Transcript
Student name: Student number:
Date: Oct 9, 2012
CHY224 LAB#3 QUESTIONS
DISTILLATION OF ORGANIC LIQUIDS
1. On the shelf in your lab, you find a bottle with a hand-written label: “pure ethyl acetate” and you measure its refractive index. The temperature in the lab is 28 °C (it’s a sweltering summer day, and the air conditioning is not working well). You measure a value of 1.3650. Does this measurement support the statement on the bottle’s label?
When the temperature increases the index of refraction (nD) and the density will decrease because the refraction index is inversely proportion with the temperature. However, the measurement had support the statement on the bottle’s label because the refraction index of pure ethyl acetate at 28oC has decreased to 1.3650 with increasing the temperature (from 23.5oC room temperature to 28oC room temperature at hot summer day).
Calculations:
nD28=nD23.5+ [0.00045oC-1x (23.5-28oC)] * (nD) of pure ethyl acetate is 1.3720
= nD23.5 -0.002025
=1.3720-0.002025
= 1.369975
2. The 91st edition of the CRC Handbook of Chemistry and Physics (available electronically through the library) provides a section with tables of data on some known azeotropes (it starts on section 6, page 155.) Use this table (and other legitimate sources) to answer the following:
a) What are the boiling points of pure ethanol and pure benzene (Give all answers in °C)?
pure ethanol b.p is 78.24oC
pure benzene b.p is 80.08oC
b) If a mixture of 50 mL of ethanol and 50 mL of benzene are distilled through a perfect fractionating column, what temperature will the first vapour passing through the top of the fractionating column be? What is the mol % composition of the first fraction of distillate?
The temperature for the first vapor passing by the top of the fractionation column will be the same as the b.p of ethanol which is 78.3oC. Because ethanol has low b.p than benzene and ethanol vapor pressure will reach 1 atm at 78.3oC.
Calculation: Determining the mol% composition of the first fraction of distillate
Substance
Molecular formula
Molecular weight(g/mol)
Density(g/cm3)
Volume (mL)
ethanol
C2H6O
46.07
0.7890
50
benzene
C6H6
78.12
0.8765
50
The mass and the mole of the ethanol were calculated using the density formula:
p=m/V
m ethanol =pV m C2H6O = 0.7890 g/cm3 * 50 mL * 1 cm3 /1mL
m C2H6O = 39.45g
n C2H6O =m C2H6O/M.W C2H6O
= 39.45g/46.07g/mole
=0.8563 mol
The mass and the mole of the benzene were computed as shown below:
p=m/V
m benzene =pV m C6H6= 0.8765 g/cm3 * 50 mL * 1 cm3 /1mL
m C6H6 = 43.83g
n C6H6 =m C6H6/M.W C6H
= 43.83g/78.12g/mole
=0.5611 mol
The mol% of the first fraction is
n total = n C2H6O + n C6H6
= 0.8563 mol + 0.5611 mol
= 1.4174 mol
x C2H6O = n C2H6O /n total
= (0.8563 mol /1.4174 mol) x 100
= 60.4%
c) Convert the mol % composition of the first fraction into volume % composition. (For the purpose of this experiment, use the approximation that the volume of a mixture of liquids is equal to the sum of the two volumes mixed – this will not always be true).
d) What is the total volume of the first fraction to distill? (Again, we are assuming perfect separation through a fractionating column, and the approximation about the volume of mixed liquids mentioned in part c)
e) After the first fraction has completed distilling, what temperature will be reached for the next fraction to boil? What is the mol % composition of the second fraction, and what is its volume?
The temperature of the next fraction will be reached at the same temperature as the boiling point of the benzene which is 80.08oC. The mol% composition of the second fraction will be calculated as shown below:
m ethanol =pV m= 0.7890 g/cm3 * 50 mL * 1 cm3 /1mL
m C2H6O = 39.45g
n C2H6O = m C2H6O/M.W C2H6O
= 39.45g/46.07g/mole
= 0.8563 mol
The mass and the mole of the benzene were computed as shown below:
p=m/V
m benzene =pV m C6H6= 0.8765 g/cm3 * 50 mL * 1 cm3 /1mL
m C6H6 = 43.83g
n C6H6 =m C6H6/M.W C6H6
= 43.83g/78.12g/mole
=0.5611 mol
So the mol% of the first fraction is
n total = n C2H6O + n C6H6
= 0.8563 mol + 0.5611 mol
= 1.4174 mol
x C6H6 = n C6H6 /n total
= (0.5611 mol /1.4174 mol) x 100
= 39.6%
The volume of the next fraction calculated as shown below:
3. Again, consult the table of azeotropes, this time with respect to the three liquids that were in the mixture you distilled in this experiment.
a) Do any two of the three liquids form a known binary azeotrope with each other? (Note: take this to mean an azeotrope listed on the table in the CRC – this table is incomplete, as there are thousands more known binary azeotropes, and azeotropes with more than two components are also possible).
b) If any binary azeotropes are expected from the mixture in this experiment: for each expected azeotrope, go through the exercise described in question 2 above, but using
the real amounts from the mixture. (I.e. what are the predicted boiling point, volume % composition of the fraction, volume of the fraction, and volume and composition of liquids remaining behind when the azeotrope has completely distilled?) – If no binary azeotropes are expected, you do not need to answer this part of the question
4. Write one clear paragraph describing the success (or lack thereof) of your distillation. Based on all available experimental data (fraction volumes, boiling points, and corrected refractive indices), was your distillation successful in isolating the three pure component liquids from the mixture? If not, was it at least as successful as theoretically possible?
Distillation is a method that has been used for many centuries to purifying or separating mixture liquids based on the differences in boiling point of the mixture. However, in this lab, the simple distillation apparatus was set up as demonstrated in appendix C4 in order to separate the component of the mixtures of three organic liquids (ethyl acetate, 2-methoxyethanol and methanol) Also, to determine the purify of the three substances by measuring its refraction index using The Abbé Refractometer apparatus as its depicted and explained in appendix B2. However, the distillation process started after obtaining 80 mL of the mixture that contain percentage by volume each of (45%ethyl acetate, 28%2-methoxyethanol, 27%methanol). Thus, the distillation apparatus heated at temperature of 65o C in order to collect the fraction since the b.p of the first liquid mixture (methanol) start at65o C till the b.p of the 2-methoxyethanol which 125 o C . The refraction indexes of multiple fractions (approximately seven fractions) were collected with the temperature range of (5-10 o C) was measured. As a result, the distillation was successful in isolating the three pure components liquids from the mixture for many reasons. Firstly, the three organic liquid compounds were separated or purified based on their boiling points as illustrated in the lab manual. For example, methanol has the lowest boiling point (65o C) in the mixture so it was the first fraction that was collected with refraction index of 1.361 at a volume of 12.5mL. After the temperature of the thermometer reached the value 77oC, Ethyl acetate was collected as the second fraction and since 2-methoxy ethanol had the higher b.p which 125oCso it was collected lastly as the third fraction with refraction index of 1.403. Secondly, the temperature changes was observed when the first fraction was completely distill the temperature between the first fraction and second was dropped down ( because there was no hot gas pass through tip of the thermometer) and then raising to the second fraction boiling point therefore the data that obtained in the experiments confirmed the correction of distillation process.
References:
The Online 92st edition of the CRC Handbook of Chemistry and Physics, 2011-2012 http://www.hbcpnetbase.com.ezproxy.lib.ryerson.ca/
CHY 224 LABORATORY MANUAL APPENDICES, Appendix B2 and C4.
