empirical formulas and molecular formulas. empirical formula simplest ratio of atoms in a formula...
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![Page 1: Empirical Formulas and Molecular Formulas. Empirical Formula Simplest ratio of atoms in a formula All ionic compounds exist as empirical formulas. Molecular](https://reader035.vdocuments.site/reader035/viewer/2022080902/56649eab5503460f94bb0de8/html5/thumbnails/1.jpg)
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•C6H12O6 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 C3HCl
•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 = C12H4Cl4
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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?