AP Chemistry 1.3: Elemental composition of pure substances

This topic contains several important calculation types: empirical formulas, molecular formulas, structural formulas, mass percent, percent-composition analysis, and combustion analysis.

A chemical formula tells you which elements are present and gives information about their relative amounts.

You need to distinguish three formula types.

Empirical formula

• Shows the simplest whole-number ratio of atoms.

Molecular formula

• Shows the actual number of atoms of each element in one molecule.

Structural formula

• Shows how atoms are connected or arranged .

Example:

Glucose:

• C₆H₁₂O₆

All subscripts can be divided by 6.

• So its empirical formula is:

• CH₂O

• That means the carbon : hydrogen : oxygen ratio is:

• 1 : 2 : 1

• It does NOT mean a glucose molecule contains only one carbon atom.

• It only shows the simplest ratio.

A molecular formula gives actual numbers.

Hydrogen peroxide:

• H₂O₂

Empirical formula:

• HO

Molecular formula:

• H₂O₂

The molecular formula is a whole-number multiple of the empirical formula.

A structural formula gives information about connectivity .

• This becomes important because different substances may have the same molecular formula but different structures.

For Unit 1, the main distinction to remember is:

• Empirical: simplest ratio

• Molecular: actual atom count

• Structural: how atoms are connected

Percent composition tells you what percentage of the total mass comes from a particular element.

• Formula:

• mass % of element = (mass of element / total mass) × 100

For a pure compound, you can use molar masses:

• mass % = (mass contributed by element in one mole / molar mass of compound) × 100

Molar mass H₂O:

• 2(1.008) + 16.00

• = 18.016 g/mol

Oxygen contributes:

• 16.00 g

• So:

• (16.00 / 18.016)(100)

• ≈ 88.81% O

Hydrogen contributes the rest:

• ≈ 11.19%

Together:

• 88.81 + 11.19 ≈ 100%

Find the percent carbon in CO₂.

Molar mass:

• C = 12.01

• O₂ = 32.00

• Total = 44.01

Carbon percent:

• (12.01 / 44.01)(100)

• ≈ 27.29% carbon

Suppose a compound contains:

• 72.0 g C

• 18.0 g H

• 28.0 g N

The key rule is:

Do NOT compare grams directly.

Chemical formulas represent mole ratios , not mass ratios.

• So first convert every element into moles.

• This is similar to a current Khan Academy Unit 1 practice problem.

Step 1: Convert Each Mass to Moles

Carbon:

• 72.0 g / 12.0 g/mol ≈ 6.00 mol

Hydrogen:

• 18.0 g / 1.00 g/mol ≈ 18.0 mol

Nitrogen:

• 28.0 g / 14.0 g/mol ≈ 2.00 mol

Step 2: Divide Everything by the Smallest

• Smallest = 2.00

C:

• 6 / 2 = 3

H:

• 18 / 2 = 9

N:

• 2 / 2 = 1

Ratio:

• 3 : 9 : 1

Empirical formula:

• C₃H₉N

If percentages are given instead of grams, use the 100 gram trick .

• Suppose:

• 40.0% C

• 6.7% H

• 53.3% O

Pretend there are exactly 100 g .

• Then:

• 40.0% → 40.0 g C

• 6.7% → 6.7 g H

• 53.3% → 53.3 g O

Now convert to moles.

C:

• 40.0 / 12.01 ≈ 3.33

H:

• 6.7 / 1.008 ≈ 6.65

O:

• 53.3 / 16.00 ≈ 3.33

Divide all by 3.33:

• C ≈ 1

• H ≈ 2

• O ≈ 1

Empirical formula:

• CH₂O

Suppose you get:

• 1 : 1.5 : 1

A chemical formula cannot have a subscript of 1.5.

Multiply the entire ratio by 2:

• 2 : 3 : 2

• So formula:

• X₂Y₃Z₂

Useful patterns:

• 1.5 → multiply everything by 2

• 1.33 or 1.67 → often multiply by 3

• 1.25 or 1.75 → often multiply by 4

Do not randomly round 1.5 to 2.

• That changes the ratio.

Suppose the empirical formula is:

• CH₂O

First determine the empirical formula mass:

• C = 12.01

• H₂ = 2.016

• O = 16.00

• Total ≈ 30.03 g/mol

• Suppose experimental data says the molecular molar mass is:

• 180.2 g/mol

Calculate:

• n = molecular molar mass / empirical formula mass

• 180.2 / 30.03 ≈ 6

• So multiply every empirical-formula subscript by 6:

• CH₂O × 6

• = C₆H₁₂O₆

Combustion analysis is one of the harder Unit 1 ideas.

• It is commonly used to determine the composition of compounds containing carbon and hydrogen , sometimes oxygen as well.

When a carbon/hydrogen compound burns completely:

Carbon atoms become part of CO₂

Hydrogen atoms become part of H₂O

• So measurements of CO₂ and H₂O allow you to work backward.

Every CO₂ molecule contains:

• 1 carbon atom

• Therefore:

• 1 mol CO₂ = 1 mol C

If you obtain 0.500 mol CO₂:

• you had:

• 0.500 mol C

Every H₂O molecule contains:

• 2 hydrogen atoms

• Therefore:

• 1 mol H₂O = 2 mol H

If combustion produces:

• 0.500 mol H₂O

• then:

• 0.500 × 2

• = 1.00 mol H atoms

For an unknown containing C, H, and possibly O:

Step 1

• CO₂ mass → moles CO₂ → moles C

Step 2

• H₂O mass → moles H₂O → moles H

Step 3

Convert moles C and H to their masses if oxygen is present.

Step 4

Find oxygen by difference:

• mass O = original sample mass − mass C − mass H

Step 5

• Convert oxygen mass → moles O.

Step 6

Divide all mole values by the smallest.

Step 7

Convert to whole-number ratio.

• That gives the empirical formula.