Mass Percent Composition
Mass percent composition expresses how much of the total mass of a compound is contributed by each constituent element. For a pure substance the composition is fixed — this is a direct consequence of the law of definite proportions — so the mass percent of any element is a characteristic constant of that compound. Comparing the measured composition of a sample against the theoretical value is the standard way to check the purity of a substance.
The defining relation, taken directly from NCERT Section 1.9, is straightforward. The mass of an element in one mole of compound is divided by the molar mass of the whole compound and scaled to a percentage:
$$\text{Mass \% of an element} = \frac{\text{mass of that element in one mole of compound}}{\text{molar mass of the compound}} \times 100$$
The numerator is found by multiplying the number of atoms of that element in the formula by its atomic mass. Because the percentages of all elements in a compound must add to 100, the last element's percentage can often be obtained by subtraction rather than by a separate calculation — a useful time-saver in the examination hall.
| Quantity | Meaning | How to obtain it |
|---|---|---|
| Mass of element in 1 mol | Contribution of one element to the molar mass | (number of atoms in formula) × (atomic mass) |
| Molar mass of compound | Sum of atomic masses of all atoms in the formula | Add every atom's atomic mass |
| Mass percent | Fraction of total mass from that element | $\dfrac{\text{element mass}}{\text{molar mass}}\times 100$ |
| Check | All percentages must total 100% | Sum of all mass percents = 100 |
Worked Examples — Percent Composition
The cleanest illustration is water, $\ce{H2O}$, whose molar mass is $18.02\ \text{g mol}^{-1}$. Water contains two hydrogen atoms (mass $2 \times 1.008 = 2.016\ \text{g}$) and one oxygen atom (mass $16.00\ \text{g}$). Applying the definition gives the two percentages, which sum to 100 as required.
Calculate the mass percent of hydrogen and oxygen in water, $\ce{H2O}$.
$$\text{Mass \% of H} = \frac{2.016}{18.02}\times 100 = 11.18\%$$
$$\text{Mass \% of O} = \frac{16.00}{18.02}\times 100 = 88.79\%$$
The two values add to $99.97\%$, i.e. $100\%$ within rounding — a built-in check that the composition is internally consistent.
A second NCERT example, ethanol $\ce{C2H5OH}$, extends this to a three-element compound. Its molar mass is $(2\times 12.01 + 6\times 1.008 + 16.00) = 46.068\ \text{g}$. The carbon mass is $2\times 12.01 = 24.02\ \text{g}$, the hydrogen mass is $6\times 1.008 = 6.048\ \text{g}$, and the oxygen mass is $16.00\ \text{g}$.
Find the mass percent of C, H and O in ethanol, $\ce{C2H5OH}$ (molar mass $46.068\ \text{g}$).
$$\text{Mass \% of C} = \frac{24.02}{46.068}\times 100 = 52.14\%$$
$$\text{Mass \% of H} = \frac{6.048}{46.068}\times 100 = 13.13\%$$
$$\text{Mass \% of O} = \frac{16.00}{46.068}\times 100 = 34.73\%$$
Total $= 52.14 + 13.13 + 34.73 = 100.00\%$.
Atoms, not molecules, in the numerator
For hydrogen in water the mass to use is $2 \times 1.008 = 2.016\ \text{g}$, accounting for both hydrogen atoms — not a single atom's mass of $1.008\ \text{g}$. A common slip is to forget the subscript multiplier and halve the hydrogen percentage.
Always multiply the atomic mass by the number of atoms of that element in one formula unit before forming the ratio.
Empirical vs Molecular Formula
Two distinct kinds of formula describe a compound, and NEET frequently tests whether candidates can tell them apart. An empirical formula represents the simplest whole-number ratio of the various atoms present in a compound. A molecular formula shows the exact number of atoms of each element present in one molecule of the compound. The molecular formula is therefore always an integral multiple of the empirical formula.
This relationship is captured by a single equation that underpins the entire subtopic, where $X$ is the empirical formula and $n$ is a whole number:
$$\text{Molecular formula} = (\text{Empirical formula})_n = X_n$$
| Compound | Molecular formula | Empirical formula | n |
|---|---|---|---|
| Glucose | $\ce{C6H12O6}$ | $\ce{CH2O}$ | 6 |
| Hydrogen peroxide | $\ce{H2O2}$ | $\ce{HO}$ | 2 |
| Benzene | $\ce{C6H6}$ | $\ce{CH}$ | 6 |
| Water | $\ce{H2O}$ | $\ce{H2O}$ | 1 |
| Sucrose | $\ce{C12H22O11}$ | $\ce{C12H22O11}$ | 1 |
When the subscripts of the molecular formula are already in their simplest ratio and cannot be reduced to smaller integers, the empirical and molecular formulas coincide and $n = 1$. Water, ammonia $\ce{NH3}$, carbon dioxide $\ce{CO2}$ and sucrose are standard examples. For ionic compounds such as $\ce{NaCl}$, $\ce{KCl}$ and $\ce{MgO}$, only an empirical (simplest-ratio) formula is meaningful, since these solids do not exist as discrete molecules.
The 5-Step Method for the Empirical Formula
If the mass percent of every element in a compound is known, its empirical formula can be derived by a fixed algorithm. The molecular formula follows whenever the molar mass is also supplied. The schematic below shows the full pipeline that this subtopic asks you to master.
| Step | Operation | Output |
|---|---|---|
| 1 | Assume a 100 g sample; each mass percent becomes that many grams | Mass of each element (g) |
| 2 | Divide each mass by the element's atomic mass | Moles of each element |
| 3 | Divide every mole value by the smallest mole value | Provisional mole ratio |
| 4 | If ratios are not whole numbers, multiply all by a small integer; write subscripts | Empirical formula |
| 5 | Compute $n = M / M_{\text{emp}}$ and multiply the empirical formula by $n$ | Molecular formula |
The assumption of a 100 g sample in Step 1 is purely a convenience. Because the empirical formula depends only on the ratio of atoms, the chosen sample size cannot affect the answer; taking 100 g simply makes each percentage equal to a mass in grams.
