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Empirical and Molecular formulas

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 Can an empirical formula also be the molecular formula?. M- 6:12:6 reduced to 1:2:1.

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Empirical and Molecular formulas

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  1. Empirical and Molecular formulas

  2. Empirical – lowest whole number ratio of elements in a compound • Molecular – some multiple of the empirical formula • Examples: CH4 C6H12O6 Na2SO4 C3H6 Can an empirical formula also be the molecular formula? M- 6:12:6 reduced to 1:2:1 E – 2:1 M -3:6 reduced to 1:2 E – 1:4 YES C12H22O11

  3. Steps for determining empirical formulas. • Assume a 100 g sample when given percents. This makes the 10.3 % Z = 10.3 g Z • Change grams into moles for each element. • Divide the all the moles by smallest number of moles to get the lowest whole number ratio. • Write the empirical formula.

  4. 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 36.11 % Ca = 36.11 g Ca x 63.89 % Cl = 63.89 g Cl x Step 3 Step 1 Step 2 = 0.9009 mol Ca = 1 mol Ca 0.9009 = 1.802 mol Cl = 2 mol Cl 0.9009 Therefore the empirical formula is CaCl2 Step 4

  5. 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 = 1.316 mol Cr 68.42 g Cr = 1 mol Cr X 2 = 2 Cr 1.316 = 1.974 mol O 31.58 g O = 1.5 mol O X 2 = 3 O 1.316 Way to much to round off so you have to get rid of the fraction. Cr2O3

  6. 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 14.37 g H x 100 g sample = 7.130 mol C = 1 mol C 7.130 Empirical formula CH2 = 14.26 mol H = 2 mol H 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 CH2 = C2H4

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