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Empirical Formulae

Empirical Formulae. “The Empirical Formula Strikes Back”. Type 1:   Determining the empirical formula of a compound from mass. Ex. 1.  What is the empirical formula for a compound if an 8.1 g sample contains 4.9 g of magnesium and 3.2 g of oxygen?

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Empirical Formulae

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  1. Empirical Formulae “The Empirical Formula Strikes Back”

  2. Type 1:   Determining the empirical formula of a compound from mass. • Ex. 1.  What is the empirical formula for a compound if an 8.1 g sample contains 4.9 g of magnesium and 3.2 g of oxygen? •      Solution:  Remember that a molar ratio is a ratio of the number of moles, not the mass.  We are not looking for a ratio of mass to mass, but of moles-moles.  Our first step is to determine how many moles of each element are found in this compound. 

  3. Given: mass of magnesium = 4.9 g           molar mass of magnesium = 24.3 g mol-1           mass of oxygen = 3.2 g           molar mass of elemental oxygen = 16.0 g mol-1 (note: use elemental, not diatomic oxygen) • Moles of Mg = 0.2 moles • Moles of O = 0.2 moles

  4. The elements are combining in a 0.20 to 0.20 ratio, or a 1:1 ratio.  This means that for every one atom of magnesium, there is one atom of oxygen in the compound.  This gives us an empirical formula of MgO.

  5. Example #2: • Calculate the empirical formula of a compound that is made from 1.67 g of cerium and 4.54 g of iodine. • Answers: • Moles of cerium: 0.0119 moles • Moles of iodine: 0.0357 moles

  6. To turn the number of moles into a simple whole number ratio, divide the smaller value into both numbers. • Answer: 1 mol Ce : 3 mol I Final empirical formula: CeI3

  7. Type 2: Determining an empirical formula from percentage composition. • Example 1. Determine the empirical formula of a compound that contains 36.5% sodium, 25.4% sulfur, and 38.1% oxygen. • Begin by assuming you have a 100 g sample. • Therefore you would have 36.5g Na, 25.4 g S, and 38.1 g O. Now you can convert to moles and solve.

  8. Moles of Na: 1.59 mol • Moles of S: 0.791 mol • Moles of O: 2.38 mol • Divide each by 0.791 mol • 2 Na : 1 S : 3 O  Na2SO3

  9. Type 3: Determining the molecular formula from the empirical formula and the molecular mass. • You may be asked to determine the molecular formula of a compound, given the empirical formula and the molecular mass.  • Solution: Divide the molecular mass by the mass shown in the empirical formula.  The result is a whole number that is used as a multiplier for all of the subscripts in the empirical formula.

  10. Example 1.  What is the molecular formula of a compound with an empirical formula of CH2 and a molecular mass of 56.0 amu? • Solution:  Find the mass of the empirical formula (CH2): • 14.0 u

  11. Next, divide that number into the molecular mass: • 56.0 u-------  = 414.0 u • Now use that number, 4, as a multiplier for the subscripts in the empirical formula: • CH2 x 4 = C4H8

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