Chemistry lab report

CCS D ay 2 S um mary Answ ers t o S um mary Q uestio ns 1. Show h ow y o ur g ro up d e te rm in ed t h e v a lu es o f a -d fo r th e fo rm ula fo r F e ( o x ) H OK 2 b c • d 2 yo ur p ro duct? In th e fir s t p art o f th e la b a tit r a tio n w as d o ne o f th e c re a te d c ry sta l s a m ple w it h . W e M n OK 4 tit r a te d th e s a m ple u ntil th e e qu iv a le nce p oin t w as r e a ch ed. F ro m th e v o lu m e o f a dded M n OK 4 we fo und th e a m ount o f m ole s o f b y m ult ip ly in g b y t h e m ola rit y .. M n OK 4 Sam ple 1 : 1 . 5 m L o f K M n O . 1 3 9 0 m o l o f K M n O3 4 × 1 L 1 0 0 0 m L × 1 L 0 . 0 0 9 9 6 4 m o l = 3 × 1 − 4 4 Sam ple 2 : 0 . 1 m L o f K M n O . 9 9 9 0 m o l K M n O3 4 × 1 L 1 0 0 0 m L × 1 L 0 . 0 0 9 9 6 4 m o l = 2 × 1 − 4 4 In the equation of the reaction the ratio of t o o xa la te is 2 :5 , s o th e n ext s te p w as to M n OK 4 mult ip ly th e m ole s o f b y ( 5 /2 ). A nd fin d th e p e rc e n t m ass b y m ult ip ly in g b y th e m ola r M n OK 4 mass o f o xa la te a nd d iv id in g b y th e g ra m s o f th e o rig in al s a m ple . Sam ple 1 . 1 3 9 0 m o l K M n O . 8 4 7 0 m o l o x a l a t e3 × 1 − 4 4 × 5 m o l o x a l a t e 2 m o l K M n O 4 = 7 × 1 − 4 . 8 4 7 0 m o l o x a l a t e . 1 2 4 6 g . 5 5 4 3 0 0 5 . 4 3 % b y m a s s7 × 1 − 4 × m o l 8 8 . 0 1 8 g ÷ 0 = 0 × 1 = 5 Sam ple 2 . 9 9 0 m o l K M n O . 4 7 5 0 m o l o x a l a t e2 × 1 − 4 4 × 5 m o l o x a l a t e 2 m o l K M n O 4 = 7 × 1 − 4 . 4 7 5 0 m o l o x a l a t e . 1 2 4 9 g . 5 2 6 8 0 0 2 . 6 8 % b y m a s s7 × 1 − 4 × m o l 8 8 . 0 1 8 o x a l a t e ÷ 0 = 0 × 1 = 5 To fin d th e a m ount o f P ota ssiu m o ne w il l u se th e d ata c o lle cte d in th e s e co nd p art o f th e la b wit h th e tit r a tio n o f th e e lu te d c ry sta l s a m ple a nd th e and c o lle ct tw o e quiv a le nce p oin ts . a O HN The fir s t e quiv a le nce p oin t is u se d to fin d th e a m ou nt o f p ota ssiu m . A t th e fir s t e quiv a le nce p oin t th e m ole s o f N aO h a dded is e qua l to t h e m ole s o f p ota ssiu m , b eca use th eir r a tio in th e e quatio n is 1 :1 . Sam ple 1 m L N a O H 0 m o l N a O H 0 m o l o f K8 × 1 L 1 0 0 0 m L × 1 L 0 . 1 m o l N a O H = 8 × 1 − 4 = 8 × 1 − 4 + 0 m o l K . 1 2 4 6 . 2 5 1 0 0 5 . 5 1 % b y m a s s8 × 1 − 4 × m o l 3 9 . 0 9 8 3 g ÷ 0 = 0 × 1 = 2 Sam ple 2 . 2 5 m L N a O H . 2 5 0 m o l N a O H . 2 5 0 m o l K8 × 1 L 1 0 0 0 m L × 1 L 0 . 1 m o l N a O H = 8 × 1 − 4 = 8 × 1 − 4 + . 2 5 0 m o l K . 1 2 4 6 . 2 5 8 0 0 5 . 8 % b y m a s s8 × 1 − 4 + × m o l 3 9 . 0 9 8 3 g ÷ 0 = 0 × 1 = 2 To fin d th e a m ount o f Ir o n, o n e m ust u se th e d ata c o lle cte d fr o m th e s e co nd e quiv a le nce p oin t. One m ust s u btr a ct th e v o lu m e o f t h e f ir s t e quiv a le nce p oin t fr o m th e fir s t e quiv a le nce p oin t. Then fr o m th e fo und m ole s o f and m ult ip ly b y ( ⅓ ) to g et th e m ole s o f ir o n. T o g et th e a O HN mass p erc e nt o ne m ust d iv id e th e m ole s o f ir o n b y th e m ass o f th e in it ia l s a m ple . Sam ple 1 7 . 5 m L m L . 5 m L 1 − 8 = 9 . 5 m L . 1 6 7 0 m o l F e9 × 1 L 1 0 0 0 m L × 1 L 0 . 1 m o l N a O H × 3 m o l 1 m o l F e = 3 × 1 − 4 . 4 1 7 0 m o l F e . 0 1 7 6 g o f F e3 × 1 − 4 × 1 m o l 5 5 . 8 4 5 g = 0 . 0 1 7 6 g F e . 1 2 4 6 . 1 4 1 9 0 0 4 . 1 9 % 0 ÷ 0 = 0 × 1 = 1 Sam ple 2 8 . 5 . 2 5 0 . 2 5 m L 1 − 8 = 1 0 . 2 5 m L . 4 1 6 7 0 m o l F e . 0 1 9 1 g F e1 × 1 L 1 0 0 0 m L × 1 L 0 . 