1. Percent composition Empirical and molecular formulas
UNIT 7 The Mole
Percent composition Empirical and molecular formulas

2. Percent composition

3. Calculating percent composition
General formula for calculating percentage:
The question, then, becomes: What is the part, and what is the whole?
For compounds, the percent composition is calculated by mass.
What is the part?
What is the whole?
% =
part
whole
x 100

4. Unknown compound Recently discovered new compound Named cookie
Known to be comprised of two elements, Cc and Do, in an unknown ratio
Chocolate chip and cookie dough, respectively
Task is two-fold:
Calculate the percent composition by mass
Identify empirical formula of compound
Empirical formula indicates the simplest whole number ratio of atoms present in compound

5. Calculating percent composition
General formula for calculating percentage:
The question, then, becomes: What is the part, and what is the whole?
When calculating the percent composition for compounds, the formula becomes:
% =
part
whole
x 100
% composition
of element
molar mass of element
molar mass of compound
x 100 =

6. For example
Calculate the percent composition of each element in sodium bicarbonate (sodium hydrogen carbonate).
Write the formula for sodium bicarbonate.
Na1+ HCO31–
NaHCO3

7. For example
Calculate the percent composition of each element in sodium bicarbonate, NaHCO3.
Calculate the molar mass of the compound. Na H C O
1 mol x 23.0 g/mol
= 23.0 g
= 1.0 g
= 12.0 g
= 48.0 g
84.0 g/mol NaHCO3 P A R T S
1 mol x 1.0 g/mol
1 mol x 12.0 g/mol
3 mol x 16.0 g/mol
WHOLE

8. For example
Calculate the percent composition of each element in sodium bicarbonate, NaHCO3.
Substitute values in the formula and solve.
23.0 g Na
84.0 g NaHCO3
x 100 =
27.4% Na
% Na =
1.0 g H
84.0 g NaHCO3
x 100 =
1.2% H
% H =
12.0 g C
84.0 g NaHCO3
x 100 =
14.3% C
% C =
48.0 g O
84.0 g NaHCO3
x 100 =
57.1% O
% O =
100.0%

9. Empirical and molecular formulas

10. definitions CH C2H2 CH2 C2H4 CH2O C6H12O6 Empirical formula
Molecular formula
Represents the simplest whole number ratio of elements in a compound
Examples
Represents the actual number of atoms in a compound
Examples CH
C2H2
CH2
C2H4
CH2O
C6H12O6
definitions

11. Empirical formula

12. Determining empirical formula
The percent composition of a sulfur oxide is 40.05% S and 59.95% O. Find the empirical formula.
When percentages are given, use the values with grams as the unit.
Find the number of moles of each element.
Conversion factor: molar mass
40.05 g S
1 mol S
mol S =
= 1.25 mol S
32.1 g S
59.95 g O
1 mol O
mol O =
= 3.75 mol O
16.0 g O

13. Determining empirical formula
Recall that the subscripts in a chemical formula indicate the number of moles of each element
If the compound contains sulfur and oxygen in the ratio of 1.25 to 3.75, then the formula could be written as
S1.25O3.75
What’s the problem here?

14. Determining empirical formula
The percent composition of a sulfur oxide is 40.05% S and 59.95% O.
Calculate the simplest mole ratio of the elements in the compound by dividing the number of moles of each element by the smallest value in the mole ratio.
The resulting factor of each calculation becomes the subscript for the element in the empirical formula.

15. Determining empirical formula
The percent composition of a sulfur oxide is 40.05% S and 59.95% O.
Ratio of S to O is 1.25 : 3.75
Smallest value: 1.25 or 3.75?
If results are not whole numbers, must multiply both by a factor that results in whole numbers.
1.25
1.25 mol
Subscript for S:
= 1
3.75 mol
1.25 mol
Subscript for O:
= 3

16. Determining empirical formula
The percent composition of a sulfur oxide is 40.05% S and 59.95% O.
Using the values in Step 2 as subscripts, write the chemical formula.
Subscript of S = 1 Subscript of O = 3
S O
empirical formula =
Calculate empirical formula for Ca3P2: 66.0% Ca and 34.0% P
SO3

17. Molecular formula

18. Molecular Formula Recall:
The molecular formula shows the actual number of atoms of each element in the compound
Molecular formula is always a whole number multiple of the empirical formula
Therefore, the molecular mass (MM of molecular formula) will be the same whole number multiple of the empirical mass (MM of empirical formula)

19. Determining molecular formula
The empirical formula of propene is CH2. What is its molecular formula if the molar mass is determined experimentally to be 42.0 grams?
Find the molar mass of the empirical formula.
Molar mass of CH2
C: 1 mol x 12.0 g/mol = 12.0 g C
H: 2 mol x 1.0 g/mol = 2.0 g H
14.0 g/mol CH2

20. Determining molecular formula
The empirical formula of propene is CH2. What is its molecular formula if the molar mass is determined experimentally to be 42.0 grams?
Compare the mass of the molecular formula to that of the empirical formula (M/E).
molar mass of molecular formula
molar mass of empirical formula
42.0 g
14.0 g =
= 3

21. Determining molecular formula
The empirical formula of propene is CH2. What is its molecular formula if the molar mass is determined experimentally to be 42.0 grams?
Multiply the subscript of each element in the empirical formula by the resulting factor (n).
Empirical and molecular formulas for glucose: 40.00% C, 6.67% H, 53.33% O; molecular mass is g/mol
molecular formula = (empirical formula)n
molecular formula = (CH2)3
molecular formula = C3H6