Chapter 6
Empirical and
Molecular Formulas
EMPIRICAL FORMULA
A chemical formula that indicates the relative
proportions of the elements in a molecule rather than
the actual number of atoms of the elements.
(An empirical formula may be obtained from percentage
composition of elements in a compound.)
MOLECULAR FORMULA
A chemical formula that indicates the actual number
of atoms of the elements in a molecule.
(Information in addition to percentage composition of
elements is needed to determine a molecular formula.)
Molecular Formulas
May Differ from Empirical Formulas
Benzene
Empirical Formula, CH
Molecular Formula, C6H6
Acetylene
Empirical Formula, CH
Molecular Formula, C2H2
Percent Composition
Mass Percent Composition of
Compounds
Percentage of each element in a compound by mass
Can be determined from
1. the formula of the compound, or
2. the experimental mass analysis of the compound
Mass Percent Composition of
Compounds
The mass percent composition, or mass percent,
of an element is the element’s percentage of the total
mass of the compound
Mass percent of element X =
Mass of X in a sample
Mass of the sample
x 100%
1. Find the mass percent of Na in NaCl
1 x (molar mass of Na)
Mass % Na =
molar mass of NaCl x 100%
1 x (22.99 g/mol)
Mass % Na =
x 100 % = 39.33%
58.44 g/mol
22.99 + 35.45 = 58.44
2. Find the mass percent of Cl in C2Cl4F2
Mass % Cl =
4 x (molar mass of Cl)
x 100%
molar mass of C2Cl4F2
4 x molar mass Cl = 4(35.45) = 141.8 g/mol
molar mass C2Cl4F2 = 2(12.01)+4(35.45)+2(19.00) = 203.8 g/mol
141.8 g/mol
Mass % Cl =
x 100% =69.58%
203.8 g/mol
Mass Percent as a Conversion Factor
3. If NaCl is 39% sodium,
find the mass of table salt containing 2.4 g of Na.
g
Na
g NaCl
100 g NaCl
39 g Na
2.4 g Na x 100 g NaCl
39 g Na
39 g Na
100 g NaCl
=
6.1538
6.2 g gNaCl
NaCl
4. Find the mass of sodium in 6.2 g of NaCl
g
NaCl
mol NaCl
1.00 mol NaCl
58.44 g NaCl
6.2 g NaCl x
mol
1.00 mol Na
1.00 mol NaCl
Na
g
Na
22.99 g Na
1.00 mol Na
1.00 mol NaCl 1.00 mol Na
22.99 g Na
x
x
= 2.4390
2.4
58.44 g NaCl 1.00 mol NaCl 1.00 mol Na
Na
gg Na
Empirical Formula
Simplest, whole-number ratio of the
atoms of elements in a compound
Can be determined from elemental analysis
Finding an Empirical Formula
from Mass Composition
1. Convert grams of elements to moles of elements.
2. Write a pseudoformula using moles as subscripts
3. Divide all by smallest number of moles
4. Multiply all mole ratios by number to make all
whole numbers
5. A 33.34 g sample of aspirin contains 20.01 g of carbon,
1.49 g of hydrogen, and 11.84 g of oxygen. Find the
empirical formula of aspirin.
20.01 + 1.49 + 11.84 = 33.34
gC
gH
gO
mol C
mol H
mol O
!
pseudoformula
CxHyOz
Manipulate
subscripts to obtain
whole-number ratio
empirical formula
CxHyOz
Calculate the moles of each element
20.01 g C x 1.000 mol C = 1.666 mol C
12.01 g C
1.48 g H x
1.000 mol H= 1.48 mol H
1.008 g H
1.000 mol O
11.84 g O x
= 0.7400 mol O
16.00 g O
Write a pseudoformula
C1.666H1.48O0.7400
C1.666H1.48O0.7400
Find the integer mole ratio
÷ 0.7400
C2.25H2.00O1.00
Multiply subscripts by factor to give whole number
(x 4)
C9H8O4
Finding an Empirical Formula
from % Composition
1. Assume 100.0 g of the compound and convert the
percentages to grams
2. Convert grams to moles
3. Write a pseudoformula using moles as subscripts
4. Divide all by smallest number of moles
5. Multiply all mole ratios by number to make all
whole numbers
6. Determine the empirical formula of stannous
fluoride, which contains 75.7% Sn (118.70 g/mol)
and the rest fluorine (19.00 g/mol)
Given: 75.7% Sn, (100 – 75.3) = 24.3% F
Therefore, in 100 g of stannous fluoride
there are 75.5 g Sn and 24.3 g of F ! !
g
Sn
mol Sn
F
mol F
g
!
pseudo
formula
empirical
formula
6. Determine the empirical formula of stannous
fluoride, which contains 75.7% Sn (118.70 g/mol)
and the rest fluorine (19.00 g/mol)
Element
Sn
F
Ratio in
Grams
Molar Mass
Ratio in
Moles
Ratio in
Moles
75.7g
X
1mol Sn
118.7g
0.6377
1
X
1mol F
19.00g
1.279
2.005
24.3g
SnF2
÷0.6377
7. Determine the empirical formula of magnetite, which
contains 72.4% Fe (55.85) and the rest oxygen (16.00)
Given: 72.4% Fe, (100 – 72.4) = 27.6% O
Therefore, in 100 g of magnetite there
are 72.4 g Fe and 27.6 g of O !!
g
Fe
mol Fe
g
O
mol O
!
pseudo
formula
empirical
formula
Practice — Determine the empirical formula of magnetite,
which contains 72.4% Fe (55.85) and the rest oxygen (16.00)
Element
Ratio in
Grams
Molar Mass
Ratio in
Moles
Ratio in
Moles
Ratio in
Moles
Fe
72.4g
X
1mol Fe
55.85g
1.296
1
3
X
1mol O
16.00g
1.725
1.33
4
O
27.6g
÷1.296
Fe3O4
x3
Molecular Formulas
The molecular formula is a multiple
of the empirical formula.
To determine the molecular formula you need to know the
empirical formula and the molar mass of the compound.
8. Find the molecular formula of butanedione if its empirical
formula is C2H3O and its molar mass (MM) is 86.03 g/mol.
Factor of 2
Molar Mass (emp. form.)
= 2 x (12.01 gC/molC) + 3 x (1.008 gH/molH) + 1 x (16.00 gO/molO) = 43.04
Molecular formula = C2H3O
x 2 = C4H6O2
g/mol
9. Benzopyrene has a molar mass of 252 g and an empirical
formula of C5H3. What is its molecular formula?
(12.01 g C/mol C, 1.01 g H/mol H)
C5 = 5(12.01 g) = 60.05 g
H3 = 3(1.01 g) = 3.03 g
C 5H 3
= 63.08 g
252
?
Molecular formula = {C5H3} x 4 = C20H12
Calculating Molecular Formulas for
Compounds: Fructose
10. Find the molecular formula for fructose (a
sugar found in fruit) from its empirical formula,
CH2O, and its molar mass, 180.2 g/mol.
The molecular formula is a whole-number multiple
of CH2O.
Based on CH2O,
empirical formula molar mass = 1(12.01) + 2(1.01) + 1(16.00)
= 30.03 g/mol
30.03 g/mol---------???-------->180.2 g/mol
Molecular Formulas
May Differ from Empirical Formulas
Formaldehyde
Empirical Formula, CH2O
Molecular Formula, CH2O
Glucose
Empirical Formula, CH2O
Molecular Formula, C 6H12O6
Fructose
Empirical Formula, CH2O
Molecular Formula, C 6H12O6