Download 4 - Emp. and Mol. Form -q

SCH3U
Unit: Quantities in Chemical Reactions
___________
Empirical Formula and Molecular Formula
Name: ___________
Date:
The Empirical Formula of a compound is the simplest formula which can represent the ratio of atoms present.
The Molecular Formula of a compound is a list of all the atoms which actually make up a compound. The numbers
of atoms present follow the ratio predicted by the empirical formula.
NOTE: Sometimes the molecular formula is the same as the empirical formula. Sometimes one empirical formula
can have more than one possible molecular formula.
Examples:
Empirical Formula
Sample Problem 1:
Molecular Formula
NH2
CH
N2H4
C6H6
CH2O
H2O
C6H12O6
H2O
Finding an Empirical Formula
A compound is 17.6% H and 82.4% N. What is the empirical formula of the compound?
Solution:
Assume that you have a mass which is a nice round number like 100 g. Therefore, there are 17.6 g H and 82.4 g N.
# moles of H
= mass of H
molar mass of H
=
17.6 g
1.0 g/mole
= 17.6 moles
# moles of N
= mass of N
molar mass of N
=
82.4 g
14.0 g/mole
= 5.89 moles
Initial Empirical Formula = N5.89H17.6
Divide through by the smallest number of moles in the ratio (in this case that’s 5.89 mol) to find the whole number
ratio of atoms in the empirical formula:
Empirical Formula
=
N5.89H17.6
5.89
= N1H3
NOTE: A summary of the steps involved in determining
empirical formula is found at the bottom of page 188.
1.
2.
3.
4.
Do #5 page 186.
A compound is 68.4% Cr and 31.6% O. What is the empirical formula of the compound?
Do #4 page 189.
Do #1, 4 page 197.
Sample Problem 2:
Finding a Molecular Formula from the Empirical Formula
and the Mass of the Empirical Formula
CH2 is the empirical formula of a certain compound whose molecular mass is 42.1 u. What is the molecular formula of this
compound?
Solution:
Mass of empirical formula, CH2 = 12.0 u of C + 2.00 u of H
= 14.0 u
Mass of molecular formula = Empirical Mass x Whole Number which represents the ratio of molecular mass to empirical
mass
Whole Number = Molecular Mass
Empirical Mass
= 42.1 u
14.0 u
= 3 This means the mass of the
molecular formula is
3 times larger than the the mass of
the empirical formula.
Because the molecular mass is 3X the value of the empirical mass, the molecular formula must be 3 X the
empirical formula of CH2.
So, the molecular formula is CH2 X 3 = C3H6.
Note: The same problem can be done in the same manner with molar masses of the molecular and empirical
g
formulas given in
.
mol
Practice:
1. A compound with a molar mass of 30.00
g
mol
has an empirical formula of CH3. Determine its
molecular formula.
2. Do #14 page 196.
Sample Problem 3:
Finding a Molecular Formula from Mass Percent (Percentage Compostion) Data
Analysis shows that the butane fuel in a butane lighter is 82.5% carbon and 17.5% hydrogen by mass. The molar
mass of butane is found to be 58.0 g. What is the molecular formula of butane?
Solution:
We need to determine the number of moles of carbon and hydrogen in the sample of butane. If we assume we have
82.5
17.5
100 g of butane, then 82.5% of the sample or
parts of it will be carbon by mass and 17.5% or
will be
100
100
hydrogen by mass. Mathematically,
Carbon
82.5
mC =
x 58.0 g sample
100
Hydrogen
17.5
mH =
x 58.0 g sample
100
= 47.8 g of carbon
m
nC =
mm
=
= 10.2 g
m
nH =
mm
=
47.8 g
g
12.0
mol
10.2 g
g
1.01
mol
= 10.0 mol of hydrogen
= 3.98 mol of carbon
In this instance, the ratio of atoms is 4 C : 10 H so the empirical formula is C4H10 .
NOTE: A summary of the steps involved in determining a molecular formula
from percent composition data is found at the top of page 193.
1.
2.
3.
Do questions 1, 2, 3 page 197.
Do questions 8, 9, 10 page 193.
Do #15 page 196.