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Entry Task: Feb 6 th - 7 th Block #2

Entry Task: Feb 6 th - 7 th Block #2. Write the question down: Provide the percent composition of Gold III chloride You have 5 minutes!. Agenda: Go over % composition ws Notes on Empirical Formulas HW: Empirical Formula ws #1. Calculate the % composition of the following compounds.

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Entry Task: Feb 6 th - 7 th Block #2

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  1. Entry Task: Feb 6th- 7th Block #2 Write the question down: Provide the percent composition of Gold III chloride You have 5 minutes!

  2. Agenda:Go over % composition wsNotes on Empirical FormulasHW: Empirical Formula ws #1

  3. Calculate the % composition of the following compounds

  4. Calculate the % composition of the following compounds

  5. Calculate the % of each metal in each of the following

  6. I can… • Determine the empirical and molecular formula for a compound from mass percent and actual mass data.

  7. Percent Composition to Empirical Formula When discovering new compounds, scientist don’t know what the formula of a compound is until they get the ratio of masses of each element. From there, they can determine its empirical formula.

  8. Let’s say I surveyed area and collected some rocks. 536g rock I chemically separated the compound that makes up the rock into elements. 50% of element X 25% of element Y 25% of element Z This means there is 2 time the amount of X to each amount of Y and Z So the empirical formula is X2YZ

  9. Percent Composition to Empirical Formula Empirical formulas- a ratio (determined by % composition and moles) to get the simplest whole number ratio to make a formula. What if a substance contains 36.84% nitrogen and 63.16% oxygen, what is the empirical formula?

  10. Empirical Formula What if a substance contains 36.84% nitrogen and 63.16% oxygen, what is the empirical formula? Assuming that you had 100 grams of this substance, then it would be 36.84 grams of nitrogen and 63.16 grams of oxygen? First: Find out how many moles there are in the given amount. 1 mole of N ------------- 36.84 g of Nitrogen = 2.63 Moles of Nitrogen --------- 14.007 g of N 1 mole of O ------------- 63.16 g of oxygen = 3.95 Moles of Oxygen --------- 15.999 g of O

  11. Empirical Formula We just figured out the ratio (moles) within the given amounts. But now we need to get the ratio between each element Second: Take the smallest mole amount and divide it into all others. 2.63 Moles of Nitrogen = 1 Moles of nitrogen 2.63 Moles of Nitrogen Need whole # 3.95 Moles of Oxygen = 1.5 Moles of oxygen X 2 2.63 Moles of Nitrogen N2O3

  12. Empirical Formula What if a substance contains 35.98% aluminum and 64.02% sulfur, what is the empirical formula? First: Find out how many moles there are in the given amount. 1 mole of Al ------------- 35.98 g of aluminum = 1.33 Moles of Aluminum --------- 26.98 g of Al 1 mole of S ------------- 64.02 g of sulfur = 1.99 Moles of Sulfur --------- 32.065 g of S

  13. Empirical Formula We just figured out its ratio within the substance, but what about to each other? Second: Take the smallest mole amount and divide it into all others. 1.33 Moles of Aluminum = 1 Moles of Aluminum 1.33 Moles of Aluminum NEED whole # 1.99 Moles of Sulfur = 1.5 Moles of sulfur X 2 1.33 Moles of Aluminum Al2S3

  14. Empirical Formula What if a substance contains 81.82% carbon and 18.18% hydrogen, what is the empirical formula? 1 mole of C ------------- 81.82 of carbon = 6.81 Moles of Carbon --------- 12.01 of C 1 mole of H ------------- 18.18 of hydrogen = 18.0 Moles of Hydrogen --------- 1.0079g of H

  15. Empirical Formula We just figured out its ratio within the substance, but what about to each other? This will give us our ratio of the compound= empirical formula. Take the smallest amount (carbon) and divide it into all others. 6.81 Moles of Carbon = 1 Moles of Carbon 6.81 Moles of Carbon NEED whole # 18.1 Moles of Hydrogen 2x = 5.3 = 2.65 Moles of hydrogen 3x = 7.95 6.81 Moles of Carbon C3H8

  16. Your Turn- Flip paper over to complete your homeworkFirst: Find out how many moles there are in the given amount.Second: Take the smallest mole amount and divided it into all others.

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