Stoichiometry and Stoichiometric Calculations: Definition, Common Mistakes to Avoid & Sample Questions

A mole is the amount of substance containing $6.022\times {10}^{23}$ particles (Avogadro's number, $N_A$ ). Molar mass is the mass of one mole of a substance in grams, numerically equal to its atomic or molecular mass. For example:

• Molar mass of ${\mathrm{O}}_2=32\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}$ .
• Molar mass of ${\mathrm{C}\mathrm{O}}_2=12+2\times 16=44\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}$ .

Number of moles $\left(n\right)$ is calculated as: $n=\frac{\text{mass}\left(\mathrm{g}\right)}{\text{molar mass}(\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l})}=\frac{\text{number of particles}}{N_A}=\frac{\text{volume at STP}(\mathrm{L},\text{for gases})}{22.4}$

NEET exam problems often involve converting between mass, moles, and particles.

A chemical equation represents reactants forming products, with coefficients indicating mole ratios. For example: $2{\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

This implies:

1. Mole ratio: $2\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{H}}_2:1\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{O}}_2:2\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{H}}_2\mathrm{O}$ .

2. Mass ratio: $4\mathrm{g}{\mathrm{H}}_2:32\mathrm{g}{\mathrm{O}}_2:36\mathrm{g}{\mathrm{H}}_2\mathrm{O}$ .

3. Volume ratio (gases, same T, P): 2 vol ${\mathrm{H}}_2:1$ vol ${\mathrm{O}}_2:2\mathrm{v}\mathrm{o}\mathrm{l}{\mathrm{H}}_2\mathrm{O}$ .

Coefficients provide stoichiometric equivalents, e.g., $2\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{H}}_2$ is equivalent to $1\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{O}}_2$ . NCERT class 11 solutions of the chapter also have questions that use these ratios for calculations.

Theoretical yield is the maximum product expected from stoichiometry. Actual yield is often lower due to side reactions. Percentage yield is: $\text{Percentage y}\text{ield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

Example: If $36\mathrm{g}{\mathrm{H}}_2\mathrm{O}$ is expected but 30 g is obtained: $\text{Percentage yield}=\frac{30}{36}\times 100=83.33\mathrm{\%}$

JEE numericals often require yield calculations.

Stoichiometric calculations involve:

1. Mass-Mass: Relate masses of reactants/products. E.g., for $2{\mathrm{N}\mathrm{a}\mathrm{H}\mathrm{C}\mathrm{O}}_3\stackrel{\phantom{\mathrm{\Delta }\mathrm{}}}{\to }{\mathrm{N}\mathrm{a}}_2{\mathrm{C}\mathrm{O}}_3+$ ${\mathrm{H}}_2\mathrm{O}+{\mathrm{C}\mathrm{O}}_2$ , calculate mass of ${\mathrm{N}\mathrm{a}}_2{\mathrm{C}\mathrm{O}}_3$ from $10\mathrm{g}{\mathrm{N}\mathrm{a}\mathrm{H}\mathrm{C}\mathrm{O}}_3$ :

$\begin{array}{rr}\text{Moles of}{\mathrm{N}\mathrm{a}\mathrm{H}\mathrm{C}\mathrm{O}}_3=& \frac{10}{84}=0.119,\mathrm{}\text{Moles of}{\mathrm{N}\mathrm{a}}_2{\mathrm{C}\mathrm{O}}_3=\frac{0.119}{2}=0.0595\\ & \text{Mass}=0.0595\times 106=6.31\mathrm{g}\end{array}$

2. Mass-Volume: Relate mass to gas volume at STP. E.g., volume of ${\mathrm{C}\mathrm{O}}_2$ from 10 g ${\mathrm{N}\mathrm{a}\mathrm{H}\mathrm{C}\mathrm{O}}_3:$

Moles of ${\mathrm{C}\mathrm{O}}_2=\frac{0.119}{2}=0.0595,\mathrm{}$ Volume $=0.0595\times 22.4=1.33\mathrm{L}$

3. Volume-Volume: Relate volumes of gaseous reactants/products. E.g., for $2{\mathrm{C}}_2{\mathrm{H}}_2+5{\mathrm{O}}_2\to$ $4{\mathrm{C}\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O},40\mathrm{m}\mathrm{L}{\mathrm{C}}_2{\mathrm{H}}_2$ needs: $\text{Vol}\text{ume of}{\mathrm{O}}_2=\frac{5}{2}\times 40=100\mathrm{m}\mathrm{L}$

JEE problems test these relationships with complex scenarios.

Example 1: Calculate the mass of NaCl decomposed by $9.8\mathrm{g}{\mathrm{H}}_2{\mathrm{S}\mathrm{O}}_4$ if $12\mathrm{g}\mathrm{N}\mathrm{a}\mathrm{H}\mathrm{S}\mathrm{O}{\mathrm{H}}_4$ and 2.75 g HCl are produced. (JEE Main)

$\mathrm{N}\mathrm{a}\mathrm{C}\mathrm{l}+{\mathrm{H}}_2{\mathrm{S}\mathrm{O}}_4\to {\mathrm{N}\mathrm{a}\mathrm{H}\mathrm{S}\mathrm{O}}_4+\mathrm{H}\mathrm{C}\mathrm{l}$

By conservation of mass: $\text{Mass of}\mathrm{N}\mathrm{a}\mathrm{C}\mathrm{l}+9.8=12+2.75⟹\text{Mass of}\mathrm{N}\mathrm{a}\mathrm{C}\mathrm{l}=4.95\mathrm{g}$

Example 2: $10\mathrm{m}\mathrm{L}{\mathrm{N}}_2$ and $25\mathrm{m}\mathrm{L}{\mathrm{H}}_2$ at $\mathrm{P},\mathrm{T}$ same  react to form ${\mathrm{N}\mathrm{H}}_3$ . Find the volume of ${\mathrm{N}\mathrm{H}}_3$ and limiting reagent. (JEE Main)

${\mathrm{N}}_2+3{\mathrm{H}}_2\to 2{\mathrm{N}\mathrm{H}}_3$

Moles ratio: 1:3:2. Volume ratio $=$ mole ratio.

$\text{Volume of}{\mathrm{H}}_2\text{needed}=3\times 10=30\mathrm{m}\mathrm{L}$

Since only $25\mathrm{m}\mathrm{L}{\mathrm{H}}_2$ , it's limiting. Volume of ${\mathrm{N}\mathrm{H}}_3$ : $\text{Volume of}{\mathrm{N}\mathrm{H}}_3=\frac{2}{3}\times 25=16.67\mathrm{m}\mathrm{L}$

• Inbalanced Equations: Incorrect use of coefficients leads to inappropriate ratios.
• Restricting Agent Inattention: In stoichiometry, another error individuals commit is considering all reactants as being utilized equally.
• Unit Blunders: Conversion of moles or STP volumes between grams and liters (22.4 L vs. 22.7 L).
• Significant Figures: Excess digits reporting into the final answers is also done by the students wrong.
• Gas Volume Application: Another mistake that students make is applying the volume ratios to non-gaseous products.

1. Mole Concept: 1 mole $=6.022\times {10}^{23}$ particles; molar mass links mass and moles.

2. Chemical Equations: Coefficients give mole, mass, and volume ratios.

3. Limiting Reagent: Determines product yield; identify via mole ratios.

4. Yield: Percentage yield compares actual to theoretical product.

5. Calculations: Mass-mass, mass-volume, and volume-volume relationships drive stoichiometry.