Percent Composition of a :Compound

Percent Composition of a : Sample

Empirical Formulas

Molecular Formulas

Percent Composition:

• Percent composition tells you the percentage of each component in a compound or sample.

• When working with %Composition, your total percentage of the components should equal 100% (give or take a little).

How can you calculate percent composition of a compound?

• To find the percent composition of a compound, you divide the total mass of each individual component by the molar mass of the compound and multiply by 100.

%composition = total mass of component molar mass of the compound

How can you calculate the percent composition of a sample?

• To find the percent composition of a sample, you must first know how much of each component the sample contains. Then you divide the mass of each component by the total mass of the sample and multiply by 100. (*Note: All components in the sample must equal the total mass of the sample!)

• %composition= total mass of component

total mass of the sample

Empirical Formulas: What are they?

• Empirical formulas are the lowest possible ratio of components in the compound.

• Ionic formulas are almost always empirical in nature.

• Covalent compounds require more information and we usually have to find the molecular formula for them.

Empirical Formulas:

• Empirical formulas can be broken down into 5 basic steps.

(I recommend you make flashcards of these steps. Write step 1, step 2, etc. on one side. On the other side you will write what you should do in that step. Shuffle them up and practice putting them in their correct order.)

Step 1

• Find the percent composition of the substance.

• If you are given the percent composition to begin with, go to step 2.

Step 2

• Assume you have a 100 gram sample

• (This is to make the math easier. You can assume another value but you must multiply it by the percentage in step 1.)

• By assuming 100g, you can remove the % sign and replace it with grams.

Step 3

• Convert the grams found in step 2 into moles. (Atomic mass – periodic table!)

Step 4

• Ratio the moles you found in step 3 and reduce (that means divide all moles by the smallest value!)

• If your numbers are not whole numbers, use a multiplier that will bring them as close to a whole number as possible.

• Example: 1: 1.3 (I would use 3 as a multiplier to make this ratio 3:4)

Step 5

• Use ratios obtained in step 4 as subscripts for your formula.

• In the example given in step 4, the ratio was 3:4. If the ratio was Fe:O, the formula would be:

Fe3O4

Molecular Formulas

• The formula for a molecular compound (covalently bonded for the most part) is called the molecular formula.

• To find molecular formulas, you first must find the empirical formula or be given the empirical formula.

How do we find the molecular formula?

• To find the molecular formula, we find the empirical formula first (or we must be given the empirical formula).

• So, in essence, we must go to steps 1-5 for empirical formulas first.

Empirical Formula Mass

• The empirical formula mass is nothing more than finding the molar mass of the empirical formula.

If molecular formulas are more complex, how do we find them?

• We must find the multiplier, “n”.

• “n” is found by dividing the molecular formula mass by the empirical formula mass.

n = molecular formula mass empirical formula mass

What do we do with the “n”?

Molecular formula= (empirical formula)n

The “n” needs to be distributed throughout the empirical formula.

If the empirical formula= CH and the “n” is 2, then the molecular formula is:

(CH)2 = C2H2 which is NOT an empirical formula anymore since the ratio is not at its lowest.

Example:

• A compound composed of hydrogen and oxygen is analyzed and a sample of the compound yields 0.59 g of hydrogen and 9.40 g of oxygen. The molecular mass of the compound is 34 g/mol. Find the empirical formula and molecular formula for this compound.

H = 0.59 gO = 9.40 g 9.99 g Total Sample

100% - 5.9% = 94.1% oxygen

Step 1: Find the percent composition

H%9.510099.9

59.0

g

g

Step 2: Assume a 100 g Sample

• Drop the percent sign and replace it with a “g” for grams.

Step 3: Convert grams to moles

5.9 g H 1 mole = 5.9 moles H

1.0 g

94.1 g 1 mole = 5.9 moles O

16 g

Step 4: Ratio the moles and reduce

Moles of H:Moles of O

5.9 moles : 5.9 moles

5.9 moles 5.9 moles

1:1

Step 5: Use mole ratio for empirical formula subscripts

• 1 hydrogen : 1 oxygen

Empirical Formula: HO

Step 6: Find the empirical formula mass

• Empirical Formula: HO

1 g + 16 g = 17 g/mol

Step 7: Divide Molecular Mass by the Empirical Formula Mass

Molecular mass (sometimes called molar mass or mass of the molecule) = 34.0 g/mol

Divide the molecular mass by the empirical formula mass to find “n”, a multiplier.

n = (molecular mass ) = 34 g/mol

(empirical formula mass) 17 g/mol

n = 2

Step 8: Multiply the subscripts in the empirical formula by “n”

Molecular Formula = (empirical formula)n

Molecular Formula = (HO)2

*distribute the 2 through the parentheses

Molecular Formula = H2O2