Chapter 4 Chemical Reactions and Chemical Quantities: Empirical & Molecular Formulas, Combustion Analysis, Stoichiometry, and Functional Groups

Empirical and Molecular Formulas

Empirical Formula

The empirical formula of a compound represents the simplest whole-number ratio of the elements present. It is derived from the mass percentages of each element in the compound, using the mole concept.

• Empirical Formula: Shows the simplest ratio of atoms in a compound.

• Molecular Formula: Shows the actual number of atoms of each element in a molecule.

• All chemical formulas must contain whole numbers of each atom, representing the simplest ratio.

Steps to Calculate the Empirical Formula:

1. Write down the elements present.

2. Write down the masses (in grams) of each element. If given percentages, assume 100 g of the compound.

3. Convert all masses to moles (using atomic masses).

4. Divide each mole value by the smallest number of moles to get the simplest ratio.

5. If the ratio is not whole numbers, multiply by a factor to obtain whole numbers.

Example: A compound is 57.47% sodium, 40.01% oxygen, and 2.52% hydrogen. Find the empirical formula by converting percentages to grams, then to moles, and simplifying the ratio.

Molecular Formula

The molecular formula gives the actual number of atoms of each element in a molecule. It is a whole-number multiple of the empirical formula.

• To determine the molecular formula, the molar mass of the compound must be known.

Steps to Calculate the Molecular Formula:

1. Determine the empirical formula.

2. Calculate the empirical formula mass.

3. Divide the molar mass by the empirical formula mass to find the n-factor.

4. Multiply the subscripts in the empirical formula by the n-factor.

Example: Ibuprofen (M = 206.3 g/mol) with a percent composition of 75.70% C, 8.80% H, and 15.50% O. Find the empirical formula, calculate its mass, and determine the molecular formula using the n-factor.

Combustion Analysis

Principles of Combustion Analysis

Combustion analysis is used to determine the empirical formula of a compound, especially for organic compounds containing C, H, and possibly O, N, or halogens. The compound is burned in excess oxygen, and the masses of CO2 and H2O produced are measured.

• Products of hydrocarbon combustion: CO2 and H2O.

• For compounds with other elements, additional products (e.g., NO2, SO2) may form.

Steps in Combustion Analysis:

1. Convert grams of CO2 to grams of C.

2. Convert grams of H2O to grams of H.

3. If necessary, subtract the masses of C and H from the sample mass to find the mass of O or other elements.

4. Convert all masses to moles.

5. Divide by the smallest number of moles to get the simplest ratio.

6. Adjust to whole numbers if needed.

Example: Combustion of 12.01 g of an acid produces 14.08 g CO2 and 4.32 g H2O. Find the empirical formula by following the steps above.

Combustion Apparatus

The combustion apparatus is used to collect and measure the masses of CO2 and H2O produced during combustion. The difference in absorber mass before and after combustion gives the amount of each product formed.

	Before combustion	After combustion
Mass of CO2 absorber	75.30 g	78.15 g
Mass of H2 absorber	49.14 g	50.11 g

Interpretation: The increase in mass corresponds to the amount of CO2 and H2O produced, which is used to calculate the empirical formula of the original compound.

Balancing Chemical Equations

Principles of Balancing

Balancing chemical equations ensures the law of conservation of mass is obeyed: the number and type of atoms on both sides of the equation must be equal.

• Coefficients are used to balance the number of atoms.

• Never change subscripts in chemical formulas to balance equations.

Steps to Balance Equations:

1. List the number of atoms of each element on both sides.

2. Adjust coefficients to balance each element.

3. If fractions appear, multiply all coefficients by a common factor to obtain whole numbers.

Example: Balance the combustion of butane: ___ C4H10 (g) + ___ O2 (g) → ___ H2O (l) + ___ CO2 (g)

Stoichiometry

Stoichiometric Calculations

Stoichiometry is the quantitative relationship between reactants and products in a balanced chemical equation. It allows calculation of the amount of product formed from a given amount of reactant and vice versa.

• Uses mole ratios from the balanced equation.

• Can convert between grams, moles, molecules, and formula units.

Steps in Stoichiometric Calculations:

1. Convert the given quantity to moles.

2. Use the mole ratio to find moles of the desired substance.

3. Convert moles to the desired units (grams, molecules, etc.).

Example: How many grams of NO are produced when 15.0 g N2 reacts?

Limiting Reagent and Percent Yield

Limiting Reagent

The limiting reagent is the reactant that is completely consumed first, thus limiting the amount of product formed. The theoretical yield is the maximum amount of product possible, while the excess reagent is left over after the reaction.

• To identify the limiting reagent, calculate the amount of product each reactant can produce; the smallest amount indicates the limiting reagent.

Example: ZnS reacts with HCl to form ZnCl2 and H2S. Given 12.11 g ZnS and 15.92 g HCl, calculate the mass of H2S formed.

Percent Yield

Percent yield measures the efficiency of a reaction:

• Actual yield: Amount of product actually obtained from an experiment.

• Theoretical yield: Maximum possible amount calculated from stoichiometry.

• Percent yield is always less than 100% due to losses and side reactions.

Percent Yield Formula:

$\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$

Example: If 2.6 g C6H6 produces 1.25 g H2O, calculate the percent yield.

Mass Percent Composition

Calculating Mass Percent

Mass percent (or weight percent) is the percentage by mass of each element in a compound.

Mass Percent Formula:

$\text{Mass Percent} = \frac{\text{mass of element in 1 mole of compound}}{\text{mass of 1 mole of compound}} \times 100\%$

Example: Calculate the percent composition of nitrogen and oxygen in NO2.

Functional Groups in Chemistry

Hydrocarbons and Functional Groups

Functional groups are specific groups of atoms within molecules that are responsible for the characteristic chemical reactions of those molecules.

• Hydrocarbons: Compounds containing only carbon and hydrogen.

• Functional groups without carbonyls: Recognized by the presence of atoms such as O, N, S, or halogens (F, Cl, Br, I).

• Functional groups with carbonyls: Contain a carbon double-bonded to oxygen (C=O), such as aldehydes, ketones, carboxylic acids, esters, and amides.

Examples: Identify the functional groups present in given molecular structures (e.g., alcohols, amines, carboxylic acids, etc.).