1. Terms to Know
 Percent composition – relative amounts of
each element in a compound
 Empirical formula – lowest whole- number
ratio of the atoms of an element in a
compound

2. An 8.20 g piece of magnesium
combines completely with 5.40
g of oxygen to form a
compound. What is the percent
composition of this compound?
1. Calculate the total mass
2. Divide each given by the total mass
and then multiply by 100%
3. Check your answer: The
percentages should total 100%

3. Answer
 The total mass is 8.20 g + 5.40 g = 13.60
g
 Divide 8.2 g by 13.6 g and then multiply by
100% = 60.29412 = 60.3%
 Divide 5.4 g by 13.6 g and then multiply by
100% = 39.70588 = 39.7%
 Check your answer: 60.3% + 39.7% =
100%

4. Calculate the percent composition
of propane (C3H8)
 1.
List the elements
 2. Count the atoms
 3. Multiply the number of atoms of
the element by the atomic mass of
the element (atomic mass is on the
periodic table)
 4. Express each element as a
percentage of the total molar mass
 5. Check your answer

5. Answer
 Total molar mass = 44.0 g/mol
 36.0 g C = 81.8%
 8.0 g H = 18.2%

6. Calculate the mass of carbon in
52.0 g of propane (C3H8)
Calculate the percent composition using
the formula (See previous problem)
2. Determine 81.8% of 82.0 g
Move decimal two places to the
left (.818 x 82 g)
3. Answer = 67.1 g
1.

7. Calculating Empirical Formulas
 Microscopic – atoms
 Macroscopic – moles of atoms
 Lowest whole-number ratio may not be the
same as the compound formula
Example: The empirical formula of
hydrogen peroxide (H2O2) is HO

8. Empirical Formulas
The first step is to find the mole-to-mole
ratio of the elements in the compound
 If the numbers are both whole numbers,
these will be the subscripts of the elements
in the formula
 If the whole numbers are identical,
substitute the number 1
Example: C2H2 and C8H8 have an empirical
formula of CH
 If either or both numbers are not whole
numbers, numbers in the ratio must be
multiplied by the same number to yield
whole number subscripts


9. What is the empirical formula of
a compound that is 25.9%
nitrogen and 74.1% oxygen?



1. Assume 100 g of the compound, so that
there are 25.9 g N and 74.1 g O
2. Convert to mole-to-mole ratio:
Divide each by mass of one mole
25.9 g divided by 14.0 g = 1.85 mol N
74.1 g divided by 16.0 g = 4.63 mol O
3. Divide both molar quantities by the
smaller number of moles

10. 




4. 1.85/1.85 = 1 mol N
4.63/1.85 = 2.5 mol O
5. Multiply by a number that converts each
to a whole number (In this case, the
number is 2 because 2 x 2.5 = 5, which is
the smallest whole number )
2 x 1 mol N = 2
2 x 2.5 mol O = 5
Answer: The empirical formula is N2O5

11. Determine the Empirical Formulas
 1.
H 2O 2
 2.
CO2
 3.
N2H4
 4.
C6H12O6
 5.
What is the empirical formula of a
compound that is 3.7% H, 44.4% C, and
51.9% N?

12. Answers
 Compound
Empirical Formula
HO
 1.
H 2O 2
 2.
CO2
CO2
 3.
N2H4
NH2
 4.
C6H12O6
 5.
HCN
CH2O

13. Calculating Molecular Formulas
 The molar mass of a compound is a
simple whole-number multiple of the
molar mass of the empirical formula
 The molecular formula may or may
not be the same as the empirical
formula

14. Calculate the molecular formula
of the compound whose molar
mass is 60.0 g and empirical
formula is CH4N.
1. Using the empirical formula, calculate the
empirical formula mass (efm)
(Use the same procedure used to calculate
molar mass.)
 2. Divide the known molar mass by the efm
 3. Multiply the formula subscripts by this value
to get the molecular formula


15. Answer
 Molar mass (efm) is 30.0 g
 60.0 g divided by 30.0 g = 2
 Answer: C2H8N2

16. Practice Problems

1) What is the empirical formula of a compounds
that is 25.9% nitrogen and 74.1% oxygen?
2) Calculate the empirical formula of a compound
that is 32.00% C, 42.66% O, 18.67% N, and 6.67%
H.
3) Calculate the empirical formula of a compound
that is 42.9% C and 57.1% O.

17. Practice Problems

4) What is the molecular formula for each compound:
a) CH2O: 90 g
b) HgCl: 472.2 g
c) C3H5O2: 146 g