Empirical and Molecular Formulas: Calculation and Practice

Empirical Formula vs. Molecular Formula

Definitions and Key Differences

The empirical formula and molecular formula are two ways to represent the composition of a chemical compound. Understanding the distinction between these formulas is essential in GOB Chemistry.

• Empirical Formula: Shows the simplest whole-number ratio of atoms of each element in a compound. It is determined from the mass percentages of the constituent elements using the mole concept.

• Molecular Formula: Indicates the actual number of atoms of each element in a molecule of the compound. It is a multiple of the empirical formula.

• The empirical formula provides the relative number of atoms, while the molecular formula gives the actual number of atoms.

• By convention, formulas must contain whole numbers of each atom and reflect the whole number ratio.

Examples:

• Empirical Formula: CH2O, CH2, CH2O2

• Molecular Formula: C6H12O6, C2H4, C2H2O4

Calculating the Empirical Formula

Step-by-Step Procedure

The empirical formula can be calculated from the masses or percentages of elements within a compound. The following steps outline the process:

1. Write down the symbols for each element in the question.

2. Write down the masses (in grams) of each element given. If percentages are provided, assume a 100 g sample to convert percentages directly to grams.

3. Convert all masses into moles by dividing each mass by the atomic mass of the element. Use at least four decimal places for accuracy.

4. Divide each mole number by the smallest value among them to obtain whole number ratios.

5. If a value is close to 0.1 or 0.9, round to the nearest whole number. If not, multiply all ratios by a factor to get whole numbers.

Formula:

• Number of moles = $\frac{\text{mass (g)}}{\text{atomic mass (g/mol)}}$

Example: Determine the empirical formula of a compound that is 68.40% chromium and 31.60% oxygen.

• Assume 100 g sample: 68.40 g Cr, 31.60 g O

• Calculate moles: $\frac{68.40}{51.996}$ for Cr, $\frac{31.60}{15.999}$ for O

• Divide by smallest mole value, round or multiply to get whole numbers

• Write empirical formula using resulting ratios

Practice Problems

Worked Examples

Practice problems help reinforce the calculation of empirical formulas from mass or percentage data.

• Practice 1: A compound that contains only carbon, hydrogen, and oxygen is composed of 48.64% C and 43.21% O by mass. What is the empirical formula of this compound? Answer: C2H5Cl

• Practice 2: Elemental analysis of a sample of an ionic compound showed 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What is the empirical formula of the compound? Answer: C2H5Cl

• Practice 3: A compound composed of carbon, hydrogen, and chlorine contains 4.19 × 1023 hydrogen atoms. If 9.00 g of the compound also contains 55.05% chlorine by mass, what is the empirical formula? Answer: C2H5Cl

Comparison Table: Empirical vs. Molecular Formula

Aspect	Empirical Formula	Molecular Formula
Definition	Simplest whole-number ratio of atoms	Actual number of atoms in a molecule
Example	CH2O	C6H12O6
Determined by	Mass or percentage composition	Molar mass and empirical formula
Use	Identifying unknown compounds	Describing molecular structure

Additional info: The practice answers provided (C2H5Cl) may be inferred or illustrative; actual empirical formula calculation should follow the outlined steps for each specific problem.