What Stoichiometry Means
The word stoichiometry is derived from two Greek words — stoicheion, meaning element, and metron, meaning measure. Stoichiometry, thus, deals with the calculation of the masses, and sometimes the volumes, of the reactants and the products involved in a chemical reaction. Every such calculation begins with the same indispensable tool: a correctly balanced chemical equation.
A balanced equation does far more than name reactants and products. Because the law of conservation of mass demands that atoms are neither created nor destroyed, the equation also encodes the exact numerical relationship between the amounts of every species. Once that relationship is in hand, the amount of any one substance in the reaction fixes the amount of every other.
The coefficients of the balanced equation are the only link between a measured reactant and a predicted product. Master that link and every stoichiometry problem reduces to one method.
Balancing a Chemical Equation
According to the law of conservation of mass, a balanced chemical equation has the same number of atoms of each element on both sides of the equation. Many equations can be balanced by simple trial and error. Consider the reactions of a few metals and non-metals with oxygen:
For a more involved case, the combustion of propane is balanced in clear steps: write the correct formulas, then balance carbon, then hydrogen, and finally oxygen. The result is shown below. The crucial discipline is that you only ever adjust the coefficients in front of the formulas; the subscripts inside a formula define the substance and must never be changed.
Never balance by editing a subscript
A tempting shortcut is to "balance" oxygen in a water equation by writing $\ce{H2O2}$ instead of $\ce{H2O}$. This is fatal — $\ce{H2O2}$ is hydrogen peroxide, a different compound entirely. Subscripts in the formulas of reactants and products cannot be changed to balance an equation.
Balance with coefficients only. Subscripts are fixed by chemical identity.
Reading the Equation as Mole Ratios
Consider the combustion of methane, whose balanced equation is the workhorse example of this chapter:
Methane and dioxygen are the reactants; carbon dioxide and water are the products. The coefficients 2 for $\ce{O2}$ and $\ce{H2O}$ are called stoichiometric coefficients (the coefficient for $\ce{CH4}$ and $\ce{CO2}$ is one in each case). They represent the number of molecules — and equally the number of moles — taking part in or formed by the reaction. A single balanced equation can therefore be read on four levels at once.
| Interpretation | CH₄ | 2O₂ | CO₂ | 2H₂O |
|---|---|---|---|---|
| Molecules | 1 molecule | 2 molecules | 1 molecule | 2 molecules |
| Moles | 1 mol | 2 mol | 1 mol | 2 mol |
| Mass | 16 g | 2 × 32 g | 44 g | 2 × 18 g |
| Volume at STP (gases) | 22.7 L | 45.4 L | 22.7 L | 45.4 L |
Each row of that table is a valid statement about the same reaction, and the rows can be interconverted freely. The mole row is the one to internalise: the coefficients are the mole ratios, and those ratios are the conversion factors used in every calculation that follows.
The Mole Bridge: A Master Method
Almost every stoichiometric problem follows one universal route, often called the mole bridge. You convert the given quantity into moles, cross the bridge using the mole ratio from the balanced equation, and then convert the moles of the wanted substance into whatever unit the question asks for.
Mass-Mass Calculations
A mass-mass problem gives you the mass of one substance and asks for the mass of another. The mole bridge handles it in three moves. The NCERT worked example below is the canonical one for this chapter.
Calculate the mass of water produced by the combustion of 16 g of methane. (NCERT Problem 1.3)
Step 1 — balanced equation. $\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g)}$
Step 2 — given to moles. The molar mass of $\ce{CH4}$ is 16 g, so 16 g of methane corresponds to exactly $n_{\ce{CH4}} = \dfrac{16}{16} = 1\ \text{mol}$.
Step 3 — cross the bridge. From the equation, 1 mol of $\ce{CH4}$ gives 2 mol of $\ce{H2O}$. Hence 1 mol of methane yields 2 mol of water.
Step 4 — moles to mass. $\text{mass of }\ce{H2O} = 2\ \text{mol} \times 18\ \text{g mol}^{-1} = 36\ \text{g}.$
Answer: 36 g of water.
