I would like some feedback on this Inorganic Chemistry assignment (solutions and dilutions)

[Funghiarrosto]

Hi all, I'm a first year biology student at university, and I've only recently started my chemistry classes. I was given an assignment on solutions and dilutions, but I'm unsure about the results I've obtained... I'm not sure if the passages I've done are correct or if this could have been done in a simpler way. Could anyone point me towards the right direction and help me correct my mistakes?

Here is the assignment with my resolution and my thought process:

 

Prepare 500mL of an aqueous solution of ethyl alcohol at 30% (v/v) and determine its molar concentration. Then, from this solution, prepare:

100mL of solution diluted 1:10

100mL of solution diluted 1:100.

MY RESOLUTION: the purity of the alcohol I used is equal to 70%, meaning that in 100mL of product, there are 70mL of ethyl alcohol. With the proportion x mL : 100 = 100mL : 70, I obtained 134mL of ethyl alcohol with 70% purity, which is the equivalent of 100mL of ethyl alcohol with 100% purity.

I then calculated the amount of ethyl alcohol that needs to be added to the solution:

30% of 500mL = 150mL of 100% pure ethyl alcohol.

If 150mL of pure ethyl alcohol are necessary to obtain a solution with the desired concentration (30% v/v), then 1,5 x 143mL of ethyl alcohol pure at 70% will be necessary to obtain an aqueous solution of ethyl alcohol at 30% (v/v).

I then calculated the amount of water necessary to reach 500mL of total solution (I did this because I also need to make this experiment at home, so I know how much water I need to add. Now that I think about it, I probably could have skipped this part and just poured water until the solution reached 500mL):

500mL – 215mL (alcohol) = 285mL of water.

I then calculated the molecular weight of ethyl alcohol C₂H₅OH:

😄 12,011g/mol

H: 1,0079g/mol

O: 15,9994g/mol

Molecular weight: 2(12,001g/mol) + 6(1,0079g/mol) + 15,9994g/mol = 46,046g/mol.

Density of ethyl alcohol with 70% purity: 885,56kg/m3 = 0,88556g/cm3. (found this online)

Mass (g) = Volume (mL) x Density (g/cm3)

Mass = 215mL x 0,88556g/cm3 = 190,40g (total weight of ethyl alcohol in the aqueous solution)

190,40g : 46,068g/mol = 4,1130 mol of ethyl alcohol.

Molar concentration: 4,1150mol : 0,5L = 8,226M. (final answer to the first part of the problem)

To obtain 100mL of solution diluted 1:10, I took 1 part of concentrated solution and added 9 parts of diluent:

(100mL : 10) x 1 = 10mL of initial concentrated solution. I then added water to these 10mL of solution to get to the desired volume of 100mL.

For the second dilution, I took 10mL of the previous 1:10 dilution and added once again diluent to get to 100mL (so I just repeated the process, but on the 1:10 solution).

Is any of this correct? am I at least on the right path? I tried to compare results with some of my classmates, but we all got different answers 😕 

Any form of help would really be greatly appreciated!!

Edited June 5, 2022 by Funghiarrosto

[Bufofrog]

47 minutes ago, Funghiarrosto said:

500mL – 215mL (alcohol) = 285mL of water.

Looks right to me.  The easiest way to calculate it is to use this equation C1 X V1 = C2 X V2.

47 minutes ago, Funghiarrosto said:

Molecular weight: 2(12,001g/mol) + 6(1,0079g/mol) + 15,9994g/mol = 46,046g/mol.

Density of ethyl alcohol with 70% purity: 885,56kg/m3 = 0,88556g/cm3. (found this online)

Mass (g) = Volume (mL) x Density (g/cm3)

Mass = 215mL x 0,88556g/cm3 = 190,40g (total weight of ethyl alcohol in the aqueous solution)

That doesn't seem right.

How about this...

500ml solution X .3 ethyl alcohol = 150 ml ethyl alcohol (100% pure).

 

Edited June 5, 2022 by Bufofrog

[Funghiarrosto]

Thank you for your help! 500ml solution x .3 ethyl alcohol = 150ml alcohol (100% pure) is the correct proportion for pure ethanol, right? The one I'm using is at 70%, not 100% purity, that's why I looked up the density of 70% ethyl alcohol and used it to calculate its mass. How can I apply the operation 500ml x .3 ethyl alcohol = 150ml ethyl alcohol to my case??

