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Ch. 7.4 Determining Chemical Formulas
Ch. 7.4 Determining Chemical Formulas
Objectives
• Define empirical formula, and explain how the term applies to
ionic and molecular compounds.
• Determine an empirical formula from either a percentage or a
mass composition.
• Explain the relationship between the empirical formula and the
molecular formula of a given compound.
• Determine a molecular formula from an empirical formula.
Ch. 7.4 Determining Chemical Formulas
• An empirical formula consists of the symbols for the elements
combined in a compound, with subscripts showing the smallest
whole-number mole ratio of the different atoms in the compound.
• For an ionic compound, the formula unit is usually the compound’s
empirical formula.
• For a molecular compound, however, the empirical formula does
not necessarily indicate the actual numbers of atoms present in
each molecule.
Empirical Formula
Example
Everyone has 2 hands and 10 fingers: H2F10
Empirical formula: HF5
Empirical Formula
Benzene
Molecular Formula: C6H6
Proportion of C to H is 1:1
Empirical Formula: CH
More than one substance may reduce to the same
empirical formula.
Benzene: C6H6
Acetylene: C2H2
Ch. 7.4 Determining Chemical Formulas
Calculation of Empirical Formulas
• To determine a compound’s empirical formula from its percentage
composition, begin by converting percentage composition to a
mass composition.
• Assume that you have a 100.0 g sample of the compound.
• Then calculate the amount of moles of each element in the
sample.
• Simplest ratio of moles
Ch. 7.4 Determining Chemical Formulas
Empirical Formula
We have a sample that is 48.6% C, 8.16% H, and 43.2% O
by weight. What is the empirical formula?
Step 1: % Composition to Moles
48.6% C = 48.6g C;
8.16% H = 8.16g H;
43.2% O = 43.2g O;
48.6g / 12.01g/mol
8.16g / 1.01g/mol
43.2g / 16.00g/mol
= 4.05 mol C
= 8.08 mol H
= 2.70 mol O
Ch. 7.4 Determining Chemical Formulas
Empirical Formula
Step 2: Mole Ratio
Divide each element by the smallest number of moles
C: 4.05 mol / 2.70 mol = 1.5
H: 8.08 mol / 2.70 mol = 2.99 RU = 3
O: 2.70 mol / 2.70 mol = 1
Ch. 7.4 Determining Chemical Formulas
Empirical Formula
Step 3: Atom Ratio
Ratio of atoms MUST be a whole number!
C = 1.5 x 2 = 3
H=3x2=6
O=1x2=2
Ch. 7.4 Determining Chemical Formulas
Empirical Formula
Step 4: Express as an Empirical Formula
C 3H 6 O 2
Ch. 7.4 Determining Chemical Formulas
Molecular Formula
Given the composition percentages and the samples Molecular
Mass.
First: We use the percent composition to Empirical Formula Rules.
Weight % → Mass → Mol → Ratio of Mol → Atoms = Empirical
Formula
Ch. 7.4 Determining Chemical Formulas
Molecular Formula
Second:
Calculate the Empirical Mass from the Empirical Formula.
Third:
Divide the Molecular Mass (given) by Empirical Mass
(MM/ EM) = n
Last: Multiply Empirical Formula subscripts by n
(EF)n = Molecular Formula
Ch. 7.4 Determining Chemical Formulas
Molecular Formula
A compound is 92.30% C and 7.80% H with a mass of
78.12g/mol. Find the empirical formula and the molecular
formula of this compound.
Ch. 7.4 Determining Chemical Formulas
Molecular Formula
Mass to mol:
C: 92.30g/ 12.01 = 7.69 mol
H: 7.80g/ 1.01 = 7.72 mol
Ratio of Mol:
C: 7.69 / 7.69 = 1
H: 7.72/ 7.69 = 1
Atoms:
C:H 1:1
Empirical Formula (EF): CH
Ch. 7.4 Determining Chemical Formulas
Molecular Formula
Empirical Mass (EM): (12.01) + (1.01)= 13.02
n = (MM/EM): 78.12 / 13.02 = 6
(EF)n: (CH) (6)= C6H6
Ch. 7.4 Determining Chemical Formulas
Molecular Formula
Sample Problem
The empirical formula of a compound of phosphorus and oxygen
was found to be P2O5. Experimentation shows that the molar mass
of this compound is 283.89 g/mol. What is the compound’s
molecular formula?