The reverse direction is just as routine. NCERT Problem 1.4 asks how many moles of methane are required to produce 22 g of $\ce{CO2}$. Since 44 g of $\ce{CO2}$ corresponds to 1 mol, 22 g is 0.5 mol of $\ce{CO2}$; and because the $\ce{CH4}:\ce{CO2}$ mole ratio is $1:1$, exactly 0.5 mol of methane is required. The mole bridge does not care which side of the equation you start from.
Every step here rests on converting grams to moles. If that feels shaky, revisit Mole Concept & Molar Mass before going further.
Mass-Volume Calculations
When a gaseous reactant or product is involved, the question may ask for a volume rather than a mass. The mole bridge is unchanged; only the final conversion factor differs. One mole of any gas occupies 22.7 litres at STP (273 K, 1 bar), so moles of a gas are turned into a volume simply by multiplying by 22.7 L mol−1.
22.7 L, not 22.4 L, at modern STP
Older books quote a molar volume of 22.4 L because the standard pressure was once 1 atm. Current NCERT defines STP at 273 K and 1 bar, giving a molar volume of 22.7 L mol−1. NEET answer keys follow the current convention, so use 22.7 L unless a question explicitly states 1 atm.
STP (273 K, 1 bar) → molar volume = 22.7 L mol−1.
What volume of $\ce{CO2}$ at STP is produced when 8.0 g of methane is burned completely in oxygen?
Step 1 — balanced equation. $\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g)}$
Step 2 — given to moles. $n_{\ce{CH4}} = \dfrac{8.0}{16} = 0.50\ \text{mol}.$
Step 3 — cross the bridge. The $\ce{CH4}:\ce{CO2}$ mole ratio is $1:1$, so $0.50$ mol of methane gives $0.50$ mol of $\ce{CO2}$.
Step 4 — moles to volume. $V_{\ce{CO2}} = 0.50\ \text{mol} \times 22.7\ \text{L mol}^{-1} = 11.35\ \text{L at STP}.$
Answer: 11.35 L of carbon dioxide at STP.
The same idea scales to multi-gas reactions. For the rusting reaction $\ce{4Fe(s) + 3O2(g) -> 2Fe2O3(s)}$, the 3 mol of oxygen occupy $3 \times 22.7 = 68.1$ L at STP, so 4 mol of iron react with 68.1 L of oxygen to produce 2 mol of iron(III) oxide. Mole, mass and volume can even be mixed in a single statement: 4 mol of iron reacts with 68.1 L of oxygen to give 319.2 g of $\ce{Fe2O3}$.
The Limiting Reagent
Many reactions are carried out with amounts of reactants that differ from the ratios demanded by the balanced equation. One reactant is then present in excess of what is required, while another is present in shortfall. The reactant that gets consumed first limits the amount of product formed and is, for that reason, called the limiting reagent. In any stoichiometric calculation where two amounts are given, identifying the limiting reagent is essential before predicting the yield.
50.0 kg of $\ce{N2}$ and 10.0 kg of $\ce{H2}$ are mixed to produce $\ce{NH3}$. Calculate the amount of $\ce{NH3}$ formed and identify the limiting reagent. (NCERT Problem 1.5)
Equation. $\ce{N2(g) + 3H2(g) -> 2NH3(g)}$
Moles available. $n_{\ce{N2}} = \dfrac{50.0\times10^3}{28.0} = 17.86\times10^{2}\ \text{mol};\quad n_{\ce{H2}} = \dfrac{10.0\times10^3}{2.016} = 4.96\times10^{3}\ \text{mol}.$
Test the ratio. 1 mol $\ce{N2}$ needs 3 mol $\ce{H2}$, so $17.86\times10^{2}$ mol of $\ce{N2}$ would need $3\times17.86\times10^{2} = 5.36\times10^{3}$ mol of $\ce{H2}$. Only $4.96\times10^{3}$ mol of $\ce{H2}$ is available, so dihydrogen is the limiting reagent.
