Download Finding the Empirical Formula of a compound

FINDING THE
EMPIRICAL
FORMULA OF A
COMPOUND
“The Ultimate Dog Tease” isn’t near as entertaining as
Empirical formula! Trust me!!
1.
The empirical formula for a compound can
only be found through experimentation where
repeated synthesis/ decomposition reactions
are performed.
2.
The empirical formula represents the smallest whole
number ratio of atoms of each element in a molecule
of a compound
e.g. The chemical formula for hydrogen peroxide is H2O2
but its empirical formula is HO. Similarly for glucose,
whose empirical formula is CH2O while its chemical
formula is C6H12O6.
If you know the percent mass composition of a
compound after it has been decomposed (of
each part it broke into), then you can find its
empirical formula using:
3.
Element %Mass
H
O
11.2
88.8
Mass in
100g
Moles
11.2g
1mol
11.2 g 
 11.1
1.01g
11.1
 2.0
5.55
88.8g
1mol
88.8 g 
 5.55
16.00 g
5.55
 1.0
5.55
Mole Ratio
Therefore, the empirical formula is H2O.
HOW’D WE DO THAT??
1.
2.
3.
4.
5.
Write down the symbol for each element in the first
column then the % mass composition for each
element. e.g.
H=11.2% and O=88.8%
Express each % as a mass out of 100 g.
Express each as a quantity in moles (Mass
Moles)
Express each as a mole ratio by comparing each to
the element with the least number of moles.
Write the empirical formula using the mole ratios as
subscripts for their respective elements.
So really…
4.
Find the empirical formula for a compound where:
N=25.9% and O=74.1% by mass.
(What do you do with mole ratios that are not whole numbers?)
Element %Mass
N
O
25.9
74.1
Mass in
100g
Moles
Mole Ratio
25.9g
1mol
25.9 g 
 1.85 1.85  1.00
14.01g
1.85
74.1g
1mol
74.1g 
 4.63 4.63  2.50
16.00 g
1.85
Therefore, the empirical formula starts as NO2.5
but becomes N2O5.
*Notice atoms/ subscripts in a formula must be
whole numbers!
1.
Since the data collected for these questions are very
accurate, then fractional mole ratios can be changed
to whole numbers by multiplying by the appropriate
whole number.
e.g.
2.5 x 2 = 5.0
2.25 x 4 = 9.0
2.2 x 5= 11.0
2.33 x 3= 7.0
Note: All values in the empirical formula must be
multiplied by the same whole number.
1.
Sometimes a compound is decomposed and the
masses of the various elements are found instead of
% mass composition. e.g. When 120.0g of a
compound containing lead and oxygen is
decomposed, the mass of lead left is 103.9g. What is
the empirical formula of the compound.
1. This problem is done the same way as above
except the masses of the elements are used
instead of the % mass composition.
5 (CONT).
2.
The rest of the procedure is the same.
Element
Mass
Pb
103.9g
O
120.0g-103.9g
= 16.1g
Moles
Mole Ratio
0.5014
103.9 g
 1.00
 0.5014
g
0.5014
207.2 mol
16.1g
1
.
01
 1.01
 2.01
g
16.00 mol
0.5014
Therefore, the empirical formula is PbO2.
2.
A compound was analyzed and found to contain 12.5g of
Ca, 10.8g of O, and 0.675g of H. What is the Empirical
Formula?
Element
Mass
Moles
Mole Ratio
12.5g
1mol
12.5 gCa 
 0.312
40.08 g
0.312
 1.00
0.312
O
10.8g
1mol
10.8 g 
 0.675
16.00 g
H
0.675g
1mol
0.675 g 
 0.668
1.01g
0.675
 2.2
0.312
0.668
 2.2
0.312
Ca
2.2 x 5 = 11… so everything else is
multiplied by 5 as well!
Empirical formula is Ca5O11H11 or Ca5(OH)11
Let’s Try some Practise!!
Hebden Pg. 93 #46 (a,c,i,k,m)
The chemical or molecular formula can be
found
if the molecular or molar mass of the
compound is
known and either the % mass composition or the mass
of each element involved in the synthesis or
decomposition.
The molecular mass is the "true ratio" of the elements
in a compound (ie C6H12O6), whereas the empirical is
the lowest whole number ratio (ie CH2O)
FINDING “MOLECULAR”
FORMULAS
4.
Find the molecular formula for a compound whose
molar mass is 26.0g and where the % mass composition
is H=7.7% and C=92.3%.
Element %Mass
H
C
7.7
92.3
Mass in
100g
Moles
Mole Ratio
7.7g
7.7 g
 7.6
g
1.01 mol
7.6
 1.00
7.6
92.3g
92.3g
 7.69
g
12.01 mol
7.69
 1.0
7.6
So the empirical formula is CH.
1.
But the empirical formula is the simplified version (it has
a molar mass of 12.01g+1.01g= 13.01g) but the real
molar mass is 26.0g… So to find the molecular mass…
6.
2.
Since the ratio of molar mass to empirical molar
mass is
26.0 g
2
13.02 g
3.
. Then the empirical formula is multiplied by two
(2) to give a Molecular formula of C2H2
Molecular Formula  True molar mass / empirical
molar mass = # you multiply the E.F by!!
TRY ANOTHER…
Ribose is an important sugar that is found in DNA and RNA.
Ribose has a molecular mass of 150g/mol, and a chemicla
composition of 40.0% Carbon, 6.67% Hydrogen, and 53.3%
Oxygen. What is the molecular formula of Ribose?
Mass in
Element %Mass
100g
C
H
O
Moles
Mole Ratio
1mol
40 gC 
 3.33
12.01g
3.33
 1.00
3.33
40.0%
40.0g
6.67%
6.60
1mol
 2.00
6.67g 6.67 gH 
 6.60
3.33
1.01g
53.3%
1mol
3.33
53
.
3
gO


3
.
33
53.3g
 1.00
16.00 g
3.33
Empirical formula is CH2O (molar mass =
30.0g/mol )
6.
2.
2.
Since the ratio of molar mass to empirical
molar mass is
150 g / mol
5
30.0 g / mol
. Then the empirical formula is multiplied
by five (5) to give a Molecular formula of
C5H10O5
6.
Let’s Try Some Practise !!!
Hebden Pg. 95 # 49,51,54,55