Chemical Reactions and Reaction Stoichiometry: Molecular Formulas, Limiting Reactants, and Stoichiometric Calculations

Chemical Reactions and Reaction Stoichiometry

Determining a Molecular Formula

The molecular formula of a compound shows the actual number of atoms of each element in a molecule, while the empirical formula gives the simplest whole-number ratio of these atoms. The molecular formula is always a whole-number multiple of the empirical formula.

• Key Point 1: The relationship between molecular and empirical formulas is given by: $\text{molecular formula} = x \times \text{empirical formula}$

• Key Point 2: To find the value of x (the multiple), use: $x = \frac{\text{molar mass}}{\text{empirical molar mass}}$

• Example: If the empirical formula is CH (empirical molar mass = 13 g/mol) and the compound's molar mass is 78 g/mol: $x = \frac{78}{13} = 6$ Molecular formula: $\mathrm{C}_6\mathrm{H}_6$

• Example: A compound contains 1.52 g N and 3.47 g O, with a molar mass of 92 g/mol.

  • Convert grams to moles: $\frac{1.52\,\mathrm{g}\,N}{14.01\,\mathrm{g}/\mathrm{mol}} = 0.108\,\mathrm{mol}\,N$ $\frac{3.47\,\mathrm{g}\,O}{16.00\,\mathrm{g}/\mathrm{mol}} = 0.217\,\mathrm{mol}\,O$

  • Empirical formula: NO2 (after dividing by smallest number)

  • Empirical molar mass: $14.01 + 2 \times 16.00 = 46.01$ g/mol

  • $x = \frac{92}{46.01} = 2$

  • Molecular formula: $\mathrm{N}_2\mathrm{O}_4$

Limiting Reactants

In a chemical reaction, the limiting reactant is the reactant that is completely consumed first, thus limiting the amount of product formed. The other reactant(s) are said to be in excess.

• Key Point 1: The limiting reactant determines the maximum amount of product that can be formed.

• Key Point 2: The limiting reactant is used in all stoichiometric calculations for product and excess reactant amounts.

• Example: In the reaction $2\mathrm{Al} + \mathrm{Fe}_2\mathrm{O}_3 \rightarrow \mathrm{Al}_2\mathrm{O}_3 + 2\mathrm{Fe}$, if 124 g Al and 601 g Fe2O3 are used:

  • Calculate moles of each reactant: $\frac{124\,\mathrm{g}\,Al}{27.0\,\mathrm{g}/\mathrm{mol}} = 4.59\,\mathrm{mol}\,Al$ $\frac{601\,\mathrm{g}\,Fe_2O_3}{160.0\,\mathrm{g}/\mathrm{mol}} = 3.76\,\mathrm{mol}\,Fe_2O_3$

  • Calculate product from each reactant: From Al: $4.59\,\mathrm{mol}\,Al \times \frac{1\,\mathrm{mol}\,Al_2O_3}{2\,\mathrm{mol}\,Al} \times 102\,\mathrm{g}/\mathrm{mol} = 234\,\mathrm{g}\,Al_2O_3$ From Fe2O3: $3.76\,\mathrm{mol}\,Fe_2O_3 \times \frac{1\,\mathrm{mol}\,Al_2O_3}{1\,\mathrm{mol}\,Fe_2O_3} \times 102\,\mathrm{g}/\mathrm{mol} = 383\,\mathrm{g}\,Al_2O_3$

  • Al gives the least product, so Al is the limiting reactant.

Limiting Reactant Calculations: Method

To determine the limiting reactant and calculate product amounts, follow these steps:

1. Calculate the amount of product (in moles or grams) formed from each reactant.

2. Identify the reactant that produces the least amount of product; this is the limiting reactant.

3. Use the limiting reactant for all further calculations.

Excess Reactant

The excess reactant is the reactant that remains after the limiting reactant is completely consumed. To find the amount left:

• Calculate the amount of excess reactant that reacts with the limiting reactant.

• Subtract the reacted amount from the initial amount: Excess mass: $\text{given mass} - \text{mass (reacted)}$ Excess moles: $\text{given moles} - \text{moles (reacted)}$

Theoretical Yield and Percent Yield

The theoretical yield is the maximum amount of product that can be formed from the limiting reactant, as calculated from stoichiometry. The actual yield is the amount of product actually obtained from the reaction. Percent yield compares these two values:

• Percent yield formula: $\text{percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$

Stoichiometric Calculations

Stoichiometry involves using balanced chemical equations to relate quantities of reactants and products. The coefficients in the equation indicate the mole ratios used for conversions.

• Key Point 1: Use the coefficients from the balanced equation to set up conversion factors between substances.

• Key Point 2: To convert grams to moles, use the molar mass; to convert moles of one substance to moles of another, use the mole ratio from the equation.

• Example: How many grams of water are produced from 1.00 g of glucose ($\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6$) in the reaction: $\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6(s) + 6\mathrm{O}_2(g) \rightarrow 6\mathrm{CO}_2(g) + 6\mathrm{H}_2\mathrm{O}(l)$

  • Convert grams of glucose to moles: $\frac{1.00\,\mathrm{g}\,C_6H_{12}O_6}{180.0\,\mathrm{g}/\mathrm{mol}} = 0.00556\,\mathrm{mol}\,C_6H_{12}O_6$

  • Use the mole ratio to find moles of water: $0.00556\,\mathrm{mol}\,C_6H_{12}O_6 \times \frac{6\,\mathrm{mol}\,H_2O}{1\,\mathrm{mol}\,C_6H_{12}O_6} = 0.0333\,\mathrm{mol}\,H_2O$

  • Convert moles of water to grams: $0.0333\,\mathrm{mol}\,H_2O \times 18.02\,\mathrm{g}/\mathrm{mol} = 0.600\,\mathrm{g}\,H_2O$

Summary Table: Key Stoichiometric Concepts

Concept	Definition	Key Formula	Example
Molecular Formula	Shows actual number of atoms in a molecule	$x = \frac{\text{molar mass}}{\text{empirical molar mass}}$	$\mathrm{C}_6\mathrm{H}_6$ from CH
Limiting Reactant	Reactant that is completely consumed first	Least product formed	Al in $2\mathrm{Al} + \mathrm{Fe}_2\mathrm{O}_3$
Excess Reactant	Reactant left over after reaction	$\text{excess mass} = \text{given mass} - \text{mass (reacted)}$	Fe2O3 in above example
Theoretical Yield	Maximum product possible	Stoichiometric calculation	234 g Al2O3
Percent Yield	Actual yield as percent of theoretical	$\text{percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$	Measured in lab

Additional info: These notes expand on the original slides and text by providing full definitions, stepwise methods, and worked examples for each concept, ensuring a self-contained study guide for General Chemistry students.