Chemical Composition: Empirical and Molecular Formulas, The Mole Concept, and Mass Percent

CH.4 Chemical Composition

Empirical and Molecular Formulas

Understanding the composition of chemical compounds is fundamental in chemistry. Two important ways to represent the composition are the empirical formula and the molecular formula.

• Empirical Formula: Shows the simplest whole-number ratio of atoms of each element in a compound.

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

For example:

• Glucose: Empirical formula = CH2O, Molecular formula = C6H12O6

• By convention, any formula must contain whole numbers of each atom and represent the most simplified form.

Calculating the Empirical Formula

1. Write down the mass percentage of each element in the compound.

2. Convert the mass percentages to grams (assuming 100 g of compound).

3. Convert the grams to moles using the molar mass of each element.

4. Divide each mole value by the smallest number of moles calculated.

5. If necessary, multiply by a factor to obtain whole numbers.

Example: A compound contains 48.65% carbon and 51.35% oxygen by mass. Find the empirical formula.

Practice Problems

• A compound contains only carbon, hydrogen, and oxygen and is composed of 48.65% C and 42.75% O by mass. What is the empirical formula?

• Elemental analysis of a sample of an ionic compound showed 2.2 g of Na, 4.5 g of Cl, and 7.3 g of O. What is the empirical formula?

• A compound composed of carbon, hydrogen, and chlorine contains 4.19 x 1022 hydrogen atoms. If 10.0 g of the compound also contains 59.0% chlorine by mass, what is the empirical formula?

Molecular Formula

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

• Once the empirical formula is determined, the molecular formula can be obtained if the molar mass is known.

Steps to Calculate the Molecular Formula:

1. Determine the empirical formula.

2. Calculate the molar mass of the compound.

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

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

Example: Cortisol (M = 362.47 g/mol), a steroid hormone, is found to contain 69.6% carbon, 9.34% hydrogen, and 22.1% oxygen by mass. What is its molecular formula?

Practice Problems

• Elemental analysis of a pure compound indicated that the compound had 72.1% C, 8.50% H, and the remainder as O. If 0.250 moles of the compound weighs 41.5 g, what is the molecular formula?

The Mole Concept

The mole is a fundamental unit in chemistry that allows chemists to count atoms, molecules, or formula units by weighing them.

• Mole: The amount of substance containing as many entities (atoms, molecules, ions) as there are in 12 g of carbon-12.

• Avogadro's Number: $6.022 \times 10^{23}$ particles per mole.

Converting Between Moles and Particles

• Number of moles = $\frac{\text{Number of particles}}{6.022 \times 10^{23}}$

• Number of particles = (Number of moles) $\times 6.022 \times 10^{23}$

Example: How many moles of chlorine gas are there in $8.33 \times 10^{22}$ molecules?

Practice: How many grams of Cl2 contain $5.25 \times 10^{22}$ molecules of Cl2? If a sample of sodium chloride, NaCl, contains 7.1 kg, what is its number of formula units?

Converting Between Mass and Moles

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

• Mass of substance = (Number of moles) $\times$ (Molar mass)

Example: How many grams of chlorine gas are there in 2.34 moles?

Practice Problems

• Calculate the number of oxygen atoms found in 783.9 g NaCl · 6 H2O.

• The density of water is 1.00 g/mL at 25°C. Calculate the number of water molecules found in $1.50 \times 10^{4}$ L of water.

• A cylindrical copper wire is used for the fences around a house. The copper wire has a diameter of 0.0750 in. How many copper atoms are found in 5.160 cm pieces? The density of copper is 8.96 g/cm3 ($V = \pi r^2 h$).

• The density of the sun is $1.41 \text{ g/cm}^3$ and its volume is $1.41 \times 10^{27} \text{ m}^3$. How many hydrogen molecules are in the sun if we assume all the mass is hydrogen gas?

Mass Percent

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

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

Example: Calculate the mass percent of carbon within sodium carbonate, Na2CO3.

Practice: Determine the percent composition of nitrogen and oxygen within nitrogen dioxide, NO2.

Summary Table: Key Formulas

Concept	Formula	Description
Empirical Formula	Smallest whole-number ratio of atoms	Determined from percent composition
Molecular Formula	n × (Empirical Formula)	n = (Molar mass)/(Empirical formula mass)
Mole	$6.022 \times 10^{23}$ particles	Avogadro's number
Mass Percent	$\frac{\text{mass of element}}{\text{mass of compound}} \times 100\%$	Percent by mass of element in compound

Additional info: Some steps and definitions were expanded for clarity and completeness. Practice problems were included to reinforce concepts and provide exam-style questions.