Mark S. Cracolice
Edward I. Peters
www.cengage.com/chemistry/cracolice
Chapter 7
Chemical Formula Relationships
Mark S. Cracolice • The University of Montana
Number of Atoms in a Formula
In writing the formula of a substance, subscript
numbers are used to indicate the number of
atoms or groups of atoms of each element in
the formula unit.
Number of Atoms in a Formula
How many atoms of each element are in a formula unit
of ammonium carbonate?
The formula of the ammonium ion is NH4+
The formula of carbonate is CO32–
The formula of the ammonium carbonate is (NH4)2CO3.
Each element in the parentheses is multiplied by two.
Total number of atoms of each element:
2 nitrogen atoms, 8 hydrogen atoms,
1 carbon atom, 3 oxygen atoms
Atomic Mass
Atomic Mass
The average mass of
atoms of an element
By definition the mass
of a carbon-12 atom
is 12 u.
Molecular & Formula Mass
Molecular Mass
The sum of the atomic
masses of each atom in
the molecule.
Formula mass
The sum of atomic
masses in the formula
unit
Molecular & Formula Mass
What is the formula mass of calcium phosphate?
Calcium ion: Ca2+
Phosphate ion: PO43–
Calcium phosphate: Ca3(PO4)2
Ca
3 × 40.08 u
= 120.24 u
P
2 × 30.97 u
= 61.94 u
O
8 × 16.00 u
= 128.00 u
Ca3(PO4)2
310.18 u
Defenition of Mole
The mole is the amount of substance that
contains as many elementary entities as there
are atoms in exactly 12 grams of carbon-12.
In 12 g of carbon-12 there are 6.022 x 10 23 atoms
1 mole of any substance = 6.022 x 10 23 units
of that substance.
When the mole is used the elementary entities must
be specified: atoms, molecules, ions…
Avogadro’s Number NA
The number of elementary units in one mole
6.02214179  1023 units/mol
The Avogadro constant is a conversion factor
between units and mole.
Conversion of Mole to Molecules
How many carbon dioxide molecules are in 2.0 moles of carbon
dioxide?
6.02 ´ 1023 molecules CO2
2.0 mol CO2 ×
=
mol CO2
1.2 × 1024 molecules CO2
Molar Mass
Molar mass of a substance is the mass in grams
of one mole of the substance.
Units: g/mol
Molar mass of an element is the mass of the
element per mole of its atoms.
Atomic mass unit and gram
The mass of one atom of carbon-12 is exactly 12
atomic mass units.
The mass of one mole of carbon-12 atoms (6.022 x
1023 atoms of carbon-12) is exactly 12 grams
(6.022 x 1023 atoms) x (12 u/atom) = 12 grams
6.022 x 1023 u = 1 gram
Molar Mass
The mass of one atom of carbon-12 is exactly 12
atomic mass units.
The mass of one mole of carbon-12 atoms is exactly 12
grams.
This leads to the conclusion:
The molar mass of any substance in grams per
mole is numerically equal to the atomic,
molecular or formula mass of that substance
in atomic mass units.
Molar Mass
Calculate the mass of one NH3 molecule and the mass
of one mole of NH3 molecules.
14.01 u + 3x1.008 u = 17.03 u
The mass of one ammonia molecule is 17.03 u.
To change from molecular mass to molar mass, change
the units from u to g/mol: 17.03 g/mol.
One mole of ammonia molecules has a mass of
17.03 g.
Molar Mass
Mass, # of Moles, # of Units
Molar mass, MM, links
mass in grams with
the number of moles.
Avogadro’s number,
NA, links the number
of moles with the
number of particles.
Mass, # of Moles, # of Units
How many molecules are in 454 g of water?

g
moles

1 mol H O
454g x
2
18.02 g H O
2
x
molecules
6.02  10 23 molecules
mol H O
= 1.52 x 1025 molecules H2O
2
H O
2
Mass↔Moles↔ Units
How many hydrogen atoms are in 1.0 kg of ammonia?
1000 g NH3
1 mol NH3
1.0 kg NH3 × kg NH
×17.03 g NH
3
3
×
6.02 ´ 1023 molecules NH3
× 3 atoms H
molecule NH3
mol NH3
=
1.1 × 1026 atoms H
Percentage Composition
Percentage
%A=
parts of A
total parts × 100%
The percentage composition of a compound is the
percentage by mass of each element in the compound.
Percentage Composition
Determine the percentage composition of calcium
fluoride Ca F2.
