EMPIRICAL FORMULAS (FORMULAE)
“Empirical Formula is the simplest representation of a compound with its
atoms shown in correct ratios of small whole numbers”
“MolecularFormula is the representation of a compound with its atoms
shown in whole numbers as actually present in a single molecule of the
substance”
EXAMPLES OF COMPOUNDS WITH SAME EMPIRICAL
FORMULAS, BUT DIFFERENT MOLECULAR FORMULAS
Compound
Actual Mol. Formula
1. Acetic Acid
Glucose
C2 H4 O2
C6 H12 O6
Empirical
Formula
C H2 O
C H2 O
2. Sorbic Acid
Acrolein
C6 H8 O2
CHO
C3 H4 O
C3 H4 O
3. Propionaldehyde
Caproic Acid
C3 H6 O
C6 H12 O2
C3 H6 O
C3 H6 O
HOW TO CALCULATE EMPIRICAL FORMULA FROM
EXPERIMENTAL SAMPLE COMBUSTION DATA
Example: Question #1-65
A hydrocarbon (i.e. compound which contains only C and H atoms) is
burned, and gives only CO2 and H2O as products. The CO2 and H2O
produced are collected and weighed.
Experimental Data
CO2 mass = 3.701 g
H2O mass = 1.082 g
Steps to Calculate Empirical Formula of Hydrocarbon:
1. Set up generic balanced equation for combustion of Hydrocarbon (Cn
Hm)
Cn Hm
+ x O2
nCO2
+
m/2 H2O
[Note, just for interest: x = n + m/4]
2.
Calculate moles of CO2 formed
Note: Moles of CO2 formed = Moles of C in original hydrocarbon compound
CO2 mass = 3.701 g
j CO2 moles = 3.701/(12.01 + 2x15.99) = 0.0841 moles = n
j moles of C in original compound = 0.0841
3.
Calculate moles of H2O formed
Note: Moles of H2O formed = ½ Moles of H in original hydrocarbon
compound
H2O mass = 1.082 g
j
H2O moles =1.082/(2x1.008 + 15.99) = 0.0601 = m/2
j Moles of H = m = 2x0.0601 = 0.1202 moles
4.
Generate Crude Empirical Formula, based on what you know.
Empirical Formula for Hydrocarbon = Cn Hm
Substitute values you have just calculated for n and m:
C0.0841 H0.1202
5.
Convert Formula to Simple Whole Numbers.
(a) Divide by smallest number = 0.0841
C H
Gives
1
1.429
(b) Convert to simple, small whole numbers:(Note: Only Chemistry
Professors might be expected to have the following weird information
stored away in the dark recesses of their minds!)
0.429 = 3/7
j 1.429 = 7/7 + 3/7 = 10/7
j C:H = 7/10
j Empirical Formula = C7H10