I'm making different concentrations of

Question

ethanol and I just wanted to check my calculations and method.

Making ethanol concentrations of:

0.40%
0.30%
0.20%
0.10%
0.05%

To make the 0.40% ethanol from stock solution using 100ml volumetric flask:

(xg ethanol)/(100ml) x 100= 0.40%
xg= 0.4g ethanol

Then 0.4g ethanol solution is added to a 100ml volumetric flask with water in it already. After the addition of 0.4g ethanol into the flask whilst weighing it on a balance.

Then I take that 0.40% solution of ethanol and dilute it with
M1V1=M2V2

To get 10ml of 0.30% ethanol solution:
Find out how much of stock is needed to dilute
(0.40%)(V1)=(0.30%)(10ml)
V1= 7.5ml ethanol

add that 7.5ml ethanol into a 10ml volumetric flask and add water to the line to get 0.3% ethanol solution

To get 10ml of 0.20% ethanol solution:
Find out how much of stock is needed to dilute
(0.40%)(V1)=(0.20%)(10ml)
V1=5 ml ethanol

add that 5 ml ethanol into a 10ml volumetric flask and add water to the line to get 0.2% ethanol solution

same process for the below concentrations..just checking my calculations:

0.1% ethanol solution:
(0.40%)(V1)=(0.10%)(10ml)
V1=2.5 ml ethanol

0.05% ethanol solution:
(0.40%)(V1)=(0.05%)(10ml)
V1=1.25 ml ethanol

I'm not sure if I can use percentage for the dilution equation, or if I can even use grams for the creation of the original solution

Thanks very much

Thanks

Answer 1

I vaguely remember us having a similar conversation previously. If you are preparing those ethanol solution to be w/v percent, then you have done it correctly. The ethanol is weighed and the total volume is in 100 mL (for the original solution). That will give you 0.40 weight/volume percent. BUT, since the density of the solution isn't too far from 1 (because it's a fairly dilute solution of 0.4 g ethanol per approximately 99.6 g H2O) it won't miss 0.40% w/w by far. If you definitely wish to prepare 0.40% w/w, you can add 0.4 g ethanol to a tared 100 mL flask, then add water to 100 g total solution. Your calculations look ok to me.

Answer 2

O.O I don't remember...It might have been a more general question on solutions though.

Ah, I was going to ask about the density but I forgot. I had to prepare 10% H2SO4 solution using stock and I was wondering about if I had to convert the grams to ml using density and I was told that it would be fine for that but not so much for ethanol.

I'm not sure why taking ml's of ethanol instead of grams would be less accurate.

Another question is that if I had a solution that I made by adding
0.85018g of K2CrO7 and 40ml of concent H2SO4, then diluting that to 100ml, what solution did I just make? (in terms of moles etc.?)
I'm used to say...0.5% x solution and 2.5M x solution, but I'm not sure how to label this solution. I made it but I just put..0.85018g K2CrO7 in 40ml of sulfuric acid and distilled H2O. (that is very strange to put on a label, I think)
So how would I label a solution with different components?

Thanks Dr.Bob

Answer 3

I wonder how you weighed the 0.85018 g? With conc H2SO4, you almost are forced to use the density if you want to measure a volume for 1 mL does not have a mass of 1 g. And you are right for ethanol, it isn't so necessary since the density of ethanol is about 0.8 or so if I remember correctly. I would label the dichromate/H2SO4 soln as
mols K2Cr2O7/100 mL in xx%H2SO4 or g K2Cr2O7/100 mL in xx%H2SO4.

Answer 4

I weighed the K2Cr2O7 on an analytical balance. (solid)
correction: needed 0.85015g
actually weighed out: 0.8508g
I actually used 0.8508g but I needed 0.85015g, I just hope it doesn't make the results funny...but I hear that it does not need to be accurate.(how it is..I don't know) (told that only the ethanol concentrations in this conway experiment need to be accurate)
And the scale doesn't measure out to that many digits.

We are using a UV/Vis spectrometer to determine the absorbance of the solution. Concentration of the ethanol solution is then determined with the beer lambert equation. A=abc since we find A and a is a constant and the cell length is known..we find c.
[saying all that because I'm not sure if the concentration of the solution will affect the readings]

Thank you

Answer 5

The only reason I asked about the balance is that most analytical balances in use in schools weigh to the nearest 0.1 mg although that may have changed since I retired. There are semi-micro balances that weigh to 5 places and micro-balances that weigh to 6 places but those aren't usually available to beginning students. And PLEASE don't call them scales. We use scales to weigh sacks of potatoes; we use analytical balances to measure mass in analytical chemistry.:-).
I am guessing that you are using the acidified dichromate solution to oxidize the alcohol and measuring the absorbance of the C=O group that is formed. As an added note about the balance, absorbance readings and spectrophotometric work usually is not accurate enough to justify four places for the oxidant (in this case the dichromate). Probably you could have used a triple beam balance just as well. Good luck on your experiment.