compound.

Solution

Step 1: Assume a 100 g sample and convert percentages to grams.

\[ \begin{align*} w(\mathrm{C})~\% = 48.6\bar{4}~\% ~\longrightarrow~ m(\mathrm{C}) &= 48.6\bar{4}~\mathrm{g} \\[1.5ex] w(\mathrm{H})~\% = 8.1\bar{6}~\% ~\longrightarrow~ m(\mathrm{H}) &= 8.1\bar{6}~\mathrm{g} \\[1.5ex] w(\mathrm{O})~\% = 43.2\bar{0}~\% ~\longrightarrow~ m(\mathrm{O}) &= 43.2\bar{0}~\mathrm{g} \end{align*} \]

Step 2: Convert the mass of each element to moles.

For this calculation, we use the standard molar masses from the periodic table.

\[ \begin{align*} n(\mathrm{C}) &= m(\mathrm{C}) ~ M(\mathrm{C})^{-1} \\[1.5ex] &= \mathrm{48.6\bar{4}~\mathrm{g}} \left ( \dfrac{\mathrm{mol}}{12.0\bar{1}~\mathrm{g}} \right ) \\[1.5ex] &= 4.04\bar{9}95~\mathrm{mol} \\[3ex] n(\mathrm{H}) &= m(\mathrm{H}) ~ M(\mathrm{H})^{-1}\\[1.5ex] &= \mathrm{8.1\bar{6}~\mathrm{g}} \left ( \dfrac{\mathrm{mol}}{1.0\bar{1}~\mathrm{g}} \right ) \\[1.5ex] &= 8.0\bar{7}92~\mathrm{mol} \\[3ex] n(\mathrm{O}) &= m(\mathrm{O}) ~ M(\mathrm{O})^{-1}\\[1.5ex] &= \mathrm{43.2\bar{0}~\mathrm{g}} \left ( \dfrac{\mathrm{mol}}{16.0\bar{0}~\mathrm{g}} \right ) \\[1.5ex] &= 2.70\bar{0}00~\mathrm{mol} \end{align*} \]

Step 3: Find the simplest whole-number ratio by dividing by the smallest mole value.

The smallest value is 2.7000 mol.

Step 4: Convert the ratio to whole numbers.

\[ \begin{equation*} \begin{gathered} \mathrm{C}_{4.04\bar{9}95}\mathrm{H}_{8.0\bar{7}92}\mathrm{O}_{2.70\bar{0}00} \\[2ex] \mathrm{C}_{\frac{4.04\bar{9}95}{2.70\bar{0}00}}\mathrm{H}_{\frac{8.0\bar{7}92}{2.70\bar{0}00}}\mathrm{O}_{\frac{{2.70\bar{0}00}}{2.70\bar{0}00}} \\[2ex] \mathrm{C}_{1.49\bar{9}98}\mathrm{H}_{2.9\bar{9}22 }\mathrm{O}_{1.00\bar{0}00} \\[2ex] \mathrm{C}_{1.5}\mathrm{H}_{3}\mathrm{O}_{1} \\[2ex] \mathrm{C}_{1.5\times 2}\mathrm{H}_{3\times 2}\mathrm{O}_{1\times 2} \\[4ex] \mathrm{C}_{3}\mathrm{H}_{6}\mathrm{O}_{2} \end{gathered} \end{equation*} \]

Why We Round to Whole Numbers

In this calculation, you’ll notice that H_2.99 ≈ H₃. Why do we round 2.99 to 3?

The Chemical Reality: Atoms are discrete particles - you can’t have 2.99 hydrogen atoms in a molecule. Nature requires whole numbers of atoms. When we determine empirical formulas, we’re looking for the simplest whole-number ratio of atoms that actually exist.

The Experimental Reality: Percent composition measurements have limited precision. Even with careful laboratory work, there’s always some experimental uncertainty. When we work backward from experimental data to find formulas, small variations in the input create slight deviations from perfect whole numbers.

The Practical Rule:

If your result is close to a whole number (within ~0.05):

2.99 → 3 atoms
0.96 → 1 atom
1.02 → 1 atom

If your result indicates simple fractions:

1.51 → 1½ atoms → This suggests multiplying ALL subscripts by 2
2.67 → 2⅔ atoms → This suggests multiplying ALL subscripts by 3
0.33 → ⅓ atom → This suggests multiplying ALL subscripts by 3

The key insight: Look for patterns that suggest simple fractions, then multiply ALL subscripts by the smallest number that gives whole numbers for every element.

Empirical formulas always have whole-number subscripts. If your ratios are close to integers (like 2.99), round to the nearest whole number (3).

Practice

What is the molar mass of a 13.43 mol sample of unknown compound if the mass of the sample was 1.278 68 kg?

Solution

Molar mass is defined as being the mass (in g) of a substance per 1 mole of that substance.

\[ \begin{align*} M(\mathrm{unknown}) &= \dfrac{m(\mathrm{unknown})}{n(\mathrm{unknown})} \\[1.5ex] &= \dfrac{1.278~6\bar{8}~\mathrm{kg}}{13.4\bar{3}~\mathrm{mol}} \left ( \dfrac{10^3~\mathrm{g}}{\mathrm{kg}} \right ) \\[1.5ex] &= 95.2\bar{1}07~\mathrm{g~mol^{-1}} \\[1.5ex] &= 95.21~\mathrm{g~mol^{-1}} \end{align*} \]

Extend

The sample was further analyzed and found to have the following formula, MCl₂, where M is a metal. What is the identity of the metal?

Solution

The molar mass of the unknown is 95.21 g mol^–1. The molar mass of two chlorine atoms is 70.9 g mol⁻¹. Find the remaining molar mass to determine the metal.

\[ \begin{align*} M(\mathrm{M}) &= M(\mathrm{unknown}) - 2~M(\mathrm{Cl}) \\[1.5ex] &= 95.2\bar{1}07~\mathrm{g~mol^{-1}} - \left [2 (35.4\bar{5}~\mathrm{g~mol^{-1}}) \right ]\\[1.5ex] &= 24.3\bar{1}07~\mathrm{g~mol^{-1}} \\[1.5ex] &= 24.31~\mathrm{g~mol^{-1}} \end{align*} \]

Finding the molar mass on the periodic table reveals that the metal is magnesium.