Every step here rests on converting mass to moles. If that conversion is shaky, revisit Mole Concept & Molar Mass first.
Worked Example — NCERT Chlorinated Compound
NCERT Problem 1.2 is the canonical application of all five steps at once, and several NEET items are direct variants of it. A compound contains $4.07\%$ hydrogen, $24.27\%$ carbon and $71.65\%$ chlorine, with a molar mass of $98.96\ \text{g}$. The table tracks Steps 1 to 3 in one view.
| Element | Mass % → mass in 100 g | Moles (÷ atomic mass) | ÷ smallest (2.021) |
|---|---|---|---|
| H | 4.07 g | $\dfrac{4.07}{1.008} = 4.04$ | $\dfrac{4.04}{2.021} = 2$ |
| C | 24.27 g | $\dfrac{24.27}{12.01} = 2.021$ | $\dfrac{2.021}{2.021} = 1$ |
| Cl | 71.65 g | $\dfrac{71.65}{35.453} = 2.021$ | $\dfrac{2.021}{2.021} = 1$ |
Determine the empirical and molecular formulas of the compound above.
Step 4 — empirical formula. The ratio $\text{H} : \text{C} : \text{Cl} = 2 : 1 : 1$, all whole numbers, so the empirical formula is $\ce{CH2Cl}$.
Step 5 — molecular formula. Empirical formula mass: $12.01 + (2\times 1.008) + 35.453 = 49.48\ \text{g}$.
$$n = \frac{M}{M_{\text{emp}}} = \frac{98.96}{49.48} = 2$$
Multiplying every subscript by $n = 2$ gives the molecular formula $\ce{C2H4Cl2}$.
Molecular Formula from Empirical Formula
Once the empirical formula is known, recovering the molecular formula is a two-stage arithmetic exercise. First, the empirical formula mass $M_{\text{emp}}$ is obtained by summing the atomic masses of every atom in the empirical unit. Second, the integer multiplier $n$ is the ratio of the experimentally known molar mass to that empirical formula mass.
$$M_{\text{emp}} = \sum (\text{atoms in empirical unit}) \times (\text{atomic mass}), \qquad n = \frac{M}{M_{\text{emp}}}$$
The value of $n$ must round to a whole number; a small deviation (such as $5.97$ or $2.04$) reflects rounding in the atomic masses or the percentages and should be taken as the nearest integer. Multiplying each subscript of the empirical formula by $n$ then yields the molecular formula.
When ratios are not clean integers
In Step 3 a ratio may emerge as $1 : 1.5$ or $1 : 1.33$. Do not round these to whole numbers prematurely — multiply every ratio by the smallest integer that clears the fraction. A $1.5$ requires multiplying by 2 (giving $2 : 3$); a $1.33$ ($\approx 4/3$) requires multiplying by 3 (giving $3 : 4$).
Rounding 1.5 down to 1 changes the compound entirely. Clear the fraction first, then round only genuine measurement noise.
Worked Example — Iron Oxide
The NCERT exercise on an oxide of iron tests the same method with a two-element compound where one percentage is found by subtraction. The oxide contains $69.9\%$ iron, and the remaining $30.1\%$ is oxygen (atomic masses $\ce{Fe} = 55.85$, $\ce{O} = 16.00$). For a 100 g sample this is $69.9\ \text{g}$ Fe and $30.1\ \text{g}$ O.
| Element | Mass in 100 g | Moles | ÷ smallest (1.252) | ×2 (clear 0.5) |
|---|---|---|---|---|
| Fe | 69.9 g | $\dfrac{69.9}{55.85} = 1.252$ | $1.00$ | $2$ |
| O | 30.1 g | $\dfrac{30.1}{16.00} = 1.881$ | $1.50$ | $3$ |
Find the empirical formula of the iron oxide (69.9% Fe, 30.1% O).
Dividing by the smaller mole value gives $\text{Fe} : \text{O} = 1 : 1.5$, which is not whole-number. Multiplying both by 2 clears the half: $\text{Fe} : \text{O} = 2 : 3$.
The empirical formula is therefore $\ce{Fe2O3}$ — haematite — consistent with iron in its $+3$ oxidation state.
This example also demonstrates the subtraction shortcut for the final percentage: only iron was measured, and oxygen was obtained as $100 - 69.9 = 30.1\%$. The same arithmetic, applied with the molar mass of the oxide ($159.7\ \text{g}$ for $\ce{Fe2O3}$), would confirm $n = 1$ since the empirical and molecular formulas of this oxide are identical.
Percentage Composition & Empirical Formula
- Mass % of an element $= \dfrac{\text{mass of element in 1 mol}}{\text{molar mass}} \times 100$; all percentages sum to 100.
- Empirical formula = simplest whole-number atom ratio; molecular formula = exact atom count per molecule.
- Relationship: molecular formula $= X_n$, where $n = M / M_{\text{emp}}$ is a whole number.
- Five steps: 100 g sample → moles → divide by smallest → clear fractions to subscripts → multiply by $n$.
- Worked anchors: $\ce{CH2Cl} \to \ce{C2H4Cl2}$ ($n=2$); iron oxide $1:1.5 \to 2:3 \to \ce{Fe2O3}$.
- When the molecular subscripts are already simplest, empirical = molecular and $n = 1$ (e.g. $\ce{H2O}$, $\ce{CO2}$).