Percent Composition

�  The relative amount of the element in a compound �  Expressed as a percent

�  % Comp will give you either the masses of the elements or the chemical formula �  Either way… you will use the following formula:

Percent Composition

Mass from the element

Total Mass (or Molar Mass) 100 %

% Composition Practice

�  If a 13.60g sample of a compound contains 8.20g Mg and 5.40g O, what is the % composition of this compound?

�  % Mg = (8.20 / 13.60) X 100 = 60.29 %

�  % O = (5.40 / 13.60) X 100 = 39.71 %

�  Your % values should equal 100%. This is how to check your answer to make sure it makes sense!

% Composition Example #2

�  Determine the % composition for glucose, C6H12O6

�  Step 1: Determine the Molar Mass:

�  6C = (6 x 12.01) = 72.06g

�  12H = (12 x 1.01) = 12.12g

�  6O = (6 x 16.00) = 96.00g

�  Total Molar Mass = 180.18g

% Composition Example #2

�  Determine the % composition for glucose, C6H12O6

�  Step 2: Calculate the % composition �  %C = (72.06 / 180.18) x 100 = 39.99 %

�  %H = (12.12 / 180.18) x 100 = 6.73 %

�  %O = (96.00 / 180.18) x 100 = 53.28 %

�  Remember: the percent should add to 100!

Molecular Formulas

�  Are the actual formulas of compounds that are NOT reduced to their lowest whole number ratio

�  Example: �  Glucose: C6H12O6

�  Benzene: C6H6

Empirical Formulas

�  Empirical means: achieved through observation or experiments

�  The lowest whole number ratio of the atoms of the different elements in a compound

�  * May or May NOT be the same as a molecular formula*

�  Example: �  Glucose: CH2O �  Benzene: CH

Empirical, Molecular, or Both?

�  N2O5

�  Both

�  H2SO4 (Sulfuric Acid) �  Molecular

�  C6H12O6 (Glucose) �  Molecular

�  H2O (Water) �  Both

�  H2O2 (Hydrogen peroxide) �  Molecular

�  C2H7

�  Both

�  C2H4

�  Molecular

�  CH4 (Methane) �  Both

�  HO �  Empirical

�  C2H8N2

�  Empirical

Which have the same Empirical Formula?

�  C2H6O2

�  No

�  C8H14O6

�  No

�  C7H11O5

�  Yes

�  C3H9O3

�  No

�  C2H5O9

�  Yes

�  C4H6O10

�  No

�  C4H12O4

�  No

Empirical Calculation Example

�  Question: A compound is found to contain 25.9% N and 74.1% O, what is the empirical formula?

�  Step 1: Assume you have 100.0 g. �  Then you will have 25.9g of N and 74.1g of O.

�  Step 2: Change masses to moles

25.9 g N

1 mol N

14.0g 1.85 mol N X = Nitrogen:

74.1 g O

1 mol O

16.0g 4.63 mol O X = Oxygen:

�  Question: A compound is found to contain 25.9% N and 74.1% O, what is the empirical formula?

�  Step 3: �  Divide both molar quantities by the smaller number of

moles

Empirical Calculation Example

1.85 mol N

1.85 1 mol N = Nitrogen:

4.63 mol O

1.85 2.5 mol O = Oxygen:

�  Question: A compound is found to contain 25.9% N and 74.1% O, what is the empirical formula?

�  Step 4: �  Multiply by a number to convert the fraction to a whole

number.

Empirical Calculation Example

1 mol N 2 2 mol N X = Nitrogen:

2.5 mol O 2 5 mol O X = Oxygen:

* Must multiply both by the

same number!

�  Question: A compound is found to contain 25.9% N and 74.1% O, what is the empirical formula?

�  Step 5: �  Show the final answer.

Empirical Calculation Example

Empirical Formula

N2O5 =