Empirical and Molecular formulas Empirical lowest whole number

Empirical and Molecular formulas

• Empirical – lowest whole number ratio of elements in a compound • Molecular – some multiple of the empirical formula • Examples: CH 4 C 6 H 12 O 6 Na 2 SO 4 C 3 H 6 E – 1: 4 M- 6: 12: 6 reduced to 1: 2: 1 E – 2: 1 M -3: 6 reduced to 1: 2 Can an empirical formula also be the molecular formula? YES C 12 H 22 O 11

Steps for determining empirical formulas. 1) Assume a 100 g sample when given percents. This makes the 10. 3 % Z = 10. 3 g Z 2) Change grams into moles for each element. 3) Divide the all the moles by smallest number of moles to get the lowest whole number ratio. 4) Write the empirical formula.

A compound was found to contain 36. 11 % calcium and 63. 89 % chlorine by mass. What is its empirical formula? What assumption did you make? 100 g sample Step 1 Step 2 Step 3 36. 11 % Ca = 36. 11 g Ca x = 0. 9009 mol Ca 0. 9009 63. 89 % Cl = 63. 89 g Cl x = 1. 802 mol Cl 0. 9009 Therefore the empirical formula is Ca. Cl 2 Step 4 = 1 mol Ca = 2 mol Cl

GOOFY MATH(. 33 = 1/3 . 5 = ½ . 67 = 2/3) A compound was found to contain 68. 42 % chromium and the rest oxygen by mass. What is its empirical formula? What assumption did you make? 100 g sample 68. 42 g Cr = 1. 316 mol Cr = 1 mol Cr 1. 316 31. 58 g O = 1. 974 mol O 1. 316 Cr 2 O 3 X 2 = 1. 5 mol O X 2 = 2 Cr =3 O Way to much to round off so you have to get rid of the fraction.

Ethene is a compound containing only carbon and hydrogen. It contains 85. 63 % carbon and the rest hydrogen by mass. It also has a molar mass of 28. 05 g/mol. What are empirical and molecular formulas of this compound? What assumption did you make? 85. 63 g C x = 7. 130 mol C 7. 130 14. 37 g H x = 1 mol C = 14. 26 mol H = 2 mol H 100 g sample Empirical formula CH 2 7. 130 Remember molecular is some multiple of the empirical. So take the molar mass of the compound and divide it by the molar mass of the empirical formula. 28. 05/14. 03 = 2 So take 2 x CH 2 = C 2 H 4