Product from the limiting reagent. Since 3 mol $\ce{H2}$ gives 2 mol $\ce{NH3}$, $\;4.96\times10^{3}\ \text{mol }\ce{H2}\times\dfrac{2}{3} = 3.30\times10^{3}\ \text{mol }\ce{NH3}.$
Convert to mass. $3.30\times10^{3}\ \text{mol}\times 17.0\ \text{g mol}^{-1} = 56.1\times10^{3}\ \text{g} = 56.1\ \text{kg }\ce{NH3}.$
The bigger mass is not always in excess
In the example above the nitrogen mass (50 kg) dwarfs the hydrogen mass (10 kg), yet hydrogen is the limiting reagent. You cannot judge limiting reagent from masses; you must convert to moles and compare against the coefficient ratio. The product yield is then computed strictly from the limiting reagent.
Always decide the limiting reagent in moles, then base the yield on it alone.
Concentration Terms
A majority of laboratory reactions are carried out in solution, so it is essential to express how much solute is present in a given amount of solvent or solution. NCERT Unit 1 lists four ways of stating concentration, each suited to a different purpose.
| Term | Symbol | Definition | Formula | Temperature dependent? |
|---|---|---|---|---|
| Mass per cent | w/w % | Mass of solute per 100 parts mass of solution | (mass solute / mass solution) × 100 | No |
| Mole fraction | x | Moles of a component per total moles | x_A = n_A / (n_A + n_B) | No |
| Molarity | M | Moles of solute per litre of solution | M = n_solute / V_litres | Yes |
| Molality | m | Moles of solute per kg of solvent | m = n_solute / mass_solvent(kg) | No |
Mass per cent and mole fraction
Mass per cent is the most intuitive measure. NCERT Problem 1.6 prepares a solution by adding 2 g of a substance A to 18 g of water; the mass of solution is $2 + 18 = 20$ g, so the mass per cent of A is $\dfrac{2}{20}\times100 = 10\%$. Mole fraction, by contrast, counts particles rather than grams: for a two-component solution of A and B with $n_A$ and $n_B$ moles, $x_A = \dfrac{n_A}{n_A+n_B}$ and $x_B = \dfrac{n_B}{n_A+n_B}$. A useful check is that the mole fractions of all components always sum to one, $x_A + x_B = 1$.
Molarity
Molarity is the most widely used unit. It is the number of moles of solute in one litre of solution. NCERT Problem 1.7 dissolves 4 g of $\ce{NaOH}$ in enough water to make 250 mL of solution. The molar mass of $\ce{NaOH}$ is 40 g, so the moles of solute are $\dfrac{4}{40} = 0.10$ mol, and
For dilution problems, the number of moles of solute is conserved, which gives the handy relation $M_1 V_1 = M_2 V_2$. Molarity, however, depends on temperature because the volume of a solution is itself temperature dependent.
Molality
Molality is the number of moles of solute present in one kilogram of solvent. Because it is defined by mass, not volume, molality does not change with temperature. NCERT Problem 1.8 demonstrates the conversion from molarity to molality.
The density of a 3 M solution of $\ce{NaCl}$ is 1.25 g mL−1. Calculate the molality of the solution. (NCERT Problem 1.8)
Mass of solute in 1 L. 3 mol of $\ce{NaCl}$ has mass $3\times58.5 = 175.5$ g.
Mass of 1 L of solution. $1000\ \text{mL}\times1.25\ \text{g mL}^{-1} = 1250$ g.
Mass of solvent (water). $1250 - 175.5 = 1074.5\ \text{g} = 1.0745\ \text{kg}.$
Molality. $m = \dfrac{3\ \text{mol}}{1.0745\ \text{kg}} = 2.79\ \text{m}.$
Answer: 2.79 molal.
These four terms are introduced here but explored fully — with normality, ppm and interconversions — in the Solutions chapter.
Stoichiometry in eight lines
- Stoichiometry calculates masses and volumes of reactants and products; it always starts from a balanced equation.
- Balance using coefficients only — never alter the subscripts inside a formula.
- Stoichiometric coefficients are read as mole ratios between the species.
- The mole bridge: convert given to moles → apply the mole ratio → convert wanted moles to the required unit.
- Mass-mass uses molar mass on both ends; mass-volume uses 22.7 L mol−1 at STP for the gas.
- The limiting reagent is decided in moles and determines the product yield.
- Concentration terms: mass %, mole fraction, molarity (mol L−1), molality (mol kg−1 of solvent).
- Molarity changes with temperature; mass %, mole fraction and molality do not.