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This 2 page Class Notes was uploaded by charlotteee on Wednesday September 9, 2015. The Class Notes belongs to 25954 at University of Rochester taught by Hafensteiner in Summer 2015. Since its upload, it has received 26 views. For similar materials see CHM Concepts, Systems, Practice 1 in Chemistry at University of Rochester.
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Date Created: 09/09/15
Jasmine Han Chemistry 131 Sept 9th 2015 WalkIn tutoring Sundays 48 pm Carlson 102 rochmywconlinecom Homework Ch 3 Practice Problems on Sapling Look over problems inside the textbook Oxygen has an atomic mass of 160048 amu What is the chance that you will pick up an atom of O that has a mass of 160048 out of a random sample of O atoms There is 0 that you will pick up an atom of O that has a mass of 160048 out of a random sample of O atoms Change Proton different element Change Neutron Isotope Change Electron ons Empirical Formula gt C3H603 Empirical formula gtsum of the mass of all the elements Molecular formula gtmoar mass Stoichiometry Example 7239 H 9145 N 2089 0 What is the empirical formula You can nd the percent of carbon by adding them all up to 100 Mass percent of carbon is 62737 Use 100 gram sample Must convert grams to moles because not all of the elements have the same molar mass and we need to compare them in ratios So there is 7239 g of H 9145 g of N 6273 g of C and 2089 g of 0 Let s say that we do it for hydrogen Convert grams into moles so 7239 g H times 1 mol H1008 g H 7182 mol H Repeat with the other elements to get 5238 mol C 06527 mol N and 1306 mol 0 Divide the lowest number which is nitrogen Carbon 523806527 8025 Hydrogen 11 Nitrogen 1 Oxygen 2 Therefore C8H11N02 which is an empirical formula Balancing Reactions conservation of mass Number of type of atoms in reactants must be same as number of type of atoms on the right Theoretical yield vs actual yield Theoretical yield is the maximum amount we can made Percent yield is the actual yield divided by the theoretical yield multiplied by 100 Theoretical yield is based on the limiting reagent
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