[studiot]

1) Molarity

500ml solution contains 150ml ethanol
1000ml solution contains  300ml ethanol = 0.300 litres

[math]{\rm{Number}}\;{\rm{of}}\;{\rm{moles}}\;{\rm{in}}\;{\rm{0}}{\rm{.3L}}\;{\rm{of}}\;{\rm{pure}}\;{\rm{ethanol  =  }}\frac{{{\rm{volume  \times  density}}\;{\rm{pure}}\;{\rm{ethanol}}}}{{{\rm{molecular}}\;{\rm{mass}}\;{\rm{ethanol}}}}{\rm{ = }}\frac{{{\rm{0}}{\rm{.300L \times 789g/L}}}}{{{\rm{46}}{\rm{.069g/mol}}}} = ?M[/math]

 

 

 

But this amount is the amount of ethanol in 1 litre of 30% aqueous solution

 

So the molarity is  ?M

 

2)

3 hours ago, Funghiarrosto said:

Prepare 500mL of an aqueous solution of ethyl alcohol at 30% (v/v) and determine its molar concentration. Then, from this solution, prepare:

100mL of solution diluted 1:10

100mL of solution diluted 1:100.

 

You do realise that diluting a solution 1part to 10 parts will result in a fraction of 1/11  or less than 10% ?

Edited June 5, 2022 by studiot

[Funghiarrosto]

2 hours ago, studiot said:

1) Molarity

500ml solution contains 150ml ethanol
1000ml solution contains  300ml ethanol = 0.300 litres

Numberofmolesin0.3Lofpureethanol=volume×densitypureethanolmolecularmassethanol=0.300L×789g/L46.069g/mol=?M

 

 

 

But this amount is the amount of ethanol in 1 litre of 30% aqueous solution

 

So the molarity is  ?M

 

2)

 

You do realise that diluting a solution 1part to 10 parts will result in a fraction of 1/11  or less than 10% ?

Thank you very much for your answer! I still do not understand why I need to refer to 150ml ethanol though.. the alcohol I'm using is not pure, so why shouldn't I use 215mL of ethanol 70%, which is the equivalent of 150mL at 100% purity?

as for number 2), what I did was taking 1part of solution and added 9 parts diluent, not 10. So 1 part solute + 9 parts solvant = 10 parts in total which is the whole solution. In the end I'll have 1part solute : 10parts solution (1part solute + 9parts solvant). Is it wrong??

[Bufofrog]

2 hours ago, Funghiarrosto said:

Thank you very much for your answer! I still do not understand why I need to refer to 150ml ethanol though.. the alcohol I'm using is not pure, so why shouldn't I use 215mL of ethanol 70%, which is the equivalent of 150mL at 100% purity?

You certainly can, I just think it is easier to find densities of pure substances in general.

2 hours ago, Funghiarrosto said:

as for number 2), what I did was taking 1part of solution and added 9 parts diluent, not 10. So 1 part solute + 9 parts solvant = 10 parts in total which is the whole solution. In the end I'll have 1part solute : 10parts solution (1part solute + 9parts solvant). Is it wrong??

That's right.

C1V1 = C2V2.  

[studiot]

10 hours ago, Funghiarrosto said:

Density of ethyl alcohol with 70% purity: 885,56kg/m3 = 0,88556g/cm3. (found this online)

Mass (g) = Volume (mL) x Density (g/cm3)

Mass = 215mL x 0,88556g/cm3 = 190,40g (total weight of ethyl alcohol in the aqueous solution)

This is where you went wrong and why it is a good idea to convert to equivalent vol of pure ethanol.

 

You have taken the density of 215ml of a liquid (that indeed has the same amount of ethanol as 150ml of pure ethanol) but it also contains 30% water since your liquid is 70% ethanol.

So some of its density ie mass will be due to the water.

So multiplying this density by the volume will not get you the mass of ethanol.

secondly you only made 500ml of solution with this 215 ml of liquid

You need to double this quantity to get moles / litre.

That is why my calc ends up at about 5M whilst yours ends up at about 8M.

You are however right, and even said  you made the volume up with 9 parts of water in your opening post  - I missed that.
I just saw the 1 : 10 which is usually interpreted as I said.

Does this help ?

[Funghiarrosto]

7 hours ago, studiot said:

This is where you went wrong and why it is a good idea to convert to equivalent vol of pure ethanol.

 

You have taken the density of 215ml of a liquid (that indeed has the same amount of ethanol as 150ml of pure ethanol) but it also contains 30% water since your liquid is 70% ethanol.

So some of its density ie mass will be due to the water.

So multiplying this density by the volume will not get you the mass of ethanol.

secondly you only made 500ml of solution with this 215 ml of liquid

You need to double this quantity to get moles / litre.

That is why my calc ends up at about 5M whilst yours ends up at about 8M.

You are however right, and even said  you made the volume up with 9 parts of water in your opening post  - I missed that.
I just saw the 1 : 10 which is usually interpreted as I said.

Does this help ?

It helped massively - thank you. That was such a dumb mystake on my part, I considered those 215mL of ethanol 70% as if they were milliliters of pure alcohol. It's all a lot clearer now.

I rewrote the problem and corrected those mistakes. This should be correct, right?

MY RESOLUTION: the purity of the alcohol I used is equal to 70%, meaning that in 100mL of product, there are 70mL of ethyl alcohol. With the proportion x mL : 100 = 100mL : 70, I obtained 134mL of ethyl alcohol with 70% purity, which is the equivalent of 100mL of ethyl alcohol with 100% purity.