1 m o l N a O H × 1 m o l F e 3 m o l N a O H = 3 × 1 − 4 × 1 m o l 5 5 . 8 4 5 g = 0 . 0 1 9 1 g F e . 1 2 4 g . 1 5 2 7 6 0 0 5 . 2 7 % 0 ÷ 0 = 0 × 1 = 1 Next o ne m ust fin d th e m ass o f w ate r. T o d o th is o ne m ust s u btr a ct th e p erc e nt m ass fr o m 1 00 to fin d th e p erc e nt b y m ass o f w ate r. Sam ple 1 5 . 4 3 % 5 . 5 1 % 4 . 9 % 5 . 8 4 5 + 2 + 1 = 9 0 0 5 . 8 4 . 1 6 % o f H O . 0 4 1 4 6 1 − 9 = 4 2 = 0 .. 0 4 1 4 6 . 1 2 4 6 . 0 0 5 1 8 8 . 0 2 . 8 7 0 m o l o f H O 0 × 0 = 0 ÷ 1 = 2 × 1 − 4 2 Sam ple 2 2 . 6 8 % 5 . 8 % 5 . 2 7 % 3 . 7 5 % 5 + 2 + 1 = 9 0 0 % 3 . 7 5 % . 2 5 % b y m a s s o f H O . 0 6 2 5 1 − 9 = 6 2 = 0 . 0 6 2 5 . 1 2 4 9 . 0 0 7 8 8 . 0 2 . 3 3 0 m o l o f H O . 0 × 0 = 0 ÷ 1 = 4 × 1 − 4 2 For th e fin al e quatio n o ne s im ply d iv id es a ll th e m asse s o f e ve ry c o m pound b y th e s m alle st mass. For o ur s a m ple o ne th e fo rm ula w e g ot w as F e ( o x ) OK 3 3 • H 2 For o ur s a m ple tw o th e fo rm ula w e g o t w as F e ( o x ) OK 3 2 • H 2 Brie f S um mary o f E xp erim en ta l P ro ced ure Oxa la te d ete rm in atio n: - Weig ht o ut tw o o f c ry sta ls , s a m ple 1 : 0 .1 46 - s a m ple 2 : 0 .1 24 - Add 6 0 m l o f D I w ate r, 6 m l o f H 2S O 4, a nd 1 m l o f 8 5% H 3P O 4 in to a fla sk - Fill a b ure tte w it h K M nO 4 - Heat th e s a m ple 1 u n til it r e ach es 8 0 C ( in it ia l a nd fin al te m pera tu re w as 8 1 fo r s a m ple 1, in it ia l te m pera tu re w as 8 0 a n d fin al te m pera tu re w as 7 8 fo r s a m ple 2 ) - Tit r a tio n o f s a m ple 1 , s ta rtin g w it h 1 m l u n til th e s o lu tio n tu rn s to p in k c o lo r - 31.5 m l o f K M nO 4 w as a dded in s a m ple 1 , a nd 3 0.1 m l in s a m ple 2 - Pota ssiu m a nd ir o n d ete rm in atio n: - Mass 0 .1 6 g fo r io n s a m ple 1 : 0 .1 64 4 g , s a m ple 2 : 0 .1 637 g H + → - Dis so lv e s a m ple 1 in 4 m l D I w ate r in 5 0 m l b eake r - Rin se c o lu m n 4 x w it h 4 m l D I w ate r u ntil it is n eutr a l - Add s a m ple 1 s o lu tio n to c o lu m n a nd c o lle ct c o lu m n in 1 50 m l b eake r - Repeat th e s a m e s te ps w it h s a m ple 2 - Calib ra te p h P ro be u sin g p H 4 a n d p H 7 - Tit r a te th e s o lu tio n c o lle cte d in 1 50 m l b eake r w it h N aO H - Reco rd d ata Specif ic R esu lt s O bta in ed Sam ple 1 Volu m e pH Fir s t e quiv a le nce p oin t 8m L 4.0 3 Seco nd e quiv a le nce p oin t 17.5 m L 10.1 Sam ple 2 Volu m e pH Fir s t e quiv a le nce p oin t 8.2 5m L 3.2 4 Seco nd e quiv a le nce p oin t 18.5 m L 9.2 Sam ple 1 T it r a tio n C urv e Sam ple 2 T it r a tio n C urv e Brie f C onclu sio ns Ove ra ll o ur fo rm ula s w e o b ta in ed f o r b oth s a m ple s w ere a nd .F e ( o x ) OK 3 3 • H 2 F e ( o x ) OK 3 2 • H 2 Som e e rro rs th at h ave m ay h ave o ccu rre d w as c o nta m in atio n o f th e c h em ic a ls . In a ddit io n, th e io n e xch ange c o uld n ot h a ve b een s to re d c o rre ctly . H ow eve r, th e m ajo r s o urc e o f e rro r th at occu rre d in o ur la b w as w hen w e w ere c o lle ctin g o ur tit r a tio n c u rv e . W e tit r a te d 1 m L o f N aO H at a tim e b ut s lo w ed d ow n to . 5 m L w hen w e g ot to th e e quiv a le nce p oin t. H ow eve r, th ere w ere still s o m e tim es a s o ne c a n s e e fr o m o ur tit r a tio n c u rv e s s h ow n a bove , e ve n a ddin g .5 m L a t a tim e w as to o m uch . T he p o in t w as n o t c o lle cte d a t th e e xa ct e quiv a le nce p oin t s o th e n um bers eit h er h ad to b e a ve ra ged o r w ere o ff fr o m th e ta rg e t v a lu e.