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
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
Divide each given by the total mass
and then multiply by 100%
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% = = 60.3%
Divide 5.4 g by 13.6 g and then multiply by 100% = = 39.7%
Check your answer: 60.3% % = 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

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
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. /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. H2O2
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 1. H2O2 HO 2. CO2 CO2 3. N2H4 NH2
4. C6H12O CH2O
5. HCN

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