Percentage Composition

1. REVIEW:
1. What is a mole?
2. What is the value of
Avogadro’s number?
3. What is the difference between
a formula mass and molecular
mass?

2. PERCENTAGE
COMPOSITION
OF ELEMENTS
IN A COMPOUND

3. PERCENTAGE COMPOSITION
Explains the ratio between the mass of each element
per 100 mass units of compound
Formula:
% mass of element = Mass of element in the compound x 100%
Mass of the compound

4. PERCENTAGE COMPOSITION
% mass of element = Mass of element in the compound x 100%
Mass of the compound
Example: Calculate the percentage composition of CO2
C = 1 x 12 = 12 g
O= 2 x 16 = 32 g
44 g (mass of the compound)

5. PERCENTAGE COMPOSITION
C = 1 x 12 = 12 g
O= 2 x 16 = 32 g
44 g (mass of the compound)
C = 12_ g x 100 = 27.27 % of C
44 g
O = 32 g x 100 = 72.73 % of O
44 g

6. SEATWORK:
PERCENTAGE COMPOSITION
Calculate the percentage composition of:
1. Al(OH)3 (Aluminum hydroxide)
2. NaCl (Sodium chloride)
3. H2O (water)

7. ANSWERS:
1. Al(OH)3 78 g
Al = 34.62 % O = 61.54 % H = 3.85 %
2. NaCl 58 g
Na = 39.66 % Cl = 60.34 %
3. H2O 18 g
H = 11.11 % O = 88.89 %

8. Empirical &
Molecular Formulas

9. EMPIRICAL FORMULA
 The formula with the smallest whole number mole ratio
of the elements
 May or may not be the same as the molecular formula
 If the two formulas are different the molecular formula will
always be a simple multiple of the empirical formula
The empirical formula for hydrogen peroxide is HO
The molecular formula for hydrogen peroxide is H2O2

10. Empirical Formula
 If the % composition is given you can assume the
total mass of the compound is 100 grams and that
the percent by mass of each element is equal to the
mass of that element in grams
 The mass of each element can be converted to
moles by dividing the molar mass of the element

11. Example
The percent composition of sulfur oxide is 40.05 % S
and 59.95 % O. Calculate its Empirical Formula
40.05 g S = 1.25 mol S
32 g/mol S
59.95 g O = 3.75 mol O
16 g/mol O

12.  How does 1.25 mol of S and 3.75 mol O transfer into
subscripts? They are not in whole numbers.
 Divide the mole values by the value of the element with the
smallest number of moles.
 1.25 mol S = 1 mol S
1.25
 3.75 mol O = 3 mol O
1.25

13. The simplest whole number mole ratio of S
atoms to O atoms is 1:3
The empirical formula for the oxide of sulfur is
SO3
The calculated mole values may not always be
whole numbers
In these cases all the mole values must be multiplied
by the smallest factor that will make them whole
numbers

14.  % to Mass
 Mass to Mole
 Divide by
smallest
 Multiply until
whole

15. EXAMPLE:
COMPUTE THE EMPIRICAL FORMULA
The chemical analysis of aspirin indicates
that the molecule is 60.00% carbon,
4.44% hydrogen, and 35.56% oxygen.
Determine the empirical formula for
aspirin.

16. ANSWER
60 mol C = 5 mol C 5 mol C = 2.25
12 2.22
4.44 mol H = 4.44 mol H 4.44 mol H = 2
1 2.22
35.56 mol O = 2.22 mol O 2.22 mol O = 1
16 2.22
C2.25H2O (C2.25H2O) x 4 C9H8O4

17. SEATWORK: COMPUTE THE
EMPIRICAL FORMULA OF THE FF
The percent composition of an
unknown substance is 75.42 %
Carbon, 6.63 % Hydrogen, 8.38 %
Nitrogen and 9.57 % Oxygen. What is
its empirical formula?

18. ANSWER
75.42 mol C = 6.29 mol C 6.29 mol C = 10.5
12 0.60
6.63 mol H = 6.63 mol H 6.63 mol H = 11
1 0.60
8.38 mol N = 0.60 mol O 0.60 mol O = 1
14 0.60
9.57 mol O = 0.60 mol O 0.60 mol O = 1
16 0.60
C10.5H11NO (C10.5H11NO) x 2 C21H22N2O2

19. Molecular Formulas
Is the formula of a compound in which the
subscripts give the actual number of each
element in the formula
Two or more substances with distinctly
different properties can have the same
percent composition and the same empirical
formula

20. Determining Molecular
Formulas
Experimentally determined molar mass of the compound = n
Mass of empirical formula of the compound
A molecular formula can be represented as the empirical formula
multiplied by an integer n.
molecular formula = (empirical formula) n
The integer is the factor by which the subscripts in the empirical
formula must be multiplied to obtain the molecular formula

21. Example for EMPIRICAL FORMULA
The empirical formula for a
compound is C21H22N2O2. Its
experimental molar mass is
334 g/mol. Determine the
molecular formula of the
compound.

22. Step 1: Find the MOLAR MASS of the
empirical formula.
C21H22N2O2
21 mol C x 12 = 252
22 mol H x 1 = 22
2 mol N x 14 = 28
2 mol O x 16 = 32____
334 g/mol

23. Step 2: Molar mass of the compound
divided by the molar mass of the
empirical formula
Experimental molar mass of the compound
Molar mass of the empirical formula
334 g/mol = 1
334 g/mol

24. Step 3: Multiply subscripts from
empirical formula by the answer you got
in step 2.
Empirical Formula = 1 (C21H22N2O2)
The MOLECULAR FORMULA is
C21H22N2O2
Again, the molecular formula may be the
same with the empirical formula.

25. SEATWORK: COMPUTE THE MOLECULAR
FORMULA OF THE FF
The empirical formula for a
compound is P2O5. Its
experimental molar mass is 284
g/mol. Determine the molecular
formula of the compound.

26. ANSWER FOR THE SEATWORK
Step 1: Compute the Molar Mass of
P2O5
P = 2 x 31 = 62
O = 5 x 16 = 80
142 g/mol

27. Step 2: Divide experimental molar mass by the molar
mass of the empirical formula
Experimental molar mass of the compound
Molar mass of the empirical formula
284 g/mol = 2
142 g/mol
Empirical Formula = 2 (P2O5)
Molecular Formula = P4O10

28. Sample Problems:
1. A sample compound with a molar mass of 34.0
g/mol is found to consist of 0.44 g H and 6.92 g
O. Calculate both the Empirical and Molecular
Formulas
2. A compound consists of 36.48 % Na, 25.41 % S,
and 38.11 % O. It has a molar mass of 252.10
g/mol. Calculate both the empirical and
molecular formulas.