Percentage Composition and Empirical Formula

A chemist analysing an unknown white powder never gets handed a formula — an instrument reports only how much of each element by mass is present. Percentage composition and empirical formula calculations are the arithmetic bridge that turns that raw mass data into the symbolic formula chemistry actually runs on. Given the mass percentages of the elements in a compound, you can work backward to the simplest whole-number ratio of atoms — the empirical formula — and, with one extra piece of information (the molar mass), scale it up to the true molecular formula. This topic sits in Unit I, "Some Basic Concepts of Chemistry," and leans entirely on skills already in hand: molar mass and the mole concept, applied in reverse. It is one of the most frequently tested calculation types in Class 11 chemistry, appearing in board exams almost every year.

The core insight

Converting each element's mass percentage to moles, dividing by the smallest mole value, and clearing any fractions gives the empirical formula, which the molar mass then scales into the molecular formula.

The core idea

Percentage composition of an element is simply the fraction of the compound's mass contributed by that element, per mole (or per 100 g), turned into a percentage. For an element present as mass $m_{\text{element}}$ in one mole of compound of molar mass $M$ :

\%\text{ of element} = \frac{m_{\text{element}}}{M} \times 100

where: $m_{\text{element}}$ is the mass of that element contained in one mole of the compound (in grams), and $M$ is the molar mass of the whole compound (in $\text{g mol}^{-1}$ ). This direction — formula known, percentages wanted — is the easy half of the topic. The harder and more important half runs backward: percentages known, formula wanted. The empirical formula is the simplest whole-number ratio of atoms of each element; the molecular formula is the true formula and is always a whole-number multiple $n$ of the empirical formula:

\text{molecular formula} = n \times (\text{empirical formula}), \qquad n = \frac{\text{molecular mass}}{\text{empirical formula mass}}

where: $n$ is a positive integer (1, 2, 3, ...), molecular mass is the compound's actual molar mass (found experimentally, e.g. from vapour density or a mass spectrometer), and empirical formula mass is the sum of atomic masses in the empirical formula as written. The standard route from percentages to the empirical formula has five fixed moves: assume $100\ \text{g}$ of compound so each percentage becomes a mass in grams; convert each mass to moles using $n = m/M$ ; divide every mole value by the smallest one to get a ratio; if that ratio is not close to whole numbers, multiply everything by a small common integer to clear the fraction; and finally write the subscripts. The figure below traces this pipeline for a two-element compound.

Figure

Flow diagram from mass percentage through grams and moles to a mole ratio and finally the empirical formula

Derivation ADetermining the empirical formula from percentage data

The goal is to turn raw mass-percentage data from an analytical instrument into a definite chemical formula using nothing beyond the mole concept, so that any two students given the same percentages arrive at the identical empirical formula through a fixed, checkable sequence of five moves.

Step 1. Assume exactly $100\ \text{g}$ of the compound. This is a convenience, not a physical claim about the sample size — it works because percentages are already fractions of 100, so each element's percentage number becomes its mass in grams directly.

Step 2. Convert each element's mass to moles using $n = \dfrac{m}{M}$ , where $M$ is that element's atomic mass. This is the step that turns mass-based data (what an instrument measures) into count-based data (what a chemical formula records), since a formula's subscripts count atoms, not grams.

Step 3. Divide every mole value obtained by the smallest mole value in the set. This produces a ratio in which at least one element has the value exactly $1.00$ , and the others are $\geq 1$ .

Step 4. Inspect the ratio. If every value is already close to a whole number, the empirical formula subscripts are those whole numbers. If one or more values sit near a common fraction — $.5$ , $.33$ / $.67$ , or $.25$ / $.75$ — multiply every value in the ratio by the smallest integer (2, 3, or 4 respectively) that clears all the fractions at once.

Step 5. Write the empirical formula using the cleared whole numbers as subscripts, listing elements in the conventional order for the compound's class. If the molar mass of the actual compound is also given, compute the empirical formula mass and use $n = M_{\text{molecular}}/M_{\text{empirical}}$ to scale up to the molecular formula.

Worked example

Problem. A compound of iron and oxygen contains $70\%$ iron and $30\%$ oxygen by mass (Fe $= 56\ \text{u}$ , O $= 16\ \text{u}$ ). Its molecular mass is separately measured as $159.6\ \text{u}$ . Find the empirical formula, confirm the molecular formula, and state the percentage composition the molecular formula predicts.

