Question

What are the empirical formula and empirical formula mass for each of the following compounds? (a) \(\mathrm{C}_{4} \mathrm{H}_{8}\) (b) \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}\) (c) \(\mathrm{P}_{4} \mathrm{O}_{10}\) (d) \(\mathrm{Ga}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) (e) \(\mathrm{Al}_{2} \mathrm{Br}_{6}\)

Short answer

(a) \( \mathrm{CH}_{2} \), 14.03 g/mol; (b) \( \mathrm{CH}_{2} \mathrm{O} \), 30.03 g/mol; (c) \( \mathrm{P}_{2} \mathrm{O}_{5} \), 141.94 g/mol; (d) \( \mathrm{Ga}_{2} \left(\mathrm{SO}_{4}\right)_{3} \), 267.4 g/mol; (e) \( \mathrm{AlBr}_{3} \), 266.68 g/mol.

Step-by-step solution

Identify Molecular Formula

Find the given molecular formula for each compound.

Determine Ratio of Atoms

Divide the subscripts of each element in the compound by the greatest common divisor of the subscripts to get the empirical formula.

Calculate Empirical Formula Mass

Sum the atomic masses of the elements in the empirical formula to get the empirical formula mass.

Example (a): \(Synthesize Empirical Formula\)

The molecular formula is \( \mathrm{C}_{4} \mathrm{H}_{8} \). The greatest common divisor of the subscripts 4 and 8 is 4, so the empirical formula is \( \mathrm{CH}_{2} \). The empirical formula mass is \( 12.01 + 2 \times 1.01 = 14.03 \) g/mol.

Example (b): \(Synthesize Empirical Formula\)

The molecular formula is \( \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3} \). The greatest common divisor of 3, 6, and 3 is 3, so the empirical formula is \( \mathrm{CH}_{2} \mathrm{O} \). The empirical formula mass is \( 12.01 + 2 \times 1.01 + 16.00 = 30.03 \) g/mol.

Example (c): \(Synthesize Empirical Formula\)

The molecular formula is \( \mathrm{P}_{4} \mathrm{O}_{10} \). The greatest common divisor of 4 and 10 is 2, so the empirical formula is \( \mathrm{P}_{2} \mathrm{O}_{5} \). The empirical formula mass is \( 2 \times 30.97 + 5 \times 16.00 = 141.94 \) g/mol.

Example (d): \(Synthesize Empirical Formula\)

The molecular formula is \( \mathrm{Ga}_{2} \left(\mathrm{SO}_{4}\right)_{3} \). Each sulfate unit \( \mathrm{SO}_{4} \) remains intact, so the empirical formula is the same: \( \mathrm{Ga}_{2} \left(\mathrm{SO}_{4}\right)_{3} \). The empirical formula mass is \( 2 \times 69.72 + 3 \times \left(32.07 + 4 \times 16.00\right) = 267.4 \) g/mol.

Example (e): \(Synthesize Empirical Formula\)

The molecular formula is \( \mathrm{Al}_{2} \mathrm{Br}_{6} \). The greatest common divisor of 2 and 6 is 2, so the empirical formula is \( \mathrm{AlBr}_{3} \). The empirical formula mass is \( 26.98 + 3 \times 79.90 = 266.68 \) g/mol.

Key concepts

Molecular Formula

The molecular formula represents the actual number of atoms of each element in a molecule. It provides detailed information about the composition of the compound.
To find the molecular formula, you count the number of each kind of atom in the molecule.

For instance, in \(\text{C}_4\text{H}_8\), there are four carbon atoms and eight hydrogen atoms. This formula helps chemists understand the exact build of the molecule. However, the molecular formula can sometimes be simplified to its basic ratio, which is known as the empirical formula.

Greatest Common Divisor

The greatest common divisor (GCD) is a key mathematical tool in chemistry for simplifying molecular formulas.
The GCD is the largest number that can evenly divide the subscripts in the molecular formula.

For example, in \(\text{C}_4\text{H}_8\), the subscripts are 4 and 8. The GCD of 4 and 8 is 4. This number helps to reduce or simplify the molecular formula to its simplest form, known as the empirical formula. The simplest ratio of \(\text{C}_4\text{H}_8\) would be \(\text{CH}_2\), achieved by dividing each subscript by 4.

Empirical Formula Mass

To find the empirical formula mass, sum the atomic masses of all atoms in the empirical formula.
This tells you the mass of each molecule based on its simplest form.

Let's take \(\text{CH}_2\) as an example. You add the atomic masses of carbon (12.01 u) and two hydrogens (1.01 u each):
\[12.01 + 2 \times 1.01 = 14.03\] grams per mole.
The empirical formula mass is very useful for comparing and understanding different compounds.

Atomic Mass

Each element has a specific atomic mass, which is the mass of a single atom of that element.
The atomic mass is usually based on the average mass of all the isotopes of an element.

Carbon, for instance, has an atomic mass of 12.01 u, while oxygen is 16.00 u. These masses are used to calculate the empirical formula mass and molecular masses.
It's vital to have a good grasp of atomic masses for solving various chemistry problems, including empirical and molecular formula calculations.

Chemical Subscripts

Subscripts in chemical formulas indicate the number of atoms of each element present in a molecule.
These small numbers after element symbols are essential for accurately representing chemical compounds.

In \(\text{C}_4\text{H}_8\), the subscript '4' means there are four carbon atoms, and '8' means there are eight hydrogen atoms.
By understanding and manipulating these subscripts using methods like finding the greatest common divisor, you can simplify molecular formulas to empirical formulas, which show the simplest whole-number ratio of atoms in a compound.