PERCENT COMPOSITION, EMPIRICAL

1
PERCENT COMPOSITION, EMPIRICAL MOLECULAR
FORMULAS
2
Percent Composition

• New food labels are required to describe the
  ingredients using percents of the daily reccom-
  mended allowance
• These numbers tell what part of
  the total of calories can be
  ob- tained from a product
• AKA percent composition

3
Percent Composition

• To get the information found on food labels the
  chemists had to know what fraction of the whole
  was each component
• Component/total and then multiply by 100
• There are a couple of procedures used to
  calculate percent compositions

4
Calculating PC given formula
What percentage of Hydrogen and Oxygen is in
Water (H2O)?
Assume you have 1 mole of water, and calculate
its molar mass
(21.008g)
(115.994g)
18.01g
5
Calculating PC given formula

• There are 2 mols of H atoms for every 1 mol of
  Water molecules
• How much do 2 mols of H atoms weigh?

H (21.008g)
2.016g H

• Percent of H in Water?

11.2
X 100
18.01 g H2O
6
Calculating PC given formula

• There is 1 mol of O atoms for every 1 mol of
  Water molecules
• How much does 1 mol of O atoms weigh?

O (115.994g)
15.994g O

• Percent of O in Water?

88.8
X 100
18.01 g H2O
7
Percent Composition

• Another method of calculating the percent
  composition is by experimental analysis.
• the overall mass of the sample is measured.
• then the sample is decomp-osed or separated into
  its component elements

8
Percent Composition

• The masses of the component elements are then
  determined and the percent composition is
  calculated as before
• by dividing the mass of each element by the total
  mass of the sample
• then multiplying by 100

9
Calculating PC given sample
Find the percent composition of a compnd that
contains 1.94g of carbon, 0.48g of Hydrogen, and
2.58g of Sulfur in a 5.0g sample of the compnd.
10
Calculating PC given sample

• Calculate the percents for each element much like
  you would calculate the percents for anything.

C 1.94g/5.0g X 100 38.8
H 0.48g/5.0g X 100 9.6
S 2.58g/5.0g X 100 51.6
11
Empirical Formulas

• Once the percent compositions are determined then
  they can be used to calculate a simple chem
  formula for the compnd
• key is to convert the percents by mass into
  amounts in moles
• Then, compare the moles using ratios to determine
  coefficients

12
Calculating Empirical Formulas
What is the empirical formula of a compound that
is 80C and 20H by mass

• Since we have been given per-cents rather than
  masses we need to make an assumption.
• Lets suppose we have a total sample that weighs
  100 g.

13
Calculating Empirical Formulas

• This allows us to say that if we had a 100 grams
  of sample,
• 80 g is Carbon
• 20 g is Hydrogen
• Now that we have a set of masses we need to
  convert them to moles
• Divide by the molar masses from the Periodic Table

14
Calculating Empirical Formulas
1 mole C
80g C
12 g C
1 mole H
20g H
1 g H

• Now calculate the simplest ratio of each by
  dividing both values by the smallest value

15
Calculating Empirical Formulas
Divide each mole value by the smaller of the two
values
C 6.7/6.71
H 20/6.7 2.98 ? 3
Ratio is 1 Cs for every 3 Hs so the formula is

CH3
16
Calculating Empirical Formulas
Determine the empirical formula of a compound
containing 25.9g of N and 74.1g of O.
Notice we have masses this time not percents,
we can convert masses directly to moles
17
Calculating Empirical Formulas
1 mol N
25.9g N
14 g N
1.85 mol
1 mol O
74.1g O
16 g O
1.85 mol
18
Calculating Empirical Formulas
Is the final answer N1O2.5? Of course not!
We need a whole number ratio
Each part of the ratio is multiplied by a number
that converts the fraction to a whole number
N2(1)O2(2.5) N2O5
19
Molecular Formulas

• The empirical formula indicates the simplest
  ratio of the atoms in the compnd
• However, it does not tell you the actual numbers
  of atoms in each molecule of the compnd
• For instance, glucose has the molecular formula
  of C6H12O6
• Empirical form would be CH2O

20
Molecular Formulas

• The empirical formula of CH2O, could be several
  compnds.
• C2H4O2 or C3H6O3 or C100H200O100
• Its more important to know the exact numbers of
  atoms involved
• The numbers of atoms define the properties of the
  compnd

21
Molecular Formulas

• The molecular formula is always a whole-number
  multiple of the emp. formula
• In order to calculate the molecular formula you
  must have 2 pieces of information
• Empirical formula
• Molar mass of the unknown compound (must be given)

22
Calculating Molecular Formulas
Find the molecular formula of a compound that
contains 56.36 g of O and 54.6 g of P.
If the molar mass of the compound is 189.5 g/mol.

1. Find the Empirical Formula
2. Find the MM of the Emp. Form.
3. Find the ratio of the 2 molar masses (Mol MM/Emp
   MM)

23

1. Find the Empirical Formula

1 mol O
56.36g O
3.5 mol O
16 g O
1.8 mol
1 mol P
1.8 mol P
54.6g P
31g P
1.8 mol
Empirical formula P1O2
24

1. Find the MM of the Emp Form.

MM of PO2 (131g P)
(216g O)
63g/mol

1. Find the ratio of the 2 molar masses (mol MM/emp
   MM)

GIVEN
189.5 g/mol
3.00
CALCULATED
25
Calculating Molecular Formulas

• So the Molecular formula is 3 times
  heavier than the Empirical formula
• Therefore, the molecular formula has 3 times more
  atoms than the emp. formula

P3(1)O2(3) P3O6