Solution:
In one mole of Ca F2
1x(40.08 g Ca) + 2x(19.00 g F)
= 78.08 g CaF2
(Solution continued on the next slide)
Percentage Composition
(40.08 g Ca)
78.08 g CaF
 100 = 51.33 % Ca
2
2x(19.00g F)
78.08 g CaF
 100 = 48.67% F
2
Check: 51.33% + 48.67% = 100.00%
Empirical Formula
Empirical Formula
The simplest ratio of atoms of the elements in a
compound.
The empirical formula of C2H4 is CH2.
Likewise, the empirical formula of C3H6 is CH2.
All compounds with the general formula CnH2n
have the same empirical formula and
therefore the same percentage composition.
Empirical Formula
Write the empirical formulas of benzene, C6H6, and
octane, C8H18.
Look for the simplest whole-number ratio of elements:
For C6H6, the 6/6 ratio can be reduced to 1/1: CH.
For C8H18, the 8/18 ratio can be divided by 2 on top and
bottom to be reduced to 4/9: C4H9.
Find Empirical Formula
To find empirical formula, you need to find the
ratio of atoms of the elements
ratio of atoms = ratio of moles of atoms
How to Find an Empirical Formula
1. Find the masses of different elements in a sample of
2.
3.
4.
5.
the compound.
Convert the masses into moles of atoms.
Determine the ratio of moles of atoms.
Express the moles ratio as the smallest possible
ratio of integers.
Write the empirical formula, using the number in the
integer ratio as the subscript in the formula.
Find Empirical Formula
What is the empirical formula of a compound that has
85.6% carbon, 14.4 % hydrogen?
Solution:
It is usually helpful to organize the calculations in a
table with the following headings:
Element
Grams
Moles
Mole
Ratio
Formula
Ratio
Empirical
Formula
Find Empirical Formula
Element Grams
C
85.6
Moles
85.6 g C
12.01 g/mol C
7.13
H
14.4
14.4 g H
1.008 g/mol H
14.3
Mole
Ratio
Formula
Ratio
7.13
7.13
Empirical
Formula
1
1
14.3
7.13
2.01
2
CH2
Find Empirical Formula
What is the empirical formula of a compound that
analyzes as 20.0% carbon, 2.2% hydrogen, and
77.80% chlorine?
Element
Grams
Moles
Mole
Ratio
Formula
Ratio
C
20.0
1.67
1
3
H
2.2
2.2
1.3
4
Cl
77.8
2.19
1.31
4
Empirical
Formula
C3H4Cl4
Molecular Formula
The molecular formula of a compound can be
found by determination of the number of
empirical formula units in the molecule.
molar mass of compound
molar mass of empirical formula
How to Find the Molecular Formula
1. Determine the empirical formula of the
compound.
2. Calculate the molar mass of the empirical
formula unit.
3. Divide the molar mass of the compound by
the molar mass of the empirical formula unit to
get n, the number of empirical formula units
per molecule.
Molecular Formula
What is the molecular formula of a compound with the
empirical formula C2H5and a molar mass of 58.12
g/mol?
The molar mass of the empirical formula unit is
2(12.01 g/mol C) + 5(1.008 g/mol H) =29.06 g/mol
The number of empirical formula units per molecule is
58.12 g/mol
29.06 g/mol
=2
(C2H5)2 = C4H10
Find Molecular Formula
• An unknown compound is found to be 40.0%
of carbon, 6.71% of hydrogen and the
remainder is oxygen. The molar mass of the
compound is 180.16 g/mol. Find the empirical
and molecular formulas of the compound.
First Find Empirical Formula
Element
Grams
Moles
Mole
Formula
Empirical
Ratio
Ratio
Formula
C
40.0
3.33
1
1
H
6.71
6.66 2
2
O
53.3
3.33
1
1
CH2O
Calculate Molar Mass/Empirical Formula Mass
The molar mass of the empirical formula unit is
(12.01 g/mol C) + 1(1.008 g/mol H) + (16.00 g/mol O) =
30.03 g/mol
The number of empirical formula units per
molecule is
180.16 g/mol = 6
30.03 g/mol
(CH2O) 6 = C6H12O6
Molecular Formula
Another way to find molecular formula is to
consider one mole of the compound (180.16 g
of compound). The numbers of moles are
also numbers of atoms in the molecule.
The masses of carbon, hydrogen and oxygen are :
40.0% x 180.16 g = 72.06 g of C
6.71% x 180.16 g = 12.09 g of H
53.3% x 180.16 g = 96.03 g of O
Molecular Formula
Element
Grams
Moles Molecular Formula
C
72.06
6
H
12.09
12
O
96.03
6
C6H12O6
Homework
7, 15, 21, 23, 43, 57, 61, 63, 65, 68, 74.