Week 9: Moles and Conversions
Week 9: Moles and Conversions CHEM 111 001
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CHEM 111 001
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This 2 page Class Notes was uploaded by Bianca Notetaker on Saturday October 24, 2015. The Class Notes belongs to CHEM 111 001 at University of New Mexico taught by smith in Summer 2015. Since its upload, it has received 78 views. For similar materials see Elements Of General Chemistry in Chemistry at University of New Mexico.
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Date Created: 10/24/15
10242015 OneNote Online Conversions Wednesday October 14 2015 1003 AM Chemical equations describe quantity relationships 0 The chemical equation describes how quantities of atoms and molecules react not mass Converting from a macroscopic quantity to moles 0 Take the amount of substance divided by the atomic mass 0 The smaller the mass on the periodic table the more moles there are Example Mass of an Conversion factor Quantity element using molar mass of atoms compound 500 g C 1 mol C 1201 g 416 mol of C Molar mass and formula mass 0 Formula mass the sum of all the atoms in the compound multiply them by their subscript o Molar mass atomic weight of an element Generalizing the mole concept 0 Moles are a convenient quantity for working with chemical reactions you can go from moles to either the mass or exact quantity 0 Massmolar mass gt moles Agt exact quantity Mole ratios 0 You can switch between molecular and atomic quantities using the ratios of the molecular formula Stoichiometry o The coefficients describe the style in which reactants combine 0 They also tell us how much product to expect httpsonenoteofficeappsivecomoonenoteframeaspxFiSDE50F6024D8D962E0395ampHemuampC5810SN2SKYWACWSHIampuienUSamprs 11 10242015 OneNote Online Friday October 16 2015 1016 AM Limiting reagent and excess 0 Reaction stops once the limiting reagent smaller number is consumed 0 You can use the given amounts of both reactants and calculate both of the products The reactant that gives you less product is limiting 0 When you39re given the amounts of more than one reactant always check that one is limiting That it39s a stoichiometric ratio 0 Start with grams convert to moles then moles of h20 then grams of h20 o In a multiple choice test the smaller number is most likely going to be the right answer Theoretical and percent yield 0 Theoretical yield is the result of the stoichiometric calculation all 0 Percent yield is the actual amount of product 0 yield actual amount of product theoretical amount of product 100 o If you get over 100 you messed up the calculation Finding empirical formulas from expiremental data 0 You need to convert both atoms to moles so multiply the given grams by 1 mil over their molar mass httpsonenoteofficeappsivecomoonenoteframeastFiSDE50F6024D8D962EO395ampHemuampC5810SNZSKYWACWSHIampuienUSamprs 11
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