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Intro Chemistry Chapter 3 Notes

by: Alexis Tate

Intro Chemistry Chapter 3 Notes 13699

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Alexis Tate

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These notes cover everything that has been covered so far in chapter 3. I have also included numerous tips and strategies to handle the problems covered in this chapter.
Intro Chemistry I
Alexander Schwab
Class Notes
General Chemistry
25 ?




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This 6 page Class Notes was uploaded by Alexis Tate on Monday September 26, 2016. The Class Notes belongs to 13699 at Appalachian State University taught by Alexander Schwab in Fall 2016. Since its upload, it has received 51 views. For similar materials see Intro Chemistry I in Chemistry at Appalachian State University.


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Date Created: 09/26/16
Chapter3: The Mole& Stoichiometry Monday, September 26, 2016  The mole o 1 mole (mol) of atoms = 6.0221423 x10 atoms *Avogadro’s number - 12  1 mol of C has a mass ofexactly12g This is a very  1 mol of C has a mass of12.0107g important number get  1 mol of O has a mass of15.9994g comfortable using it!!! You’ll justbe expected  Molecular Mass (aka Molecular Weight) 23 o 1 mol of H 2olecules has 2times 1 mol of H atoms to use 6.022 x10  H 2 2x 1.00794g =2.01588g most of the time This number represents This number came from how many atoms you the periodic table have of a specific element o Formula Mass canbecalculatedthe sameway  i.e.CaCl2,PCl5,NaI o Example:  What is the molar mass of sugar(C 6 12)?6 C = 6x (12.01g) Add thesetogether to H =12 x (1.0079g) O = 6 x(15.999g) getmolar mass (g/mol) 180.155 g/mol  What is the mass of 0.350 mol of sugar? - *Molar Mass inmost problems will be your conversion tool to get grams to moles or viceversa 0.350 mol 180.155 g = 63.1 g 1 mol  What is the mass of one sugarmolecule ingrams? -*Avogadro’s number inmost problems will be your conversion tool to getfrom moles to whatever you’re solving for. (i.e.molecules, atoms, etc.) 180.155 g 1 mol = 2.992 x 10-22g mol 6.022 x 10 molecules  How many moles of sugarare there in10.0 g? 10.0g mol =0.0555 mol 180.155g  How many molecules of sugar? 0.0555 mol 6.022 x 10 molecules = 3.34 x 10 molecules 1 mol  You’re given0.350 mol of sugar.How many O atoms arein this sample? 23 0.350 mol 6 O 6.022 x 10 = 1.26 x 10 Oatoms molecules 1 C6 12 6 1 mol +  You need 7moles of Na ions.How many grams of sodium sulfatedo you need to supply theseions? Molar Mass: Na = 2 x(22.990g) S = 1 x(32.06g) O =4 x (15.9994g) 142.04 g/mol 7 mol Na + 1 Na 2O 4 142.04 g = 497 g,so 500 g becauseof + 2 Na + mol onesig fig (7mol ofNa ) * Differencebetween finding TOTAL number ofmoleculesand TOTAL number ofatoms- When calculating formolecules usuallyall you have to do is multiply Avogadro’s number by the number of moles of the compound. For atoms, you need to add up the total number of atoms in a compound and then multiply that total by Avogadro’s number. This phrasing may help * In anelement likeCl,there are Avogadro’s number of atoms 23 (6.022 x 10 atoms make up Cl) * In one mole of a compound likesugar,there are Avogadro’s number of molecules (6.022 x 10 molecules of sugarmake up one mole of sugar) o Example Problem:  Find the total number of molecules in2 moles for sugar. 23 = 1.2044 x 10 moleculesso, 2 mol 6.022 x 10 molecules 24 1 mol 1 x 10 moleculesb/c of1sig fig (2 mol)  Find the total number of atoms in 2 moles for sugar. - we alreadyknow the total number of molecules intwo moles of sugar - to getto atoms we need to add the total number of atoms inone molecule of sugar(6C + 6O + 12H =24 atoms) 2 mol 6.022 x 10 molecules 24 atoms of C 6 12 6 1 mol 1 molecule of C 6 12 6 = 2.89056 x 10 atomsso, 3x 10 atomsb/c of1 sig fig (2mol)  Mass Percent Composition o Describes the fraction of a particular element ina compound based on mass o Defined as: Total mass ofelement atoms Formula weight X 100 =% element o Example:  What is the mass percentage of C,H, and O in sugar? Mass of C Total mass ofsugar X 100 =% C 