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Percent Composition Indicates relative % of each element in a compound Total % of the components ~ 100%

Examples 1.Determine the % of children in a group consisting of 10 men, 8 women, and 9 children = 27 total people 9 children x 100 = 33.3% 27 people

Examples 2.Find the % of aluminum in aluminum oxide. 3.Find the % of nitrogen in ammonium nitrate, a compound used in fertilizers. 4.Copper(II) sulfate pentahydrate is a blue compound used to make colored pigments, insecticides, and electric batteries. Determine the % water.

Empirical Formulas 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 …..is the EF= MF?

Steps to determining the EF 1.Find the percent composition of the substance. 2.Assume you have a 100 gram sample assuming 100g, you can remove the % sign and replace it with grams 3. Convert the grams found in step 2 into moles.

Steps to Determine the EF 4. Ratio the moles you found in step 3 5. Divide all moles by the smallest value this determines the relative molar amount of each of the atoms— ~ whole number 6. If the numbers are not whole numbers multiply by a factor that will yield a whole number Example: 1: x 3 = 3:4

Steps to Determine the EF Use molar ratios obtained as subscripts in the formula. Example 1 What is the EF of a compound that contains 53.73% Fe and 46.27% S? ( Fe 2 S 3 ) Example 2 What is the EF of a compound that contains 90.7% Pb and 9.33% O? ( Pb 3 O 4 )

Molecular Formulas Molecular formulas can be found from the EF 1.Find the empirical formula and mass 2.Find the relationship between the EF mass and the molecular mass n = molecular formula mass empirical formula mass 3. (Empirical formula) n = Molecular formula EF = CH 4 n = 2 C 2 H 8

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 g O = 9.40 g 9.99 g Total Sample 100% - 5.9% = 94.1% oxygen Step 1: Find the percent composition

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 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 = H 2 O 2