1 / 7

Unit: Chemical Quantities

Day 6 - Notes. Unit: Chemical Quantities. Determining the Empirical Formula from Percent Composition. After today, you will be able to…. Calculate the empirical formulas for various compounds given their percent makeup. Empirical Formula Determination from % Composition.

holt
Download Presentation

Unit: Chemical Quantities

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Day 6 - Notes Unit: Chemical Quantities Determining the Empirical Formula from Percent Composition

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

  3. Empirical Formula Determination from % Composition To determine empirical formulas from % composition follow these steps: • Assume 100.0g of the compound (change % into grams) • Convert g moles for each element using molar mass • Divide each mole answer by the smallest mole answer • If you do not have a whole number, multiply by the smallest whole # to get a whole # ratio. • Skip to this step if you already have a whole number ratio from #3. These whole #s are the subscripts for each element.

  4. To sum up these steps here is a poem for my favorite little chemists: “Percent to mass Mass to Moles Divide by small Multiply by whole!”

  5. Example: What is the empirical formula for a compound that is 21.2% N, 6.10% H, 24.3% S, and 48.4%O? 1 mol N 14.01gN 1.51 21.2gN 6.10gH 24.3gS 48.4gO x = = 2 0.758 0.758 0.758 0.758 1 mol H 1.01gH 6.04 = x 7.97~ = 8 N2H8SO4 1 mol S 32.07gS 0.758 = x = 1 3.03 1 mol O 16.00gO = x = 4

  6. Example: What is the empirical formula for a compound that is 69.8% Fe and 30.2% O? 1 mol Fe 55.85gFe 1.25 69.8gFe 30.92gO x = = 1 (2) = 2 1.25 1.25 1.89 1 mol O 16.00gO (2) 1.5 3 = x = = Fe2O3

  7. HOMEWORK: Time to Practice!Complete WS 6

More Related