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Abu Zakariyah

preparing and purifiying 1 bromobutane

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:D

im trying to make 1 bromobutane from butanol, i know how to extract it and make it etc

but i cant seem to work out how much of the alcohol i need to use,

 

i need to be able to calculate the volume of butanol to be used to prepare 5g of 1 bromobutane, the yeild will be 85 percent, and 8g of NaBr and 10cm of conc H2SO4 are an excess,

 

i just need som1 to help me work out how to find the volume of C4H90H (butanol)

:P

:D

Edited by slave

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PropellerAds

selam,

 

 

10 cm is cm cube ?

 

and what is abd

and do you have equation or should i find it?

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yes its cubed mainly dealing with liquid here,

 

this is the main reaction to

 

C4H90H + HBr ------> C4H9Br + H20

 

this is part of the extraction:

 

NABr + H2SO4 -----> NAHS04 + HBR

 

The reactions are:

(1) Sodium Bromide reacts with Sulphuric Acid to form Hydrogen Bromide and Sodium Hydrogen Sulphate.

NaBr + H2SO4 > HBr + NaHSO4

(2) Hydrogen Bromide is oxidised to Bromine molecules as concentrated Sulphuric Acid is a very good oxidising agent. The Sulphuric Acid reacts to form Sulphur Dioxide gas.

HBr + H2SO4 > Br2 + 2SO2 (g)

(3) Hydrogen Bromide dissociates and the Bromide ion from it attacks the Carbon atom with the -OH function group in Butan-1-ol and displaces the -OH function group forming a Bromo function group and a hydroxide ion, which then associates itself with another H+ ion to form water.

CH3CH2CH2CH2OH + Br- > CH3CH2CH2CH2Br + OH-

(4) A molecule of Sulphuric Acid attacks the lone pair on an -OH function group, releasing a molecule of water, and a mixture of Butoxybutane and But-1-ene is formed, along with the regenerated Sulphuric Acid.

CH3CH2CH2CH2OH + H2SO4 > CH3CH2CH=CH2 + H2O + H2SO4

or

2 CH3CH2CH2CH2OH + H2SO4 > CH3(CH2)3O(CH2)3CH3 + H2O + H2SO4

 

(obviously the wanted substance will be separated)

density of bromobutane is 1.3

density of buntan 1 ol is 0.81

Edited by slave

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lol abd was a typo,

 

the main question is

 

>calculate the volume of butan-1-ol to be used to prepare 5g of 1 bromobutane assuming a yeild of 85 percent for the prepatartation and that other reagents are in excess (assuming that 8g of sodiam bromide and 10cm3 of cocn sulphuric acid are an excess and maby be used in your prepartion)

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i figured this,

 

since 1 mole of butanol gives us 1 mole of bromobutane we can work it out easily,

 

butanol=74g and bromobutane give is 137g

 

since we need 5g

 

we do this

 

74/137 x 5 = 2.7g of butanol.

 

the density of butanol is 0.81

 

i dont no the forumala to to find volume using density and mass :D ????

Edited by slave

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5gr %85 C4H9Br this became 5.89 gr by mass...

 

if we have 5.89 gr

 

74 / 137 * 5.89 = 3.2

 

(d) density = m/v (gr/mlt)

 

0.81 = 3,2/v

 

v=3.9mlt

 

i didnt understand, y the density of bromobutane is given , the answer seems like this i hope i didnt skip any part

btw something wrong w/ the yield part usually does not give the yields of product but reactant of the rxn i solved w/ respect to the product i hope thats true..

Edited by elif74

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:D

brother/sister

i dont understand how you got 5.89 the 5g is the mass of bromobutane not the alchohol,

 

how comes u didnt use 2.7 either,

 

this is what i have done now,

 

forumala: %yeild = actul yeild in grams/ theoretical yeild in gram X 100%

 

(85 X 2.7)/100= 2.3g

 

 

(d) density = m/v (gr/lt)

are you sure your forumla is correct units? beacuase were dealing in test tube levels LT is faar to much. ml OR CM3 seems more applicable.

well using the density forumla i got

 

0.81 X 2.3 = 1.9cm3

which sounds just about right for a test tube level experiment.

what do you think??

jazakallah khair for your outstanding help! :D

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selam,

 

your sis in Islam

you may use whatever you want i mean lt or ml or cm3 .thats depends on your usage...and it should be given that 0.81 s unit..

but i am sure its ml here... :D

Edited by elif74

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ok whole answer...

