Need help with 2a and 2b!!! LABORATORY REPORT FORM: 6. CSI Buies Creek NAME: Sec

Question

Need help with 2a and 2b!!! LABORATORY REPORT FORM: 6. CSI Buies Creek NAME: Section Lab Partner: roue UNKNOWN #: H o 00 1. Look up the densities of ethyl acetate and water from a reliable source and record below WITH UNITS: density of ethyl acetate: 09024/mL density of water DMt4lmL C a6t) 2. Extractions rely on differences in solubility of the components that are being separatea The amount of a material that will transfer into the solvent in an extraction can be calculated using a partition coefficient. This is a ratio of the solubility of the material in each solvent that can be used to predict outcomes of extractions. A certain organic compound, Z, has the distribution coefficient (partition coefficient), Ko 10.45 when partitioned between ether and water. Therefore, the equation used to determine the separation is: conC-wet;"Wt.in k,=conc.inwater Vol.ofwase vol mL of water are extracted with 50.00 mL of ether. SHOW CALCULATIONS. [HINT: I then (4.23-x) wt. in water.] t. in ether, a. Use the equation above to calculate the weight and percent of Z recovered if 4.23 g of Z in 100.00

Explanation / Answer

2a) kp=10.45=(Wt in ether/Volume of ether)/(Wt in water/Volume of water)

If x=wt in ether,4.23-x=wt in water

then,10.45=(x/50ml)/((4.23g-x)/100ml)

or,(4.23g-x)/100ml)*10.45=x/50ml

or,44.203g-10.45x=2x

or.12.45x=44.203g

x=44.203g/12.45=3.550

Thus,x=3.550g goes into ether and can be extracted from the organic solvent,by evaporating the solvent.

percent Z recovered=(mass recovered/total mass)*100=(3.55g/4.23g)*100=83.936%

grams of Z=3.550g,percent of Z=83.936%

b) first extraction

If x=wt in ether,4.23-x=wt in water

then,10.45=(x/25ml)/((4.23g-x)/100ml)

or,(4.23g-x)/100ml)*10.45=x/25ml

or,44.203g-10.45x=4x

or.14.45x=44.203g

x=44.203g/14.45=3.059

Thus,x=3.059g goes into ether and can be extracted from the organic solvent,

grams of Z1=3.059g

percent of Z1=(3.059g/4.23g)*100=72.317%

Thus remaining Z, in water=4.23g-3.059g=1.171g

Thus ,again 25ml ether is added for second step extraction,

If wt in ether=y

wt in water=1.171g-y

10.45=(y/25ml)/((1.171g-x)/100ml)

or,(1.171g-y)/100ml)*10.45=y/25ml

or,12.237g-10.45y=4y

or.14.45y=12.237g

y=12.23g/14.45=0.847g

Thus,y=0.847 g goes into ether and can be extracted from the organic solvent,

grams of Z2=0.847g

percent of Z2=(0.847g/1.171g)*100=72.331%

total grams of Z=0.847g+3.059g=3.906g

total Z percent recovered=(3.906/4.23)*100=92.340 %

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