empirical and molecular formulas part 2: calculating the empirical formula
TRANSCRIPT
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.