I did a lab for Organic Chemistry
We did a simple distillation of an unknown alcohol to determine the boiling point, which was found to be 74*C
The four possibilities for the unknown were methanol, ethanol, 1-propanol and 2-propanol. So from the boiling point the unkown must be ethanol...
We then did a fractional distillation of this alcohol mixed with water. The alcohol was distilled off after 4.0 mL (25.0 mL of mixture was used).
To find:
1.
beginning percent composition of unknown/H20
would this be :
4.0/25.0*100%
2.
Does it form any azeotrops, list them:
does this mean binary with water etc., ternary...? or am I supposed to write some other description?
3.
If it did form an azeotrops Calculate the beginning % composition of the mixture
confused about this...
Any help would be greatly appreciated!
I skipped your earlier post. I don't understand the question. For example, "The alcohol was distilled off after 4 mL." Does that mean that all of the alcohol was gone after 4 mL or does it mean that 4 mL of the mixture was distilled BEFORE any alcohol began to appear? I'm confused by that and the question then asks about the beginning percent composition.
With regard to azeotropes, yes, ethanol/water form an azeotrope. It is very close to 95%alcohol/5% water.(see
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1. To find the beginning percent composition of the unknown/water mixture, you can use the equation:
For Unknown: X (percent composition) * V (volume) = Unknown (mass)
For Water: Y (percent composition) * V (volume) = Water (mass)
Since we know that 4.0 mL of alcohol was distilled off, the remaining volume in the distillation flask is 25.0 mL - 4.0 mL = 21.0 mL.
Assuming that the volume of alcohol and water do not change significantly during distillation, the composition of the mixture can be calculated as follows:
Unknown: X * 21.0 mL = 4.0 mL (unknown mass)
Water: Y * 21.0 mL = 21.0 mL - 4.0 mL (water mass)
Simplifying the equations, we get:
X * 21.0 = 4.0
Y * 21.0 = 21.0 - 4.0
Solving for X:
X = 4.0 / 21.0 * 100%
2. An azeotrope is a mixture of two or more substances that forms a constant boiling point. In this case, an azeotrope would be formed if the mixture of ethanol and water had a boiling point that was different from the boiling point of pure ethanol or pure water.
In order to determine if the mixture forms an azeotrope, you would need to compare the boiling point of the mixture to the boiling points of pure ethanol and pure water. If the boiling point of the mixture is different, then it does not form an azeotrope. If the boiling point of the mixture is the same as either pure ethanol or pure water, then it does form an azeotrope.
3. If the mixture forms an azeotrope, you would calculate the beginning percent composition of the mixture by using a similar approach as mentioned in question 1.
Assuming the volume of alcohol and water do not change significantly during distillation, you can use the equation:
Unknown: X (percent composition) * V (volume) = Unknown (mass)
Water: Y (percent composition) * V (volume) = Water (mass)
Where the volume V is the initial volume of the mixture (25.0 mL).
You would then solve for X and Y to find the percent composition of the mixture at the beginning.
1. To find the beginning percent composition of the unknown/water mixture, you would divide the volume of alcohol (4.0 mL) by the total volume of the mixture (25.0 mL) and then multiply by 100%. So the calculation would be:
(4.0 mL / 25.0 mL) * 100% = 16%
The beginning percent composition of the unknown alcohol in the mixture is 16%.
2. An azeotrope is a mixture of two or more substances that has a constant boiling point and composition. In this case, you need to determine if the mixture of ethanol and water forms an azeotrope. To do this, you would compare the boiling point of the mixture to the boiling points of pure ethanol and pure water.
The boiling point of pure ethanol is 78.4°C, and the boiling point of pure water is 100.0°C. If the boiling point of the mixture is different from the boiling points of ethanol and water, then it does not form an azeotrope. However, if the boiling point of the mixture is the same as the boiling points of ethanol and water, then it does form an azeotrope.
Based on the information given, since the boiling point of the unknown alcohol was found to be 74°C, which is lower than the boiling point of pure ethanol, it suggests that the alcohol is ethanol. Since the boiling point of pure ethanol is lower than that of water, it indicates that the mixture does not form an azeotrope.
3. Since the mixture does not form an azeotrope, you do not need to calculate the percent composition of the mixture at the beginning.