empirical and molecular formulas part 2: calculating the empirical formula

Empirical and Molecular FormulasPart 2: Calculating the empirical formula

Objectives:

• When you complete this presentation, you will be able to• determine the empirical formula of a compound from the

percent composition of that compound• determine the molecular formula of a compound from the

empirical formula and molar mass of the compound

Introduction

• Chemical formulas that have ...• the lowest whole number ratio of atoms in the compound

are empirical formulas.• the total number of atoms in the compound are molecular

formulas.

Introduction

• If we know the chemical formula of a compound, then we can find the percent composition of each of the atoms in the compound.

• We use the molar mass of the compound and the average atomic masses of the atoms in the compound.

percent composition = 100%✕atomic mass of atoms

molar mass of compound

Finding the Empirical Formula

• If we know the percent composition of a compound, then we can find the empirical formula of the compound.• First, we assume that we have 100 g of the compound and

find the mass of each atom in the compound.• Second, we find the number of mols of each atom.• Third, we find the lowest whole number ratio of mols.

Finding the Empirical Formula

For example, we have a compound with 52.2% C, 13.1% H, and 34.7% O.

Step 1: Assume 100 g of compound

mC = 52.2 g, mH = 13.1 g, and mO = 34.7g

Step 2: Find the number of moles of each atomnC =

mC

MC

=52.2 g12.0

g/mol

= 4.35 mol

nH =mH

MH=

13.1g1.01

g/mol

= 13.0 mol

nO =mO

MO

=34.7 g16.0

g/mol

= 2.17 mol

Finding the Empirical Formula

For example, we have a compound with 52.2% C, 13.1% H, and 34.7% O.

Step 1: Assume 100 g of compound

mC = 52.2 g, mH = 13.1 g, and mO = 34.7g

Step 2: Find the number of moles of each atom

nC = 4.35 mol, nH = 13.0 mol, and nO = 2.17 mol

Step 3: Find the lowest whole number ratio of mols• We find the proper ratios by dividing the lowest number

of mols into the higher number of mols.

• We will be dividing nO into nC and nO into nH.

Finding the Empirical Formula

For example, we have a compound with 52.2% C, 13.1% H, and 34.7% O.

Step 1: Assume 100 g of compound

mC = 52.2 g, mH = 13.1 g, and mO = 34.7g

Step 2: Find the number of moles of each atom

nC = 4.35 mol, nH = 13.0 mol, and nO = 2.17 mol

Step 3: Find the lowest whole number ratio of mols

=nC

nO

=4.35 mol2.17 mol

21

=nH

nO

=13.0 mol2.17 mol

61

Finding the Empirical Formula

For example, we have a compound with 52.2% C, 13.1% H, and 34.7% O.

Step 1: Assume 100 g of compound

mC = 52.2 g, mH = 13.1 g, and mO = 34.7g

Step 2: Find the number of moles of each atom

nC = 4.35 mol, nH = 13.0 mol, and nO = 2.17 mol

Step 3: Find the lowest whole number ratio of mols

This gives the empirical formula: C2H6O

=nC 2nO 1

=nH 6nO 1

Finding the Empirical Formula

For example, we have a compound with 44.9% K, 18.4% S, and 36.7% O.

Step 1: Assume 100 g of compound

mK = 44.9 g, mS = 18.4 g, and mO = 36.7g

Step 2: Find the number of moles of each atomnK =

mK

Mk=

44.9 g39.1

g/mol

= 1.15 mol

nS =mS

MS

=18.4 g32.1

g/mol

= 0.573 mol

nO =mO

MO

=36.7 g16.0

g/mol

= 2.29 mol

Finding the Empirical Formula

For example, we have a compound with 44.9% K, 18.4% S, and 36.7% O.

Step 1: Assume 100 g of compound

mK = 44.9 g, mS = 18.4 g, and mO = 36.7g

Step 2: Find the number of moles of each atom

nK = 1.15 mol, nS = 0.573 mol, and nO = 2.29 mol

Step 3: Find the lowest whole number ratio of mols• We find the proper ratios by dividing the lowest number

of mols into the higher number of mols.

• We will be dividing nS into nK and nS into nO.

Finding the Empirical Formula

For example, we have a compound with 44.9% K, 18.4% S, and 36.7% O.

Step 1: Assume 100 g of compound

mK = 44.9 g, mS = 18.4 g, and mO = 36.7g

Step 2: Find the number of moles of each atom

nK = 1.15 mol, nS = 0.573 mol, and nO = 2.29 mol

Step 3: Find the lowest whole number ratio of mols

=nK

nS

=1.15 mol

0.573 mol

21

=nO

nS

=2.29 mol

0.573 mol

41

Finding the Empirical Formula

For example, we have a compound with 44.9% K, 18.4% S, and 36.7% O.

Step 1: Assume 100 g of compound

mK = 44.9 g, mS = 18.4 g, and mO = 36.7g

Step 2: Find the number of moles of each atom

nK = 1.15 mol, nS = 0.573 mol, and nO = 2.29 mol

Step 3: Find the lowest whole number ratio of mols

This gives the empirical formula: K2SO4

=nK 2nS 1

=nO 4nS 1

Summary

• There is a three step process to finding the empirical mass of a compound when we know the percent composition.• First, we assume that we have 100 g of the compound and

find the mass of each atom in the compound.• Second, we find the number of mols of each atom.• Third, we find the lowest whole number ratio of mols.