Step 1. Assume $100\ \text{g}$ of compound: mass of Fe $= 70\ \text{g}$ , mass of O $= 30\ \text{g}$ .

Step 2. Convert to moles: $n(\text{Fe}) = \dfrac{70}{56} = 1.25\ \text{mol}$ and $n(\text{O}) = \dfrac{30}{16} = 1.875\ \text{mol}$ .

Step 3. Divide both by the smaller value, $1.25$ : Fe gives $1.25/1.25 = 1.00$ and O gives $1.875/1.25 = 1.50$ .

Step 4. The ratio $1 : 1.5$ is not whole-number yet; multiply both by 2 to clear the half: Fe becomes $2.00$ and O becomes $3.00$ . Empirical formula: $\text{Fe}_2\text{O}_3$ , empirical formula mass $= 2(56) + 3(16) = 160\ \text{u}$ .

Step 5. Find $n$ : $n = M_{\text{molecular}}/M_{\text{empirical}} = 159.6/160 \approx 1$ , so the molecular formula is the same as the empirical formula, $\text{Fe}_2\text{O}_3$ . Checking the percentage composition it predicts: $\%\text{Fe} = (2 \times 56)/160 \times 100 = 70.0\%$ and $\%\text{O} = (3 \times 16)/160 \times 100 = 30.0\%$ , exactly reproducing the given data. The empirical and molecular formula are both $\text{Fe}_2\text{O}_3$ .

Competitive stretch

If instead the same $\text{Fe}_2\text{O}_3$ ratio were obtained for a compound whose molecular mass is measured near $320\ \text{u}$ , then $n = 320/160 = 2$ , giving molecular formula $\text{Fe}_4\text{O}_6$ — a reminder that the empirical formula alone never fixes the molecular formula; the molar mass is not optional data, it is what selects $n$ .

Common traps

Stopping at the empirical formula when the question asks for the molecular formula — always check whether a molar mass is given, since that signals the extra multiplication by $n$ .
Rounding a mole ratio too aggressively, such as rounding $1.5$ straight to $2$ instead of recognising the whole set needs multiplying by 2; a ratio ending near $.33$ , $.5$ , $.67$ , or $.25$ hints at multiplying by $3$ , $2$ , $3$ , or $4$ respectively.
Confusing empirical formula mass with molecular mass in $n = M_{\text{molecular}}/M_{\text{empirical}}$ — swapping numerator and denominator gives a fraction instead of an integer, an instant sign that something was inverted.
Treating percentages as if they were already mole ratios — percentages are mass-based and must be converted to moles via atomic mass before any ratio is taken.

Ask the tutor: if your mole ratio in an empirical-formula problem doesn't land close to a whole number even after multiplying by 2 or 3, ask the AI tutor to double-check your mole calculations step by step.

Why do we always assume exactly 100 g of the compound?

It is a bookkeeping shortcut, not a claim about the real sample. Percentages are already defined as parts per hundred, so if you imagine having 100 g of the compound, each element's percentage number converts directly into its mass in grams with no extra arithmetic. Any other assumed mass would work too, since only the ratio of moles at the end matters, but 100 g makes the percent-to-gram step a free conversion.

My mole ratio came out as 1 to 1.33 to 2. What do I multiply by?

A value ending near $.33$ or $.67$ signals a hidden thirds, so multiply the entire ratio by 3: $1 \to 3$ , $1.33 \to 4$ , $2 \to 6$ , giving empirical formula subscripts $3, 4, 6$ . The general rule is to recognise the decimal remainder as a simple fraction ( $.25 = 1/4$ , $.33 = 1/3$ , $.5 = 1/2$ , $.67 = 2/3$ ) and multiply by that fraction's denominator so every value in the set lands on (or very near) a whole number simultaneously.

If I already know the empirical formula, do I still need the molecular mass to find the molecular formula?

Yes, always. The empirical formula only fixes the ratio of atoms, not the actual count — $\text{CH}_2\text{O}$ describes formaldehyde, glucose, and several other molecules equally well, since they all share the ratio 1 carbon to 2 hydrogen to 1 oxygen. The experimentally measured molecular mass is the one extra piece of information that pins down which multiple $n$ of the empirical formula is the real molecule, via $n = M_{\text{molecular}}/M_{\text{empirical}}$ .

Percentage Composition and Empirical Formula

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads