Chptr Chmicl Equtions nd Stoichiomtry Homwork (This is VERY importnt chptr) Chptr 27, 29, 1, 9, 5, 7, 9, 55, 57, 65, 71, 75, 77, 81, 87, 91, 95, 99, 101, 111, 117, 121 1 2 Introduction Up until now w hv n tlking out compounds nd lmnts nd how to writ nd undrstnd wht thy mn. Chmistry in mor thn just tht. Whn w think o chmistry, w think o chmicl rctions. In ordr to undrstnd chmicl rction, w nd to study stoichiomtry nd th chmicl qution. Stoichiomtry: th quntittiv study o chmicl rctions. Chmicl qution: writtn rprsnttion o chmicl rction, showing th rctnts nd products, thir physicl stts, nd th dirction in Th Mol Whn two chmicls rct with ch othr, w wnt to know how mny toms or molculs o ch r usd do tht ormul o th product cn stlishd. This mns w nd wy to count toms, ut HOW? Th solution to this prolm is to din convnint unit o mttr tht contins know numr o prticls. 1 mols 6.0221676 10 2 prticls This vlu is commonly known hs Avogdro s numr. Rmmr tht on mol lwys contins th sm numr o prticls, no mttr wht th sustnc. which th rction procds. Molr Th mss o on mol o toms o ny lmnt is th molr mss, which is numriclly qul to th tomic mss, ut in grms. Exmpl: 1 mol o sodium wighs 22.99 g This not only tru or toms, ut lso or molculs, ut w us th molculr mss instd o th tomic mss. To dtrmin th molculr mss o molcul, w dd up th tomic msss o th lmnts tht mk up th molcul. Exmpl: CCl W hv 1 cron nd chlorins 12.001 g/mol + (5.527 g/mol) 15.81 g/mol Thror, i w hv 1 mol o CCl it would wigh 15.81 Molculs, Compounds, nd Th Mol Thinking out mols with rspct to molculs nd compounds is not ll tht dirnt thn with toms. I I hv 1 mol o CH : 1. How mny CH molculs do I hv? I hv 6.022 10 2 molculs o CH 2. How much dos it wigh? 16.02 g. How mny cron toms do I hv? 6.022 10 2. How mny hydrogn toms do I hv? 6.022 10 2 2.09 10 2 g. 5 6
MOLES r th CENTER prticls ) Multiply y Avogdro s Numr ) Divid y Avogdro s Numr c) Multiply y th molculr wight o X d) Divid y th molculr wight o X mols o X c d Vol. o n idl gs @ STP ) Multiply y 22. L / mol ) Divid y 22. L / mol 7 Wht is th mss, in grms, o 1.5 mols o silicon? 2.128 g How mny mols r rprsntd y 5 g o sulur? How mny toms? 1.2 mols 8.5 10 2 toms EXAMPLES At irst ths concpts r diicult to undrstnd, ut working prolms is th st wy to com comortl with ths concpts. Clcult th molr mss o CCO (limston). 100.08 g/mol I you hv 5 g o CCO, how mny mols do you hv? How mny molculs? How mny toms o oxygn?.5 mols 2.7 10 2 molculs 8.20 10 2 tom o O 8 Prcnt Composition Empiricl nd Molculr Formuls Prcnt Composition Th prcntg o th mss o compound rprsntd y ch o its constitunt lmnts. According to th lw o constnt composition, ny smpl o pur compound lwys consists o th sm lmnts comind in th sm proportion y mss. Exmpl: Wht is th prcnt composition o N nd H in mmoni (NH )? Prcnt N in NH o N in 1 mol o NH 1.01 g N 100 o 1 mol o NH 17.0 g NH 82.27% Prcnt H in NH o H in 1 mol o NH (1.008) g H 100 o 1 mol o NH 17.0 gnh 17.76% 9 W hv discussd molculr ormuls Empiricl ormuls A molculr ormul showing th simplst possil rtio o toms in molcul. From prcnt compositions w cn dtrmin mpiricl nd molculr ormuls y th ollowing procdur. % A g A x mols A x mol A % B g B x mols B Find mol y mol B rtio Rtio givs ormul A x B y 10 Exmpl Eugnol is th ctiv componnt o oil o clovs. It hs molr mss o 16.2 g/mol nd is 7.1% C nd 7.7% H; th rmindr is oxygn. Wht r mpiricl nd molculr ormuls or ugnol? Whr to strt?! 11 Exmpl Eugnol is th ctiv componnt o oil o clovs. It hs molr mss o 16.2 g/mol nd is 7.1% C nd 7.7% H; th rmindr is oxygn. Wht r mpiricl nd molculr ormuls or ugnol? Whr to strt?! 1. Wht is th % o oxygn? 100% - (7.1% + 7.7%) 19.9% 2. How mny grms o C, H, nd O? Assum you hv 100 g o th compound thror your prcntgs r th numr o grms o C, H, nd O. 7.1 g o C 7.7 g o H 19.9 g o O. Chng th grms into mols. 7.1 g o C 1 mol o C / 12.011g o C 6.09 mols o C 7.7 g o H 1 mol o H / 1.0079 g o H 7.1 mols o H 19.9 g o O 1 mol o O / 15.999 g o O 1.22 mols o O. Find th mol rtios (divid th lrg mount(s) o mols y th smllst mount o mols). 6.09 mols o C / 1.22 mols o O 5 mols o C to 1 mol o O 7.7 mols o H / 1.22 mols o O 6 mols o H to 1 mol o O 5. So wht dos this mn? C 5 H 6 O is th mpiricl ormul 6. How to gt th molculr ormul? Molculr wight o th mpiricl ormul. 82 g/mol Divid th molculr wight o ugnol y th molculr wight mpiricl ormul. 16.2 g/mol / 82 g/mol 2 So thr r 2 units o th mpiricl ormul. 12 (C 5H 6O) 2 C 10H 12O 2
Hydrtd Compounds I ionic compounds r prprd wtr solution nd thn isoltd s solids, th crystls otn hv molculs o wtr trppd in th lttic. Compounds in which molculs o wtr r ssocitd with th ions o th compound r clld hydrtd compounds. Exmpls: NiCl 2 6H 2 O CSO 2H 2 O But wht out th molr mss? (Do I dd th wtr?) YES! NiCl 2 6H 2 O 27.7 g/mol 129.6 g/mol rom NiCl 2 108.1 g/mol rom 6H 2 O Dtrmining th Units o Hydrtion Exmpl: Th ollowing rction tks plc. 1.02 g o CuSO x H 2 O + ht 0.65 g CuSO +? g H 2 O Find th units o hydrtion. Stp 1: dtrmin th mss o wtr 1.02 g o CuSO x H 2 O 0.65 g o CuSO 0.69 g o wtr Stp 2: dtrmin th numr o mols 0.69 g o wtr 1 mol o wtr / 18.02 g o wtr 0.0205 mols o wtr 0.65 g CuSO 1 mols CuSO / 159.6 g CuSO 0.0010 mols CuSO Stp : dtrmin th molr rtio 0.0205 mol H2O 0.0010 mol CuSO So, w know tht w strtd with 1.02 g o CuSO 5 H 2 O CSO 2H 2 O 172.1 g/mol 1 1 5.00 mol H2O 1.00 mol CuSO Chmicl Eqution A writtn rprsnttion o chmicl rction, showing th rctnts nd products, thir physicl stts, nd th dirction in which th rction procds. Exmpl: 2 Al(s) + Br 2 (l) Al 2 Br 6 (s) So it tks: 2 toms (or mols) o solid luminum molculs (or mols) o liquid romin rctnts To mk: 1 molcul (or mol) o solid Al 2 Br 6 product(s) Th givn rction is n xmpl o lncd qution, which is our nxt topic. Exmpl: F(s) + O 2 (g) 2 F 2 O (s) Wht r th stoichiomtric coicints? I you usd 8000 toms o F, how mny molculs o O2 r rquird to consum th iron compltly? molculs o O 8000 toms o F toms o F 2 6000 molculs o O2 15 Blncing Chmicl Equtions W hv sn wht lncd chmicl qution looks lik. Blncing chmicl qutions is n ppliction o oth th modrn tomic thory nd th lw o consrvtion o mttr. Th numr o on kind o tom on th rctnts sid must qul to th numr o th sm kind o tom on th product sid. 16 Exmpl Blncing Chmicl Equtions Exmpl: writ th lncd qution or th complt comustion o propn (C H 8 ). Comustion is th procss o urning compound in th prsnc o oxygn (O 2 ) to orm CO 2 gs nd H 2 O liquid. This is th unlncd chmicl qution. Blnc th numr o C toms How mny on th lt? How mny o th right? 1 So, w nd to dd thr C toms to th right, how? Chng th stoichiomtric coicint. Blnc th numr o H toms How mny on th lt? 8 How mny on th right? 2 So, w nd to dd 6 H toms to th right, y chnging th stoichiomtric coicint, ut nd up with totl o 8 H toms on th right. Blnc th numr o O toms How mny on th lt? 2 How mny on th right? 10 So, w nd to dd 8 O toms to th lt nd nd up with totl o 10 on th lt. Mor Exmpls Comustion o utn (C H 10 ). 2C H 10 (g) + 1 O 2 (g) 8CO 2 (g) + 10 H 2 O(l) Comustion o pntn (C 5 H 12 ). C 5 H 12 (g) + 8O 2 (g) 5 CO 2 (g) + 6 H 2 O(l) C H 8 (g) + 5O 2 (g) CO 2 (g) + H 2 O(l) 17 18
Figur.12: Hydrogn nd nitrogn rct to orm mmoni ccording to th qution N 2 + H 2 2NH. Th POWER o Blncd Chmicl Eqution With lnc chmicl qution, w cn do mny things. Dtrmin how much (mols & grms) o our products w will mk with known mount o rctnts. Dtrmin how much rctnts (mols & grms) w will nd to crt known mount o product. For xmpl w wnt to mk 6 g o H 2 O(l) rom th comustion o mthn (CH ), how much mthn nd oxygn will w nd? CH (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(l) 6 g o H 2 O.56 mols o H 2 O.56 mols o H 2 O 1 mol o CH / 2 mols o H 2 O 1.78 mols o CH.56 mols o H 2 O 2 mol o O 2 / 2 mols o H 2 O.56 mols o O 2 How did I do tht? 19 20 From Chptr MOLES r th CENTER prticls Vol. o n idl gs @ STP Using th Blncd Eqution Vol. o n idl gs @ STP Vol. o nd idl gs @ STP ) Multiply y Avogdro s Numr ) Divid y Avogdro s Numr c) Multiply y th molculr wight o X d) Divid y th molculr wight o X mols o X c d ) Multiply y 22. L / mol ) Divid y 22. L / mol 21 prticls ) Multiply y Avogdro s Numr ) Divid y Avogdro s Numr c) Multiply y th molculr wight o X or Y d) Divid y th molculr wight o X or Y g mols o X mols o Y c x d x h g) Multiply y th stoichiomtric coicint o Y nd divid y th stoichiomtric coicint o X. h) Multiply y th stoichiomtric coicint o X nd divid y th stoichiomtric coicint o Y. c y d y prticls ) Multiply y 22. L / mol ) Divid y 22. L / mol 22 Exmpl I 5 g o propn (C H 8 ) is urnd, wht mss o oxygn (in grms) is rquird or complt comustion nd wht msss o cron dioxid nd wtr (in grms) r ormd? C H 8 (g) + 5 O 2 (g) CO 2 (g) + H 2 O(l) 5 g o C H 8 1 mol o C H 8 /.10 g o C H 8 10. mols o C H 8 10. mols o C H 8 5 mols o O 2 / 1 mol o C H 8 51.5 mols o O 2 2.0 g/mol 168 g o O 2 10. mols o C H 8 mols o CO 2 / 1 mol o C H 8 0.9 mols o CO 2.0 g/mol 160 g o CO 2 10. mols o C H 8 mols o H 2 O / 1 mol o C H 8 1.2 mols o H 2 O 18.0 g/mol 72 g o H 2 O 2 Limiting Rctnt Limiting Rctnts Th rctnt prsnt in limitd supply tht dtrmins th mount o product ormd. In othr words, on o our rctnts in chmicl rction will run out irst nd this dictts th mount products ormd. Exmpl: 225 g o SiCl is mixd with 225 g o Mg, which compound is th limiting rctnt nd how much Si (s) will producd. SiCl (l) + 2 Mg(s) Si(s) + 2 MgCl 2 (s) 225 g o SiCl 1 mol o SiCl / 169.90 g o SiCl 1.2 mols o SiCl 1.2 mols o SiCl 1 mol o Si(s) / 1 mol o SiCl (l) 1.2 mols o Si(s) 225 g o Mg 1 mol o Mg / 2.050 g o Mg 9.26 mols o Mg 9.26 mols o Mg 1 mol o Si(s) / 2 mol o Mg.6 mols o Si(s) So, SiCl (l) is th limiting rctnt cus it only will orm 1.2 mols o Si(s). 1.2 mols o Si(s) 28.0855 g / 1 mol 7.1 g o Si(s). 2
Prcnt Yild Prcnt Yild Th ctul yild o chmicl rction s prcntg o its thorticl yild. ctul yild Prcnt y ild 100% thorticl yild Exmpl: mthnol (CH OH) dcomposs to orm hydrogn nd cron monoxid gs. I 125 g o mthnol dcomposs wht is th thorticl yild o hydrogn gs? I you otind only 1.6 g o hydrogn, wht is th prcnt yild o th gs? CH OH(l) 2 H 2 (g) + CO(g) 125 g o CH OH 1 mol o CH OH / 2.02 g o CH OH.90 mols o CH OH(l).90 mol o CHOH(l) 2 mols o H 2 (g) / 1 mol o CHOH(l) 7.8 mols o H 2 (g) 7.8 mols o H 2 (g) 2.0158 g o H 2 (g) / 1 mol o H 2 (g) 15.72 g o H 2 (g) 1.6 / 15.72 100 86.51 % yild 25