1. Empirical Formula  %  g  g  mol  mol / mol

2. Percentage Composition Mg magnesium 24.305 12 Cl chlorine 35.453 17 Mg 2+ Cl 1- MgCl 2 1 Mg @ 24.305 amu = 24.305 amu 2 Cl @ 35.453 amu = 70.906 amu 95.211 amu 25.52% Mg 74.48% Cl (by mass...not atoms) It is not 33% Mg and 66% Cl

3. Empirical and Molecular Formulas A pure compound always consists of the same elements combined in the same proportions by weight. PERCENT BY WEIGHT Therefore, we can express molecular composition as PERCENT BY WEIGHT. Ethanol, C 2 H 6 O 52.13% C 13.15% H 34.72% O

4. Empirical Formula of a Hydrocarbon CxHyCxHy g H 2 O g CO 2 mol H 2 O mol CO 2 mol C mol H burn in O 2 1 mol CO 2 44.01 g x 1 mol H 2 O 18.02 g x 2 mol H 1 mol H 2 O x 2 mol C 1 mol CO 2 x Empirical formula Kotz & Treichel, Chemistry & Chemical Reactivity, 3 rd Edition, 1996, page 224

5. Empirical Formula A sample weighing 250.0 g is analyzed and found to contain the following: 27.38% sodium 1.19% hydrogen 14.29% carbon 57.14% oxygen Determine the empirical formula of this compound. Step 1) convert %  gram Step 2) gram  moles Step 3) mol / mol Assume sample is 100 g. 27.38 g Na 1.19 g H 14.29 g C 57.14 g O / 1.19 mol = 1 Na / 1.19 mol = 1 H / 1.19 mol = 1 C / 1.19 mol = 3 O NaHCO 3

6. Empirical Formula Quantitative analysis shows that a compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical formula of this compound. = 1.408 mol Na = 0.708 mol S = 2.812 mol O / 0.708 mol = 2 Na = 1 S = 4 O Na 2 SO 4 32.38% Na 22.65% S 44.99% O 32.38 g Na 22.65 g S 44.99 g O sodium sulfate Step 1) %  gStep 2) g  mol Step 3) mol mol Na 2 SO 4

7. Empirical & Molecular Formula A 175 g hydrocarbon sample is analyzed and found to contain ~83% carbon. The molar mass of the sample is determined to be 58 g/mol. Determine the empirical and molecular formula for this sample. Determine the empirical formula of this compound. Step 1) convert %  gram Step 2) gram  moles Step 3) mol / mol Assume sample is 100 g. / 6.917 mol = 1 C / 6.917 mol = 2.5 H (2.4577 H) CH 2.5 Then, 83 g carbon and 17 g hydrogen. C2H5C2H5 MM molecular = 58 g/mol 58/29 = 2 Therefore 2(C 2 H 5 ) = C 4 H 10 MM empirical = 29 g/mol 2 C @ 12 g = 24 g 5 H @ 1 g = 5 g 29 g butane

8. Molar Mass vs. Atomic Mass 6.02x10 23 H 2 = _______ H 2 = _____ H 2 O = ________ H 2 O = _____ MgSO 4 = ________ MgSO 4 = _____ (NH 4 ) 3 PO 4 = _____ (NH 4 ) 3 PO 4 = ________ Percentage Composition Empirical vs. Molecular Formula  %  g  g  mol  mol mol Empirical Formula 2 amu 18 amu 120 amu 149 amu 2 g 18 g 120 g 149 g % = x 100 % part whole (lowest ratio)