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Empirical Formula

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

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Empirical Formula

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

  2. 24.305 35.453 Mg Cl 12 17 magnesium chlorine Percentage Composition (by mass...not atoms) 25.52% Mg Mg2+ Cl1- 74.48% Cl MgCl2 It is not 33% Mg and 66%Cl 1 Mg @ 24.305 amu = 24.305 amu 2 Cl @ 35.453 amu = 70.906 amu 95.211 amu

  3. Empirical and Molecular Formulas A pure compound always consists of the same elements combined in the same proportions by weight. Therefore, we can express molecular composition as PERCENT BY WEIGHT. Ethanol, C2H6O 52.13% C 13.15% H 34.72% O

  4. Empirical Formula of a Hydrocarbon 1 mol CO2 44.01 g x 2 mol C 1 mol CO2 x burn in O2 g CO2 mol CO2 mol C mol H Empirical formula CxHy g H2O mol H2O 2 mol H 1 mol H2O x 1 mol H2O 18.02 g x Kotz & Treichel, Chemistry & Chemical Reactivity, 3rd Edition , 1996, page 224

  5. / 1.19 mol = 1 Na / 1.19 mol = 1 H NaHCO3 / 1.19 mol = 1 C / 1.19 mol = 3 O 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 27.38 g Na 1.19 g H 14.29 g C 57.14 g O Assume sample is 100 g. Determine the empirical formula of this compound. Step 1) convert %  gram Step 2) gram  moles Step 3) mol / mol

  6. / 0.708 mol 32.38 g Na 22.65 g S 44.99 g O / 0.708 mol / 0.708 mol 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. sodium sulfate = 2 Na 32.38% Na 22.65% S 44.99% O = 1.408 mol Na Na2SO4 Na2SO4 = 0.708 mol S = 1 S = 2.812 mol O = 4 O Step 1) %  g Step 2) g  mol Step 3) mol mol

  7. / 6.917 mol = 1 C CH2.5 / 6.917 mol = 2.5 H (2.4577 H) 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. 2 C @ 12 g = 24 g 5 H @ 1 g = 5 g 29 g Step 1) convert %  gram Step 2) gram  moles Step 3) mol / mol Assume sample is 100 g. Then, 83 g carbon and 17 g hydrogen. MMempirical = 29 g/mol C2H5 MMmolecular = 58 g/mol 58/29 = 2 Therefore 2(C2H5) = C4H10 butane

  8. part whole % = x 100 % 6.02x1023 Molar Mass vs. Atomic Mass 2 g H2 = _____ H2 = _______ 2 amu 18 g H2O = _____ H2O = ________ 18 amu 120 g MgSO4 = _____ MgSO4 = ________ 120 amu 149 g (NH4)3PO4 = _____ (NH4)3PO4 = ________ 149 amu Percentage Composition Empirical Formula • %  g • g  mol • mol • mol Empirical vs. Molecular Formula (lowest ratio)

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