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Empirical Formulas and Molecular Formulas
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Empirical Formula Simplest ratio of atoms in a formula All ionic compounds exist as empirical formulas. Molecular compounds are not guaranteed to have the simplest ratio of atoms.
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Molecular Formula Some whole number multiple of the empirical formula.
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Empirical Formula C 6 H 12 O 6 is the molecular formula of glucose. What is its empirical formula?
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Problem Solving A compound contains only carbon, hydrogen, and chlorine. A sample is known to contain 49.67%C, 1.39%H, and 48.92%Cl. The molecular weight of the compound is 289.90g/mol. What are the EF and MF of the compound?
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Problem Solving Make the assumption that you have 100g of the compound. How many grams of C would be in that particular sample? Of H? Of Cl?
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Problem Solving 49.67gC x 1molC = ? 12.01gC 1.39gH x 1molH = ? 1.01gH 48.92gCl x 1molCl = ? 35.45gCl
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Problem Solving 4.14molC 1.38molH 1.38molCl Now, you will divide each of the moles by the smallest # of moles that you got.
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Problem Solving 4.14molC = ? 1.38mol 1.38molH = ? 1.38mol 1.38molCl = ? 1.38mol
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Problem Solving 3C 1H 1Cl This gives you your ratio of atoms in the EF.
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Problem Solving There are special rounding rules at this point. If the numbers are <.2, round down. If they are >.8, round up.
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Problem Solving If they are anywhere in between.2 and.8, you must multiply all of the numbers by a factor that will make them roundable.
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Problem Solving EF is C 3 HCl You will need to determine the molar mass of the EF (also known as the empirical weight, or EW).
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Problem Solving EW = 72.49g/mol Since the MF is always some whole number multiple of the EF, the molar mass of the MF (the MW) will always be some multiple of the EW.
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Problem Solving Divide the MW by the EW to determine the multiple of the MF. MW = 289.90g/mol = ? EW 72.49g/mol
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Problem Solving MW = 4 EW Thus, the MF is 4 times the EF. MF = C 12 H 4 Cl 4
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Problem Solving Try another… A compound consists 43.6%P and 56.4%O and has a molecular weight of 283.88g/mol. What are its EF and MF?
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