Transcript Empirical and Molecular Formulas

Empirical and Molecular
Formulas
Empirical Formula
• If we are given an unknown substance
how could we determine its chemical
formula?
• We can determine its percent composition
• Then we can determine its empirical
formula: the simplest whole-number
ratio of atoms in a compound
• However, the empirical formula does not necessarily
give information about the number of atoms in a
molecule (molecular formula).
Molecular Formula
Empirical Formula
C2H2
C4H4
CH
C6H6
• The empirical formula gives the combining ratio in its
simplest form
• The molecular formula gives the same ratio but with
the actual number of atoms
Note: The empirical formula and the molecular formula can
be the same in some cases. E.g. H2O, CH4, CO2
Calculating Empirical Formula
E.g. What is the empirical formula of a substance that is 40.0%
carbon, 6.70% hydrogen, and 53.3% oxygen by mass?
Given:
Assume 100 g
mC = 40.0 g
mH = 6.70 g
mO = 53.3 g
Divide each element by the lowest quantity
3.33 mol C = 1
6.63 mol H = 1.99
3.33 mol O = 1
3.33
3.33
3.33
Mole ratio of C:H:O = 1:2:1
Convert mass to moles
nC = 40.0 g X 1 mol
12.01 g
nO = 53.3 g X 1 mol
nC = 3.33 mol
16.00 g
nH = 6.70 g X 1 mol
1.01 g
nH = 6.63 mol
nO = 3.33 mol
 The empirical formula of the
substance is CH2O
Calculating Molecular Formula
•
•
The empirical formula tells us the simplest ratio of
atoms in a molecule, but it does not tell us the
actual number of atoms in a molecule
However, if we know molar mass and the
empirical formula of a compound we can
determine its molecular formula
1. Divide the molar mass of the compound by the molar
mass obtained from the empirical formula
2. Multiply the subscripts by this number
E.g. What is the molecular formula of a compound that has a
molar mass of 34 g/mol and the empirical formula HO?
Given:
Empirical Formula: HO
MMcmpd = 34 g/mol
MMHO = 1.01 + 16.00
= 17.01 g/mol
Molecular Formula = MMcmpd
MMHO
= 34 g/mol
17.01 g/mol
= 2
 The molecular formula of the compound is H2O2
E.g. What is the molecular formula of a compound that has a
molar mass of 45.00 g/mol and the empirical formula CH3?
Given:
Empirical Formula: CH3
MMcmpd = 45.00 g/mol
MMCH3 = 12.01 + 3(1.01)
= 15.04 g/mol
Molecular Formula = MMcmpd
MMCH3
= 45.00 g/mol
15.04 g/mol
= 3
 The molecular formula of the compound is C3H9
Calculating Molecular Formula
• We can calculate molecular formula if
empirical formula and molar mass is given
• We can also calculate molecular formula if
percent composition and molar mass is
given.
E.g. What is the molecular formula of a compound with a
molar mass of 58.0 g/mol and a percent composition of
82.5% C and 17.5% H?
Given:
mH = 0.175 X 58.0 g
% C = 82.5%
% H = 17.5%
= 10.2 g
nH = 10.2 g X 1 mol
MM = 58.0 g/mol
m = 58.0 g (Assume 1 mole)
1.01 g
= 10.1 mol
mc = 0.825 X 58.0 g
Mole ratio C:H = 4:10
= 47.8 g
nc = 47.8 g X 1 mol
12.01 g
= 3.98 mol
 The molecular formula is C4H10
Homework
• P. 186 #2-6
• P. 188 #1-6
• P. 193 # 8,9
Hydrates
• Solid ionic compounds are formed from dehydrating ionic
solutions
• Some of the liquid water molecules may remain between
molecules of the solids, creating a hydrate
• When a hydrate is heated and water is evaporated it is
referred to as anhydrous
CaCl2  2 H 2O( s ) heat

 CaCl2( s )  2 H 2O( g )
hydrate
heat

 anhydrous  water vapour
hydrate
Hydrate Example
A 50.0 g sample of a hydrate of barium hydroxide was
heated. The anhydrous solid weighed 27.2 g.
a) Determine the percent composition of water in the
hydrate.
b) Calculate the molecular formula of the hydrate. Name
the hydrate.
Hint: Start with a BCE
Ba(OH)2xH2O Solve for x