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!
To answer your questions, let's go step by step:
1. To find the beginning percent composition of the unknown alcohol/water mixture, you need to calculate the ratio of the amount of alcohol to the total amount of the mixture. As you mentioned, 4.0 mL of alcohol was distilled off, and the total mixture was 25.0 mL. Thus, the calculation would be:
4.0 mL / 25.0 mL * 100% = 16% (rounded to the nearest whole number)
So the beginning percent composition of the unknown alcohol in the mixture is approximately 16%.
2. Azeotropes are mixtures that have a constant boiling point, meaning the composition of the vapor and liquid phases remain constant during distillation. To determine if any azeotropes formed in the fractional distillation of the alcohol/water mixture, you need to compare the boiling points of the components.
In this case, the boiling point of ethanol is known to be 78.4 °C, and you found that the boiling point of the unknown alcohol was 74 °C. Since the boiling point of ethanol is higher, it means that the unknown alcohol is not forming an ethanol-water azeotrope.
Therefore, in this specific case, there is no azeotrope formed with ethanol and water.
3. If an azeotrope had formed, you would need additional information to calculate the beginning percent composition of the mixture. The composition of an azeotrope is different from the starting composition of the mixture, so it requires specific data or equations to determine the exact composition.
However, since no azeotrope formed in this case, there is no calculation needed for the beginning percent composition of the mixture. You already determined in question 1 that the beginning percent composition of the unknown alcohol in the mixture is approximately 16%.
I hope this clarifies your questions! Let me know if there's anything else I can assist you with.