Some Basic Concepts of Chemistry • Topic 2 of 3

Stoichiometry & Equations

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It rests on the mole concept and the balanced chemical equation, which obeys the law of conservation of mass — the number of atoms of each element must be identical on both sides.

Before any calculation, the formula of a compound is established from experiment. The percentage composition gives the mass percent of each element: $\%\,element = \dfrac{\text{mass of element in 1 mol}}{\text{molar mass}}\times100$. From this we derive two formulae:

  • The empirical formula is the simplest whole-number ratio of atoms of each element. It is found by dividing each element's mass percent by its atomic mass to get mole ratios, then dividing by the smallest value.
  • The molecular formula gives the actual number of atoms in one molecule. It is a whole-number multiple $n$ of the empirical formula, where $n = \dfrac{\text{molar mass}}{\text{empirical formula mass}}$.

For example, glucose has empirical formula $CH_2O$ (empirical mass $30$) and molar mass $180$, so $n = 180/30 = 6$ and the molecular formula is $C_6H_{12}O_6$.

A balanced equation such as $N_2 + 3H_2 \rightarrow 2NH_3$ reads in moles: $1\,mol$ $N_2$ reacts with $3\,mol$ $H_2$ to give $2\,mol$ $NH_3$. These coefficients are the stoichiometric ratios used to convert between any two species: convert the known mass to moles, apply the mole ratio, then convert back to mass or volume.

In practice reactants are rarely mixed in the exact stoichiometric ratio. The limiting reagent is the reactant that is completely consumed first; it determines the maximum amount of product. The other reactant, present in excess, is left over. To identify it, compute the moles of each reactant divided by its coefficient — the smallest value is the limiting reagent. All product calculations are based on the limiting reagent, never on the excess reactant.

Limiting reagent: 2 H2 + O2 gives 2 H2OMix 3 H2 with 1 O2 : which runs out first?Available:3 mol H21 mol O2Reaction needs 2 H2 per 1 O22 H2 + O2 used to 2 H2O1 mol H2 left overO2 is LIMITING(used up first)H2 is in EXCESS
1
Worked Example
A compound contains $40\%$ carbon, $6.7\%$ hydrogen and $53.3\%$ oxygen by mass. Find its empirical formula.
Solution
  1. Mole ratio: C $= 40/12 = 3.33$; H $= 6.7/1 = 6.7$; O $= 53.3/16 = 3.33$.
  2. Divide by smallest ($3.33$): C $= 1$, H $= 2.01\approx2$, O $= 1$.
  3. Simplest ratio C:H:O $= 1:2:1$.

Answer: Empirical formula is $CH_2O$.

2
Worked Example
The empirical formula of a compound is $CH_2O$ and its molar mass is $180\,g\,mol^{-1}$. Find the molecular formula.
Solution
  1. Empirical formula mass $= 12 + 2 + 16 = 30$.
  2. $n = \dfrac{\text{molar mass}}{\text{empirical mass}} = \dfrac{180}{30} = 6$.
  3. Molecular formula $= (CH_2O)_6 = C_6H_{12}O_6$.

Answer: $C_6H_{12}O_6$ (glucose).

3
Worked Example
Calculate the mass of $CO_2$ produced when $12\,g$ of carbon is burnt completely: $C + O_2 \rightarrow CO_2$.
Solution
  1. Moles of C $= 12/12 = 1\,mol$.
  2. From the equation, $1\,mol$ C gives $1\,mol$ $CO_2$.
  3. Mass of $CO_2 = 1\times44 = 44\,g$.

Answer: $44\,g$ of $CO_2$.

4
Worked Example
$3\,mol$ of $H_2$ react with $1\,mol$ of $O_2$: $2H_2 + O_2 \rightarrow 2H_2O$. Identify the limiting reagent and the moles of water formed.
Solution
  1. Divide by coefficients: $H_2 = 3/2 = 1.5$; $O_2 = 1/1 = 1.0$.
  2. Smallest value is for $O_2$, so $O_2$ is the limiting reagent.
  3. $1\,mol$ $O_2$ produces $2\,mol$ $H_2O$; $H_2$ used $= 2\,mol$, leaving $1\,mol$ $H_2$ in excess.

Answer: $O_2$ is limiting; $2\,mol$ of water are formed.

5
Worked Example
Calculate the percentage of nitrogen in ammonium nitrate, $NH_4NO_3$.
Solution
  1. Molar mass $= 14 + 4 + 14 + 48 = 80\,g\,mol^{-1}$.
  2. Mass of nitrogen $= 2\times14 = 28\,g$.
  3. $\%\,N = \dfrac{28}{80}\times100 = 35\%$.

Answer: $35\%$ nitrogen.

6
Worked Example
When $50\,g$ of $CaCO_3$ is heated: $CaCO_3 \rightarrow CaO + CO_2$. What mass of $CaO$ is obtained?
Solution
  1. Moles of $CaCO_3 = 50/100 = 0.5\,mol$.
  2. $1\,mol$ $CaCO_3$ gives $1\,mol$ $CaO$, so $CaO = 0.5\,mol$.
  3. Mass of $CaO = 0.5\times56 = 28\,g$.

Answer: $28\,g$ of $CaO$.

Key Points

  • A balanced equation conserves atoms; its coefficients are mole ratios linking reactants and products.
  • Empirical formula = simplest whole-number atom ratio; molecular formula = n x empirical formula, n = molar mass / empirical mass.
  • Percentage composition: % element = (mass of element in 1 mol / molar mass) x 100.
  • The limiting reagent is fully consumed and fixes the maximum product yield; the excess reagent is left over.
  • To find the limiting reagent, divide each reactant's moles by its stoichiometric coefficient — the smallest result limits the reaction.
Tap an option to check your answer0 / 4
Q1.In the reaction $N_2 + 3H_2 \rightarrow 2NH_3$, how many moles of $NH_3$ form from $2\,mol$ $N_2$ (with excess $H_2$)?
Explanation: $1\,mol$ $N_2$ gives $2\,mol$ $NH_3$, so $2\,mol$ $N_2$ gives $4\,mol$ $NH_3$.
Q2.The empirical formula of a compound is $CH$ and its molar mass is $78\,g\,mol^{-1}$. Its molecular formula is:
Explanation: Empirical mass of $CH = 13$; $n = 78/13 = 6$; molecular formula $= C_6H_6$.
Q3.If $1\,mol$ $N_2$ is mixed with $1\,mol$ $H_2$ for $N_2 + 3H_2 \rightarrow 2NH_3$, the limiting reagent is:
Explanation: Per coefficient: $N_2 = 1/1 = 1$, $H_2 = 1/3 = 0.33$. The smaller value is $H_2$, so $H_2$ limits.
Q4.The percentage by mass of oxygen in water ($H_2O$) is:
Explanation: $\%\,O = (16/18)\times100 = 88.9\%$.