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 Calculate the % of each metal in each of the following I can Determine the empirical and molecular formula for a compound from mass percent and actual mass data. Percent Composition to Empirical Formula When discovering new compounds, scientist dont 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. Lets say I surveyed area and collected some rocks. I chemically separated the compound that makes up the rock into elements. 536g rock 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 X 2 YZ Percent Composition to Empirical Formula What if a substance contains 36.84% nitrogen and 63.16% oxygen, what is the empirical formula? Empirical formulas- a ratio (determined by % composition and moles) to get the simplest whole number ratio to make a formula. 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 grams of nitrogen and grams of oxygen? First: Find out how many moles there are in the given amount g of Nitrogen 1 mole of N g of N = 2.63 Moles of Nitrogen g of oxygen 1 mole of O g of O = 3.95 Moles of Oxygen Empirical Formula We just figured out the ratio (moles) within the given amounts. But now we need to get the ratio between each element 2.63 Moles of Nitrogen = 1 Moles of nitrogen Second: Take the smallest mole amount and divide it into all others Moles of Nitrogen 3.95 Moles of Oxygen = 1.5 Moles of oxygen 2.63 Moles of Nitrogen Need whole # N2O3N2O3 X 2 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 g of aluminum 1 mole of Al g of Al = 1.33 Moles of Aluminum g of sulfur 1 mole of S g of S = 1.99 Moles of Sulfur Empirical Formula We just figured out its ratio within the substance, but what about to each other? 1.33 Moles of Aluminum = 1 Moles of Aluminum Second: Take the smallest mole amount and divide it into all others Moles of Aluminum 1.99 Moles of Sulfur = 1.5 Moles of sulfur 1.33 Moles of Aluminum NEED whole # Al 2 S 3 X 2 Empirical Formula of carbon 1 mole of C of C = 6.81 Moles of Carbon of hydrogen 1 mole of H g of H = 18.0 Moles of Hydrogen What if a substance contains 81.82% carbon and 18.18% hydrogen, what is the empirical formula? 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 Moles of Carbon = 1 Moles of Carbon Take the smallest amount (carbon) and divide it into all others Moles of Carbon 18.1 Moles of Hydrogen = 2.65 Moles of hydrogen 6.81 Moles of Carbon C3H8C3H8 NEED whole # 2x = 5.3 3x = 7.95 Your Turn- Flip paper over to complete your homework First: Find out how many moles there are in the given amount. Second: Take the smallest mole amount and divided it into all others.