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Chapter 3 part 2

by: Kyle A. Headen

Chapter 3 part 2 CHEM 1307

Kyle A. Headen

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Rest of Chapter 3
Experimental Principles of Chemistry 1
Class Notes
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This 6 page Class Notes was uploaded by Kyle A. Headen on Wednesday September 28, 2016. The Class Notes belongs to CHEM 1307 at Texas Tech University taught by Whittlesey in Fall 2016. Since its upload, it has received 32 views. For similar materials see Experimental Principles of Chemistry 1 in Chemistry at Texas Tech University.


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Date Created: 09/28/16
Writinga Formulafroma Name: Forbinary nonmetalcompounds: Thenumerical prefixes will tell you howmanyof each element is present. Ifthere is noprefix, assumethat there is oneof that element. Forionic compounds: You mustknowthecharges on all oftheions. Write downtheions with their charges, anddetermine thesmallest numberofeach ion that is needed so that thetotalnumberofpositivecharges is equal to thetotalnumberofnegative charges. Foracids: Figureout thecharge on theanion. Write downthenumberofH+ ions that are needed to balance thenumberofnegative charges on the anion. Mass Percent: Knowingthemass percent of an element in acompoundis usefulfordetermining theformulaofthe compound. Themass percent ofan element can be calculated fromtheformula, as follows: 1) Find thetotalmass ofeverything in theformula(themolarmass). 2) Foreach element, dividethe totalmass ofthat element in the compoundbythemolarmass of thecompound. This gives the massfraction foreach element. 3) Multiply each massfraction by 100toget themasspercent foreach element. Example: C20H24N2O2 (Quinine,usedinthe treatmentof malaria) C: (20)(12.011)=240.22 H: (24)(1.00794)=24.19056 N: (2)(14.0067)=28.0134 O: (2)(15.9994)=31.9988 Total molar mass forC20H24N2O2is324.42gmol-1 %C: 240.22/324.42x100=74.05% %H: 24.19056/324.42x100=7.46% %N: 28.0134/324.42x100=8.63% %O: 31.9988/324.42x100=9.86% Empirical Formula: Theformulaofan unknowncompoundcanbedetermined in thefollowingway: I. Measurethemass ofa sampleofthe unknowncompound. II. Break downthe compoundusingchemical methodsand determine themass (ormass percent) of each element present. III. Convert themassof each element into moles (numbersofatoms)ofthat element. IV. Determine thesimplest wholenumberrelationship between the elements in thecompound. This wholenumberrelationship is called the"empirical formula”. We usethemolarmasses to convert whatwemeasure in thelaboratory (massesofthe elements in the compound)towhatwe wouldlike to know(thenumber of atomsof each element in the compound). In manyproblems youare given the masspercents of theelements present and youmust determinethe formula. Problem: A puresampleof anew chemical compoundwasanalyzed and wasfoundto havethefollowing percentages, by weight, oftheelements C, H, and Cl: C 43.01%; H 2.58; Cl 54.41%. What is the empirical formulaofthis compound? Solution: Themass percentages aretrue forany size sampleofthis compound,andthe formulathat weare trying to find is trueforany sizesample, so wemay chooseany size that makesthe arithmetic easier. Since percentages are"parts perhundred,"wewill choose100g ofcompoundforthecalculation. Stepsto findthe EmpiricalFormula: Step 1: Calculate thenumberofgrams (mass)ofeach element present. Step 2: Convert themassesto moles (numberofatoms)of each element. Step 3: Adjustthenumbersby multiplying bya commonfactorsothat wehavea whole-number relationship between them.  Step 1 Calculate themass ofeach element present. A 100g samplehas 43.01gofcarbon. 100g x 43.01/100 = 43.01g A 100g samplehas 2.58gof hydrogen. A 100g samplehas 54.41gofchlorine.  