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E mpirical F ormulas

E mpirical F ormulas. The empirical formula of a chemical compound is the simplest whole number ratio of atoms of each element present in a compound.

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E mpirical F ormulas

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  1. Empirical Formulas The empirical formulaof a chemical compound is the simplest whole number ratio of atoms of each element present in a compound. The term empiricalrefers to the process of elemental analysis, a technique of analytical chemistry used to determine the relative amounts of each element in a chemical compound. AKA: Simplest Formula In contrast, the molecular formula identifies the actual number of each type of atom in a molecule, and the structural formula also shows the structure of the molecule. Examples: C4H10 (molecular) → ? Empirical C12H22O12 (molecular) → ? Empirical C2H2 (molecular) → ? Empirical CH4 (molecular) → ? Empirical C2H5 C6H11O6 CH CH4

  2. 1 Empirical Formulas Example Problem A compound consists of 82.66% carbon and 17.34% hydrogen. Find the empirical formula of this compound. Step 2 Convert grams to moles Step 1 Change % to grams 1 mol C 82.66 g C 17.34 g H = 6.882 mol C Hint: Keep at least 4 or 5 sig. figs. here 12.011 g C 1 mol H = 17.20 mol H 1.00794 g H Current Formula Ratio: But… we can’t have 6.882 carbon atoms! C6.882 H17.20

  3. 1 Empirical Formulas Example Problem A compound consists of 82.66% carbon and 17.34% hydrogen. Find the empirical formula of this compound. Step 3 Divide by Smallest C6.882 H17.20 Keep 2 decimals C1.00 H2.50 = 6.882 6.882 We’re closer but we still don’t have all whole numbers. Step 4 Find the Least Common Multiple C1.00 H2.50 Change the decimal to a fraction 0.5  ½ X 2 X 2 Use the denominator of the fraction as the multiplier for the formula ratio Empirical Formula ! C2H5

  4. 2 Empirical Formulas Example Problem A compound consists of 18.983 g carbon and 3.717 g hydrogen. Find the empirical formula of this compound. Step 1 Convert grams to moles 1 mol C 18.983 g C 3.717 g H = 1.5805 mol C 12.011 g C 1 mol H = 3.6877 mol H 1.00794 g H Current Formula Ratio: C1.5805 H3.6877

  5. 2 Empirical Formulas Example Problem A compound consists of 18.983 g carbon and 3.717 g hydrogen. Find the empirical formula of this compound. Step 2 Divide by Smallest C1.5805 H3.6877 C1.00 H2.33 = 1.5805 1.5805 We’re closer but we still don’t have all whole numbers. Step 3 Find the Least Common Multiple C1.00 H2.33 Change the decimal to a fraction 0.33  ⅓ X 3 X 3 Use the denominator of the fraction as the multiplier for the formula ratio Empirical Formula ! C3H7

  6. Molecular Formulas The molecular formula can be determined using the empirical formula if the molecular mass is known. The relationship is: The molecular formula identifies the actual number of each type of atom in a molecule. It is a whole number multiple of the empirical formula. Molecular Formula = Empirical Formula

  7. 3 Example Problem Molecular Formulas To be able to use this formula  Molecular Formula = Empirical Formula If the empirical formula for a substance is found to be C3H7 and it’s molecular mass is known to be 86.2 g/mol, what is the molecular formula? The empirical mass must be determined Step 1 Calculate Empirical Mass (3) 12.011 g/mol C (7) 1.00794 g/mol H 43.089 g/mol C3H7

  8. 3 Example Problem Molecular Formulas Step 2 Substitute values Molecular Formula = Empirical Formula 86.2 g/mol C3H7 If the empirical formula for a substance is found to be C3H7 and it’s molecular mass is known to be 86.2 g/mol, what is the molecular formula? Molecular Formula = 43.089 g/mol = C6H14 C3H7 x 2 Molecular Formula = Molecular Formula

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