empirical and molecular formulas part 2: calculations using percent composition
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Empirical and Molecular Formulas
Part 2: Calculations using percent composition
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 atomsmolar mass of
compound
Finding the Empirical FormulaIf 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➙ 52.2 g C, 13.1 g H, and 34.7 g O
Step 2: Find the number of moles of each atom
52.2 g
nC =
mC MC
= 12.0 g/mol
=4.35 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➙ 52.2 g C, 13.1 g H, and 34.7 g O
Step 2: Find the number of moles of each atom
13.1 g
nC =
4.35 mol
34.7 g52.2 g
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➙ 52.2 g C, 13.1 g H, and 34.7 g O
Step 2: Find the number of moles of each atom
13.1 g
nC =
4.35 mol
34.7 g52.2 g
nH =
mH MH
= 1.01 g/mol
=13.0 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➙ 52.2 g C, 13.1 g H, and 34.7 g O
Step 2: Find the number of moles of each atom
13.1 g
nC =
4.35 mol
34.7 g52.2 g
nH =
13.0 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➙ 52.2 g C, 13.1 g H, and 34.7 g O
Step 2: Find the number of moles of each atom
13.1 g
nC =
4.35 mol
34.7 g52.2 g
nH =
13.0 mol
nO =
mO MO
= 16.0 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➙ 52.2 g C, 13.1 g H, and 34.7 g O
Step 2: Find the number of moles of each atom
13.1 g
nC =
4.35 mol
34.7 g52.2 g
nH =
13.0 mol
nO =
2.17 mol
nOnO2.17 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➙ 52.2 g C, 13.1 g H, and 34.7 g O
Step 2: Find the number of moles of each atom
13.1 g
nC =
4.35 mol
34.7 g52.2 g
nH =
13.0 mol
nO =
2.17 molStep 3: Find the lowest whole number ratio of
mols*Remember, the element with the lowest number of mols goes in the denominator.
nC4.35 mol
nH13.0 mol
2.17 mol
nC
= =21 = =
61
nO
This gives an empirical formula of:
C H O
2 6
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➙ 44.9 g K, 18.4 g S, and 36.7 g O
Step 2: Find the number of moles of each atom
44.9 g
nK =
mK MK
= 39.1 g/mol
=1.15 mol
Finding the Empirical Formula
Step 1: Assume 100 g of compound➙ 44.9 g K, 18.4 g S, and 36.7 g O
Step 2: Find the number of moles of each atom
18.4 g
nK =
1.15 mol
36.7 g44.9 g
For example, we have a compound with 44.9% K, 18.4% S, and 36.7% O.
Finding the Empirical Formula
Step 1: Assume 100 g of compound➙ 44.9 g K, 18.4 g S, and 36.7 g O
Step 2: Find the number of moles of each atom
18.4 g
nK =
1.15 mol
nS =
mS MS
= 32.1 g/mol
=0.573 mol
For example, we have a compound with 44.9% K, 18.4% S, and 36.7% O.
Finding the Empirical Formula
Step 1: Assume 100 g of compound➙ 44.9 g K, 18.4 g S, and 36.7 g O
Step 2: Find the number of moles of each atomnK =
1.15 mol
nS =
0.573 mol
For example, we have a compound with 44.9% K, 18.4% S, and 36.7% O.
Finding the Empirical Formula
Step 1: Assume 100 g of compound➙ 44.9 g K, 18.4 g S, and 36.7 g O
Step 2: Find the number of moles of each atomnK =
1.15 mol
36.7 g
nS =
0.573 mol
nO =
mO MO
= 16.0 g/mol
=2.29 mol
For example, we have a compound with 44.9% K, 18.4% S, and 36.7% O.
Finding the Empirical Formula
Step 1: Assume 100 g of compound➙ 44.9 g K, 18.7 g S, and 36.7 g O
Step 2: Find the number of moles of each atomnC =
4.35 mol
nH =
13.0 mol
nO =
2.29 mol
For example, we have a compound with 44.9% K, 18.4% S, and 36.7% O.
K S O
4
nK =
1.15 mol
nO =
2.29 mol
nS =
0.573 mol
nS0.573 mol
nK1.15 mol
nOnO2.29 mol
Finding the Empirical Formula
Step 1: Assume 100 g of compound➙ 44.9 g K, 18.7 g S, and 36.7 g O
Step 2: Find the number of moles of each atom
Step 3: Find the lowest whole number ratio of mols*Remember, the element with the lowest number of mols
goes in the denominator.
nC
= =21 = =
1This gives an empirical formula of:
2
For example, we have a compound with 44.9% K, 18.4% S, and 36.7% O.
nS0.573 mol
4
=K2SO4
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.