AP®ChemistryCollege BoardStudy GuidesUnit 1: Atomic Structure & PropertiesElements & MixturesElemental Composition

Elemental Composition (College Board AP® Chemistry)

Study Guide

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Written by: Martín

Reviewed by: Stewart Hird

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

Examiner Tips and Tricks

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

Examiner Tips and Tricks

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

straight n equals fraction numerator 8.0 space straight g space of space Ca over denominator 40.08 space straight g space mol to the power of negative 1 end exponent space of space Ca end fraction

straight n equals 0.20 space mol space of space Ca 

straight n equals fraction numerator 14.2 space straight g space of space Cl over denominator 35.45 space straight g space mol to the power of negative 1 end exponent space of space Cl end fraction

straight n equals 0.40 space mol space of space Cl 

Divide answers by the smaller value

 equals fraction numerator 0.2 over denominator 0.2 end fraction

= 1

 begin mathsize 14px style equals fraction numerator 0.4 over denominator 0.2 end fraction end style

= 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=86.0 g of C12.01 g mol-1 of C 

n=7.16 mol C

 n=14.0 g of H1.008 g mol-1 of H

n=13.8 mol H

Divide answers by the smaller value

 =7.167.16

= 1

 =13.87.16

= 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 space number space equals space fraction numerator molar space mass space of space the space molecular space formula over denominator molar space mass space of space the space empirical space formula end fraction space 

  • 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 space number space equals space fraction numerator 180 space straight g space mol to the power of negative 1 end exponent over denominator 30.026 space straight g space mol to the power of negative 1 end exponent end fraction

whole space number space equals space 5.99

whole space number space almost equal to space 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

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    Martín

    Author: Martín

    Expertise: Chemistry Content Creator

    Martín, a dedicated chemistry teacher and tutor, excels in guiding students through IB, AP, and IGCSE Chemistry. As an IB Chemistry student, he came from hands-on preparation, focusing on practical exam techniques and rigorous practice. While at Universidad San Francisco de Quito, his academic journey sparked a passion for computational and physical chemistry. Martín specializes in chemistry, and he knows that SaveMyExams is the right place if he wants to have a positive impact all around the world.

    Stewart Hird

    Author: Stewart Hird

    Expertise: Chemistry Lead

    Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.