Elemental Composition | College Board AP Chemistry Revision Notes 2022

Formula Units

Pure substances and chemical formula

Atoms are the small building blocks of matter
One of the broadest classifications of matter are pure substances
Pure substances can be classified into elements or compounds
Elements are composed of only one kind of atom
- Aluminum (Al) or oxygen (O₂)
Compounds have a unique combination of atoms and they can be split into molecular or ionic
- This combination of atoms can be represented with a chemical formula, in which atomic symbols identify the elements in the compound and subscripts are used to show how many atoms are present

Molecular compounds

A molecular compound is composed of individual molecules.
A molecule is formed when two or more atoms are bonded together with a covalent bond
- Water is a molecular compound because its atoms are bonded with covalent bonds
- The chemical formula of water is H₂O

A molecule of water (H2O)

molecule-of-water

A molecule of water (H₂O) showing 2 hydrogen atoms, 1 oxygen atom and the covalent bonds

Ionic compounds

An ionic compound consists of atoms or ions held together in a formula unit
A formula unit is the arrangement of atoms or ions in fixed proportions by an ionic bond
- Salt (sodium chloride) is an ionic compound because its ions are held together in a fixed arrangement by ionic bonds
- The chemical formula of sodium chloride is NaCl

Formula unit of sodium chloride (NaCl)

formula-unit-of-sodium-chloride

A formula unit of sodium chloride (NaCl) showing the regular but alternating arrangement of sodium ions and chloride ions

The Law of Definite Proportions

Elements interact with other elements to form compounds
The Law of Definite Proportions states that the ratio of the masses between the elements in a compound is always the same
Therefore, a pure compound has the same composition and properties regardless of where it comes from

Law of Definite Proportions

understanding-the-law-of-definite-proportions-with-two-samples-of-h2o-with-different-masses

Understanding the Law of Definite Proportions with two samples of H₂O with different masses

Exam Tip

Most of the exam questions regarding the Law of Definite Proportions involve calculating the ratio between the masses of the elements in different compounds. This information can be used to identify if two unknown compounds are the same or not

Empirical Formula

Molecular and empirical formulae

The chemical formula of a molecular compound is also known as molecular formula
The molecular formula shows the actual number of atoms in a molecule
- Eg: the molecular formula of butene is C₄H₈
The empirical formula is the simplest formula and shows the simplest ratio of the atoms in a compound
- Eg: the empirical formula of butene is CH₂
The chemical formula of an ionic compound is both empirical and molecular formula at the same time

Exam Tip

The calculation of empirical and molecular formulae are assessed continuously. Expect at least one question related to this topic either in Section I or Section II of your exam

Calculating the empirical formula

The empirical formula shows the smallest ratio between the atoms in molecule or a formula unit
It can be calculated from the masses of each element in a sample of compound
If percentages are given instead of masses, assume 100 g of sample and then transform all your % to g

Worked example

Calculate the empirical formula of a compound which contains 8.0 g of calcium and 14.2 g of chlorine

Answer:

	Calcium	Chlorine
Write down the mass of each element in the sample	8.0	14.2
Calculate the moles using the molar mass using n = m/M	$n = \frac{8.0 g of Ca}{40.08 g {mol}^{- 1} of Ca}$ $n = 0.20 mol of Ca$	$n = \frac{14.2 g of Cl}{35.45 g {mol}^{- 1} of Cl}$ $n = 0.40 mol of Cl$
Divide answers by the smaller value	$= \frac{0.2}{0.2}$ = 1	$= \frac{0.4}{0.2}$ = 2
Use the answers as subscripts	CaCl₂

Worked example

Calculate the empirical formula of a compound which contains 86.0% of carbon and 14.0% g of hydrogen

Answer:

	Carbon	Hydrogen
Write down the % of each element in the sample	86.0%	14.0%
Assume 100 g of sample and transform % to g	86.0 g	14.0 g
Calculate the moles using the molar mass using n = m/M	$n = \frac{86.0 g of C}{12.01 g {mol}^{- 1} of C}$ $n = 7.16 mol C$	$n = \frac{14.0 g of H}{1.008 g {mol}^{- 1} of H}$ $n = 13.8 mol H$
Divide answers by the smaller value	$= \frac{7.16}{7.16}$ = 1	$= \frac{13.8}{7.16}$ = 1.93 ≈ 2
Use the answers as subscripts	CH₂

Calculating the molecular formula

The molecular formula is a “factor” of the empirical formula
It can be calculated by dividing the molar mass of the molecular formula and the molar mass of the empirical formula
This calculation will result in a whole number greater or equal than 1

$whole number = \frac{molar mass of the molecular formula}{molar mass of the empirical formula}$

Finally, all the subscripts from the empirical formula need to be multiplied times the whole number, obtaining the molecular formula

Worked example

The empirical formula of an unknown compound is CH₂O, and the molar mass of the compound is 180 g mol^-1. Determine the molecular formula of the unknown compound

Answer

Step 1. Calculate the molar mass (M) of the empirical formula
- M of empirical formula = (C x 1) + (H x 2) + (O x 1)
- M of empirical formula = (12.01 x 1) + (1.008 x 2) + (16.00 x 1)
- M of empirical formula = 30.026 g mol^-1
Step 2. Calculate the whole number by dividing M of the molecular formula by M of the empirical formula. If the whole number has decimal places, round it to the nearest integer

$whole number = \frac{180 g {mol}^{- 1}}{30.026 g {mol}^{- 1}}$

$whole number = 5.99$

$whole number \approx 6$

Step 3. Multiply the subscripts of the empirical formula times 6
- Molecular formula = 6 x CH₂O
- Molecular formula = C₆H₁₂O₆
The molecular formula of the unknown compound is C₆H₁₂O₆

Elemental Composition (College Board AP Chemistry)

Revision Note

Formula Units

Pure substances and chemical formula

Molecular compounds

A molecule of water (H2O)

Ionic compounds

Formula unit of sodium chloride (NaCl)

The Law of Definite Proportions

Law of Definite Proportions

Exam Tip

Empirical Formula

Molecular and empirical formulae

Exam Tip

Calculating the empirical formula

Worked example

Worked example

Calculating the molecular formula

Worked example

You've read 0 of your 0 free revision notes

Get unlimited access

Join the 100,000+ Students that ❤️ Save My Exams

Author: Martín