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Drill – 2/7/11

Drill – 2/7/11. What is the molar mass of BF 3 ? What is the molar mass of B 2 F 6 ? What is the ratio between these values?. Empirical Formulas. How would a scientist determine the identity of a substance that they made?. Percent Composition. Mass of element in compound X 100%

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Drill – 2/7/11

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  1. Drill – 2/7/11 • What is the molar mass of BF3? • What is the molar mass of B2F6? • What is the ratio between these values?

  2. Empirical Formulas

  3. How would a scientist determine the identity of a substance that they made?

  4. Percent Composition • Mass of element in compound X 100% Total mass of compound • Repeat for each element in compound • Will add up to 100

  5. Empirical Formula • The symbols for the elements combined in the compound, with subscripts showing the smallest whole-number ratio of the different atoms in the compound. • Ionic = formula unit • Covalent = reduced molecular formula

  6. If you know the percent composition of a compound, then you can determine its empirical formula. • We do this by first assuming we have a 100.0 g sample so that the percentage values can be converted to grams. • We then convert the grams to moles (to account for atoms having different molar masses) • To correct for our assumption of 100.0 g we then divide each number of moles by the smallest number.

  7. What is the empirical formula if a compound contains 32.4% Na, 22.6% S, & 45.0% O? • Mass composition (in 100.0g) = • 32.4 g Na, 22.6 g S, 45.0 g O • Composition in moles = • 32.4 g Na x (1 mol Na / 22.99 g Na) = 1.41 mol Na • 22.6 g S x (1 mol S / 32.07 g S) = 0.705 mol S • 45.0 g O x (1 mol O / 16.00 g O) = 2.81 mol O • Smallest whole-number mole ratio • 1.41 mol Na / .705 = 2.00 mol Na • .705 mol S / .705 = 1.00 mol S • 2.81 mol O / .705 = 3.99 mol O

  8. Percent Composition Mass Composition Composition in moles Smallest whole-number mole ratio

  9. A compound’s empirical formula can also be used to determine its molecular formula. • x (empirical formula) = molecular formula • x = whole-number factor

  10. Determine the molecular formula of the compound with an empirical formula of CH and a mass of 78.11 amu. • Molecular formula mass = 78.11 amu • Empirical formula mass = C + H = 12.01 amu + 1.01 amu = 13.02 amu x (empirical formula) = molecular formula x (13.02 amu) = 78.11 amu x = 78.11 amu / 13.02 amu = 5.999 So… CH  C6H6

  11. In the green workbook: • Pg 67 #1 b • Pg 77 #1 a & b • Pg 79 #1 a & b • Pg 89 #1 a & b • Pg 94 #1 a & b

  12. Drill – 2/8/11 • What is the percent composition of each element in Fe2O3?

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