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Do Now. What is the empirical formula for a compound that is 50.61% copper, 11.16% nitrogen, and 38.23% oxygen?. Do Now. What is the empirical formula for a compound that is 50.61% copper, 11.16% nitrogen, and 38.23% oxygen?. SMALLEST?. 1 mol Cu 63.55gCu. 0.796 mol. 50.61g Cu 11.16g N

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  1. Do Now What is the empirical formula for a compound that is 50.61% copper, 11.16% nitrogen, and 38.23% oxygen?

  2. Do Now What is the empirical formula for a compound that is 50.61% copper, 11.16% nitrogen, and 38.23% oxygen? SMALLEST? 1 mol Cu 63.55gCu 0.796 mol 50.61g Cu 11.16g N 38.23g O = 1 x 2 = 0.796 1 mol N 14.00g N 0.797 mol = = 1 x 2 0.796 1 mol O 16.00 g O 2.014 mol x 2 2.5 = = 0.796 Cu2N2O5

  3. Check HW Time • I’m passing out answer keys to HW • You have 5 mins to check your work, then we are moving on to todays work. • Remember: you NEED to practice these calculations to understand them. • This time is to CHECK/ CLARIFY your work, not copy my work =/

  4. Day 7 - Notes Unit: Chemical Quantities Determining the Molecular Formula from the Empirical Formula

  5. After today, you will be able to… • Calculate the empirical formulas for various compounds given their percent makeup

  6. Empirical Formula vs. Molecular Formula • Empirical Formula: The simplest whole-number ratio of atoms in the compound. Ex. Iso-octane, the empirical formula is C4H9 • Molecular Formula: Specifies the exact number of atoms of each element in one molecule. Ex.Iso-octane, the molecular formula is C8H18

  7. Molecular Formula Determination from the Empirical Formula To determine molecular formulas from the empirical formula follow these steps: • First determine the empirical formula • Use the formula: • Multiply the subscripts in the empirical formula by this whole number to get the molecular formula Molar mass of molecular formula Molar mass of empirical formula

  8. Example: What is the molecular formula for a compound that is 74.0% C, 8.70% H, and 17.3% N and has a molar mass of 243.39g? 1 mol C 12.01gC 6.16 70.4gC 8.70gH 17.3gN Empirical: C5H7N x = = 5 5C=12.01(5) = 60.05 1 mol H 1.01gH 8.61 7H=1.01(7) = 7.07 = x 1.23 1.23 1.23 = 7 1N=14.01(1) = +14.01 1 mol N 14.01gN 1.23 81.07 1 = x = C15H21N3 = 3 x C5H7N 243.39 81.07

  9. Time to Practice!1. Complete HW 72. Complete Review Packet for Moles (Day 1-7)3. Complete Quiz Re-doDO NOT PUT YOUR STUFF AWAY YOU HAVE STUFF TO DO

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