6(12.011g) 180.155g/mol X 100 =40.002% Mass ofH X 100 =% H Total mass ofsugar 12(1.0079g) X100 = 6.7136% 180.155g/mol % O = 100 –%C – %H % O = 100 –40.002 –6.7136 % O = 53.284% * Iknow in the notes itwas rounded to four sig figs butitshould be five due to all ofthe masses being fivesig figs  How many grams of carbon arein 5.00 g of sugar? For anyone who has trouble dealing with problems likethese: - for any problem always startoff with the givennumber (5.00g) - think of what this value means (we have 5.00 g of sugar) - seewhat you need to get to for your final answer (grams of carbon) - inmost problems more than likelyyou’re going to use molar mass and/or a “mole bridge” (using moles of something to get to moles of something else) to get to your answer - Finallymultiply across the top and divide that value by the bottom - Don’t forget sig figs. 5.00 g C6 12 6 1 mol C6 12 6 6 mol C 12.011g C 180.155g C 6 12 6 1 mol C 6 12 6 1 mol C = 200. g C Mass Percent: 200. g C X 100 = 40.0% C 500. g C6 12 6  Elemental Analysis o Unknown compounds are decomposed  Constitutive elements are quantified or relative amounts of these elements are determined o Combustion analysis forhydrocarbons  Classic CHanalysis o Example of Combustion Analysis  Let’s sayyou place 2.200 g of anorganic compound which should contain C,H, and O. CHanalysis tells us thereare 0.8247 g of C and 0.2768 g of H. What is the empirical formula for this compound? *For any question dealing with finding the empirical formula, always get eachindividual element in the compound to moles. 0.8247 g C 1 mol C 12.011 g C = 0.06866 molC 0.2768 g H 1 mol C = 0.2746 molH 1.0079 g H - All of the elements masses mustequal 2.200 g so, Mass ofO =2.200g –0.8247g C – 0.2768g H= 1.098]5g 1.0985g O 1 mol O = 0.06866 We’ll keep the 15.999g O (0.0687) whole valuefor molO accuracybut keep in mind sig figs. *After getting each element to moles,divide each answer by the element with the smallestnumber of moles. In this particular problem moles of O = moles of C so itdoes not matter which number you pick. 0.06866 mol C =1 C 0.06866 mol C 0.06866mol O = 1 O 0.06866 mol C 0.2746 mol H =3.999 H,soapproximately 4 0.06866 mol C *For any decimals always round to the nearest whole number unless itis in the middle between two numbers (~1.5) TheempiricalformulaisCH O. 4 o Elemental Analysis Example:  Other analysis techniques yieldresults as mass percents o For example:  Analysis on anunknown compound finds itis 2.055% H, 32.693% S, and 65.252% O by mass.What is the empirical formula? *When dealing with mass percents, it is safetoassumeyou have a 100 g becausepercents add up to equal 100. By doing this,each percent composition for eachelement canbe used as a mass. 2.055g H 1 mol H = 2.039 molH 1.0079g H 32.693g S 1 mol S 32.06g S = 1.0197 molS 65.252g O 1 mol O = 4.0785 molO 15.999g O 1.0197 mol S = 1S 1.0197 mol S 2.039 mol H = about 2H 1.0197 mol S 4.0785 mol O = about 4O 1.0197 mol S TheempiricalformulaisH SO 2sul4uric acid–very strong acid)  Chemical Reactions (Rxns): o Chromium reacts with hydrochloric acid(HCl –very strong acid,dissolves in solutions)to produce chromium(III) chloride and hydrogen gas.  Write the chemical equation: Cr (s) + HCl (aq)  CrC3 (aq) + 2 (g) * Trickto remember diatomic elements: HalogenHON. All of the halogens up to Iodine are diatomic and soare Hydrogen, Oxygen,and Nitrogen.  Balancethe chemical equation 2 Cr(s) + 6HCl (aq)  2 CrCl3(aq) + 3H 2g)  Mass ofProducts to Mass ofReactants o Example:  You drop 10.0 g of chromium into an enormous vat of HCl.Predict the amount of chromium(III) chloride that would be made. - This problem will require a “mole bridge” and molar mass - You cantell becausethe problem gives you the mass of anelement and is asking you to find the mass of a compound - This problem alsoinvolves the chemical equation that was described above. This is where you will get your values for the “mole bridge”.(It may help to think of this as a ratio) Molar Mass:Cr= 51.996 Cl = 3(35.45) 158.35 g/mol 10.0 g Cr 1 mol Cr 2 mol CrCl 3 158.35g CrCl 3 51.996 g Cr 2 mol Cr 1 mol = 30.5 g CrCl3


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