 

NaBr + H2SO4 ---> NaHSO4 + HBr

8gr 10cm3

 

 

C4H9OH + HBr ---> C4H9Br + H2O (this is our main eqxn)

0,81 gr/ml 1,3gr/ml

V=? 5 gr

74 gr/mole 137 gr/mole

 

it is not necessary to use 5,89

lets do via your way

74/ 137 * 5 = 2,7 gr C4H9OH ok!

 

2,7 gr is in use, thats what i understand correct me if i am wrong..

if we have 2,7 gr in USE, and this is %85 we should have more than 2,7 gr ok?

i mean we have some gr. of %85 should be 2,7. if this some gr. is "x" the eqxn will be :

 

x * % 85 = 2,7 gr

x = 3,2 same conclusion with former one....

i dont explain 2u the 5,89 bec. may confuse you....

 

and your last eqxn

0,81 = 3,2 /v

v= 3,2/0,81 whatever ml

 

i hope this makes you clear :D

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:D EID MUBARAK 2 you,

 

i think your geting the yeild part wrong,

yes it would be less then 2.7 beacuase the yeild is 85%

 

a freind of mine got a similar answer to the 1 i worked out, but the density forumla is not the same,

you told me to do density X the mass,

but this person did MASS/DENSITY,

can you check this out and see if its correct or was my answer of 1.9 correct,

 

this is what she did:

 

Calculations:

 

C4H9OH + HBr → C4H9Br + H2O

 

Molar mass of C4H9OH = 74g.

Molar mass of C4H9Br = 137g.

Density of C4H9OH = 0.81g/cm3.

 

VOLUME OF BUTAN-1-OL TO BE USED:

1 mole of C4H9Br = 1 mole of C4H9OH

137g of C4H9Br = 74g of C4H9OH

5g of C4H9Br = 5 x 74/137= 2.7g of C4H9OH.

 

At 85% yield, mass of C4H9OH = 2.7 x 85/100= 2.296g of C4H9OH.

Density = Mass/Volume

 

Volume = Mass/Density

 

Volume of C4H9OH = 2.296g/0.81 g /cm3 = 2.83cm3

 

which 1 is right 1.9 or 2.8 :D

Edited by slave

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bro , i didnt understand tell me where i told density X the mass

 

you write Density = Mass/Volume Volume = Mass/Density

this is true thats what i said before...i looked at it again again ya this is what i did say???

 

i may be a bit confused at yield part bec. i didnt understand exactly that part but i;ll check again inshaAllah...

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what i am confused about yield part is bec. of this:

 

C4H9OH + HBr --> C4H9Br + H2O

2.7g C4H9OH , C4H9Br gr 5g

 

to have 5gr bromobutane we should have 2,7 gr butanol , butanol cant be less than 2,7. if it is we cant produce 5gr bromobutane but less than it...if something is not hundred percent, to reach 2,7 gr we should take more than it..how can you explain this to me.....

Edited by elif74

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:D

(85 X 2.7)/100= 2.3g

(d) density = m/v (gr/lt)

are you sure your forumla is correct units? beacuase were dealing in test tube levels LT is faar to much. ml OR CM3 seems more applicable.

well using the density forumla i got

 

0.81 X 2.3 = 1.9cm3

which sounds just about right for a test tube level experiment.

what do you think??

jazakallah khair for your outstanding help!  :D

 

ok i got it, you did wrong not me, bro...

you write d = m / v and than you did

0,81 * 2,3 = 1,9 (d*m=v this is not true..)

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nasel sen...

ok i got it, you did wrong not me, bro...

you write d = m / v and than you did

0,81 * 2,3 = 1,9 (d*m=v this is not true..)

yep i;m so silly i arragned the forumal wrong, I didnt mean you told me to do density X mass,

i arranged the forumala in the wrong way,

 

its ment to be VOLUME= MASS/ DENSITY

 

teshekur

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what  i am confused about yield part is bec. of this:

 

C4H9OH + HBr --> C4H9Br + H2O

2.7g C4H9OH , C4H9Br gr  5g

 

to have 5gr bromobutane we should have 2,7 gr butanol , butanol cant be less than 2,7. if it is we cant produce 5gr bromobutane but less than it...if something is not hundred percent,  to reach 2,7 gr we should take more than it..how can you explain this to me.....

:D

honstly i have no idea... all i did is use the percentage yield forumla :

%yeild = actul yeild in grams/ theoretical yeild in gram X 100%

:D

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:D

 

just to discuss , not to offend you bro....

