This document provides information and examples for calculating percent composition, empirical formulas, and molecular formulas of compounds. It defines key terms like percent composition and empirical formula. It then works through examples of calculating the percent composition of magnesium and oxygen that form a compound, the percent composition and mass of carbon in propane, and determining empirical formulas from elemental percentages or mole ratios. The document explains how to calculate molecular formulas from empirical formulas and molar masses. Finally, it provides practice problems for the reader to work through.

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