Step 2 Calculate themoles of each element present. 43.01gC x 1 molC/12.01115C= 3.581molC 2.58g Hx 1mol H/1.00797 = 2.560molH 54.41gCl x1 molCl/35.453=1.535molCl  Step 3 Adjustthenumbersofmoles to get awhole-numberrelationship. Therelationship between theelements will bemaintained if you multiply ordivideall ofthem by the samenumber. First, divideall the numbersby thesmallest one. 3.581/1.535=2.333C 2.560/1.535=1.668H 1.535/1.535=1Cl Forevery 1 Clthere are very close to 2(1/3)Cand 1 (2/3)H. We can get wholenumbers forall of theelements if wemultiply all of them by3. (3)(2.333)= 6.999C = 7 C (3)(1.668)= 5.004H = 5 H (3)(1)=3.000Cl= 3 Cl Theempirical formulaforthecompoundis C7H5Cl3,which is thesimplest whole-numberrelationship between theatoms ofeach element. In somesituationsyoumay be given themasses oftheelements present and youmust determinethe formula. A 13.0100gsampleofa new chemical compoundwasanalyzed and wasfoundto havethefollowing masses oftheelements C and H: C 4.1100g H 0.6895g Therest ofthe compoundwasmadeup ofoxygen. What is theempirical formulaofthis compound? Step 1: We already havethemasses of each element present except foroxygen, soweadd themasses of thecarbon andhydrogen andsubtract that sumfrom thetotalmass to findthe oxygen. Step 2: We then convert thosemasses to moles (numberof atoms)ofeach element. Step 3: Oncewedetermine howmany moles ofeach element are present, weadjustthosenumbers so that wehaveawhole-numberrelationship between them. Step 1 Calculate themass ofeach element present. 4.1100gC + 0.6895gH = 4.7995gC +H 13.0100gtotal- 4.7995gC +H =8.2105gO Step 2 Calculate themoles of each element present. 4.1100gC x1 molC/12.01115 = 0.342182molC 0.6895gHx 1molH/1.00797 =0.6840481molH 8.2105gOx 1 molO/15.9995 =0.5131722molO Step 3 Adjustthenumbersofmoles to get awhole-numberrelationship. First,divide all thenumbersby the smallest one. 0.342182/0.342182 =1 C 0.6840481/0.342182 =1.9999H 0.5131722/0.342182 =1.4999O Forevery oneC there are 2 Hand 1 (1/2)O. We can get wholenumbers forall of theelements if wemultiply all of them by2. (2)(1) = 2 C (2)(2) = 4 H (2)(1.5) = 3 O Theempirical formulaforthecompoundis C2H4O3. MolecularFormula: Theempirical formulais thesimplest whole-numberrelationship between the atomsof thethedifferent elements in thecompound However, themolecules ofthecompoundmay bemadeup ofmany moreatoms,still in thesame proportionto each other. Forexample, thechemically different compoundsC2H4,C3H6,C4H8,andC5H10allhavetheempirical formulaCH2 (all havetwohydrogensforevery carbon). Theformulaofany of thesemolecules could bewritten as a multipleof theempirical formula,i. e. C2H4 could bewritten (CH2)2,andC3H6wouldbe(CH2)3,andin general anycompoundwith this empirical formulacould bewritten as (CH2)n,wheren is an integer. Ifthe molecular massis known(fromexperiment) andthe empirical formulais known,thevalueofn (andhence themolecular formula)can beobtained by dividing themolecular mass bythe massof the empirical formula. Thenumbern is thenumberof empirical formula"units"it takes to makeup a molecule. Stepsfor MolecularFormula: (1) Find themolarmass fortheempirical formula. (2) Divide themolecular massby themass forthe empirical formulato get n. (3) Multiply theempirical formulaby n to get the molecularformula. Example: Theempirical formulaof thecompoundis C2H4O3andits molecular massis 228.158. Whatis the molecular formula? Step 1 Themolar massfortheempirical formulaC2H4O3is: (2)(12.01115)+(4)(1.00797)+(3)(15.9995)=76.05268molarmass Step 2 Themolar massforthemolecular formuladivided bythe molarmassfortheempirical formulais: 228.158/76.05268=2.9999995=3 Step 3 Themolecular formulais (C2H4O3)3,whichis C6H12O9.


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