 

ok let me arrange the yield formula:

 

% yield = actual gram / theo. gram * 100

 

85 = 2,7 / theo. gram * 100

 

theo gr. = 2,7 * 100 / 85

 

theo gr. = 3,2 gr this is what i found in any ways....

 

i just arranged the formula actual gr means the gram that we have,

theoretical means the gram that we should have .this is what i understood...

can you find the definition of the formula, or do you have any solved questions...

now it is hard for me to search through my old books...

 

btw it is mübarek to you...

 

and iyiyim teshekkür...

 

learning chem or turkish B)

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:D

% yield = actual gram / theo. gram * 100

 

85 = 2,7 / theo. gram * 100

 

theo gr. = 2,7 * 100 / 85

 

theo gr. = 3,2 gr this is what i found in any ways....

 

that isnt what i did,

 

i did 2.7 X 85 / 100

 

can you find the definition of the formula, or do you have any solved questions...

"you can't post links until you reach 50 posts_www.iun.edu/~cpanhd/C101webnotes/quantchem/thtclandpctyld.html"]examply of yeild calculation[/url]

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ok, i looked at one more example either...these are just to confirm me...

first, let me explain what the yield is....

 

we usually solve the problems theoretically..i mean if we have one mole butanol , the product will be one mole bromobutane.....

 

one mole reactant --->one mole product....this is what you usually do....

but many times we cant get one mole product i mean not one mole bromobutane but less than this....

 

this can be thought for mass like mole , ok?

 

i mean one mole bromobutane such gr....we expect such gram theoretically but have just less than that this will be the ACTUAL gr. ok?

 

for example: for one mole of bromobutane we expect 137 gr. bromobutane as a product, but we cant have , we will have less than this gr....

 

the yield is always less than %100, bec. we expect the product -not reactant- theoretically %100, but cant have bec. of the media of the rxn....(u may c them the link you gave)

 

this is important we should look at the product not reactant while the yield is subject

 

 

ok lets take our example:

 

% yield = actual gr (thats we have 5gr. of product)/theoretical gr. of product (thats we expect to have but not) *100

 

%85 = 5gr / theo. gr * 100

 

theoretically we should have 5,89 gr. but we have just 5gr...

 

5,89 is how much mole (for bromobutane)

5,89 / 137 = 0,04 mole this is theoretically what we should have if this is the case we will have 0,04 mole of butanol .....i mean

 

C4H9OH + HBr ---> C4H9Br + H2O

 

if we have 0,04 mole bromobutane thats what we reach theoretically....and we should have 0,04 mole of butanol either...

how many gr. of butanol we have if we have 0,04 mole ;

0,04 mole * 74 = 3,2 (if you take just 0,04 you conclude 2,96 but if you take whole numbers after coma you reach 3,2)

 

d = m/v

 

0,81 gr/ml = 3,2 gr / v

v= almost 4 ml

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:D  EID MUBARAK 2 you,

 

i think your geting the yeild part wrong,

yes it would be less then 2.7 beacuase the yeild is 85%

 

Calculations:

 

    C4H9OH + HBr ? C4H9Br + H2O

 

        Molar mass of C4H9OH = 74g.

      Molar mass of C4H9Br = 137g.

      Density of C4H9OH = 0.81g/cm3.

 

VOLUME OF BUTAN-1-OL TO BE USED:

    1 mole of C4H9Br    = 1 mole of C4H9OH

    137g of C4H9Br    =        74g of C4H9OH

          5g of C4H9Br  =    5 x 74/137= 2.7g of C4H9OH.

 

At 85% yield, mass of C4H9OH = 2.7 x 85/100= 2.296g of C4H9OH.                       

Density = Mass/Volume                                                                                                       

 

Volume = Mass/Density                                                                                                       

 

Volume of C4H9OH = 2.296g/0.81 g /cm3 =  2.83cm3

 

which 1 is right 1.9 or 2.8  :D

 

 

2,7 is true but not yield part, i think i made it clear to you the yield part.....

 

i dont know, are you going to uni or college, but if i am wrong as a chemistry teacher, i should hide myself :D :P :P

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aslam wa alikum,

the answer does look correct, the yeild calculation thats how you do it, i;m studying chemistry at collage, ooo now that i know your a chemistry teacher expect more quetions from me :D.

jazakallah khair.

i need some help on calculations involving equilbiram. i need to study further into it before i ask for questions dont want to waste any1s time.

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