I then calculated the amount of ethyl alcohol that needs to be added to the solution:

30% of 500mL = 150mL of 100% pure ethyl alcohol.

If 150mL of pure ethyl alcohol are necessary to obtain a solution with the desired concentration (30% v/v), then 1,5 x 143mL of ethyl alcohol pure at 70% will be necessary to obtain an aqueous solution of ethyl alcohol at 30% (v/v).

I then calculated the amount of water necessary to reach 500mL of total solution:

500mL – 215mL (alcohol 70%) = 285mL of water.

I then calculated the molecular weight of ethyl alcohol C₂H₅OH:

C : 12,011g/mol

H: 1,0079g/mol

O: 15,9994g/mol

Molecular weight: 2(12,001g/mol) + 6(1,0079g/mol) + 15,9994g/mol = 46,046g/mol.

This is where the correction comes in: By adding 215mL of alcohol 70%, I get the concentration required of alcohol (30% v/v). That is because 70% of 215mL of non-pure ethyl alcohol are equal to 150mL of pure ethanol. So from now on I’ll use the values of pure alcohol.

Density of pure ethyl alcohol: 789kg/m3 = 0,789g/cm3.

Mass (g) = Volume (mL) x Density (g/cm3)

Mass = 150mL x 0,789g/cm3 = 118,35g (total weight of PURE ethyl alcohol in the aqueous solution)

118,35g : 46,068g/mol = 2,5690 mol of ethyl alcohol.

Molar concentration: 2,5690mol : 0,5L = 5,138M.

For the molar concentration, you said I need to double the amounts so I can calculate it over 1L, so it’d be 5,138mol / 1L = 5,138M. The result is still the same, do I need to double the amounts because molarity must be calculated over 1L of solution, or is it still correct if I calculate it with the actual values I got in the problem??

To obtain 100mL of solution diluted 1:10, I took 1 part of concentrated solution and added 9 parts of diluent:

(100mL : 10) x 1 = 10mL of initial concentrated solution. I then added water to these 10mL of solution to get to the desired volume of 100mL.

For the second dilution, I took 10mL of the previous 1:10 dilution and added once again diluent to get to 100mL (so I just repeated the process, but on the 1:10 solution).

[studiot]

7 hours ago, Funghiarrosto said:

For the molar concentration, you said I need to double the amounts so I can calculate it over 1L, so it’d be 5,138mol / 1L = 5,138M. The result is still the same, do I need to double the amounts because molarity must be calculated over 1L of solution, or is it still correct if I calculate it with the actual values I got in the problem??

Well I was replying to your question to bufofrog

On 6/5/2022 at 3:32 PM, Funghiarrosto said:

How can I apply the operation 500ml x .3 ethyl alcohol = 150ml ethyl alcohol to my case??

 

Which your repeated acouple of times.

In fact I did double the volume in my original post to get from 500ml to 1000ml as the easiest way to get to moles/litre.

But sure you could work on actual volumes if you prefer  -  It's just more mistake prone and longer winded.

 

Bufofrog also offered you a very handy formula that can be applied in many cases   That is C1V1 = C2V2

This is much used by pharmacists who do a lot of fiddling about with concentrations like you have been doing.

Here are a couple of pages from a pharmaceutical calculations course that also show what to do when you can't directly apply the formula.

I don't know what you wil be getting up to in Biology, but the most general situation is when you are mixing two different strengths of solution etc.

There is one formula that works in all cases.

[math]\frac{{{\rm{Resultant}}\;{\rm{Concentration - Weaker}}\;{\rm{Concentration}}}}{{{\rm{Stronger}}\;{\rm{Concentration - Resultant}}\;{\rm{Concentration}}}}{\rm{ = }}\frac{{{\rm{Amount}}\;{\rm{of}}\;{\rm{Stronger}}}}{{{\rm{Amount}}\;{\rm{of}}\;{\rm{Weaker}}}}[/math]

This formula works whether you use w/w, w/v, v/v %, P ie all units so long as you are consistent.

 

 

 

 

[Funghiarrosto]

14 minutes ago, studiot said:

Here are a couple of pages from a pharmaceutical calculations course that also show what to do when you can't directly apply the formula.

I don't know what you wil be getting up to in Biology, but the most general situation is when you are mixing two different strengths of solution etc.

There is one formula that works in all cases.

ResultantConcentration−WeakerConcentrationStrongerConcentration−ResultantConcentration=AmountofStrongerAmountofWeaker

This formula works whether you use w/w, w/v, v/v %, P ie all units so long as you are consistent.

 

 

 

 

Thank you very much for these valuable resources - I have a much clearer understanding of this subject now. I'll try to implement that formula in all of my future excercises.

And thank you to both of you for correcting my mistakes! Without your help I'd still be stuck on that problem 

[studiot]

1 minute ago, Funghiarrosto said:

Thank you very much for these valuable resources - I have a much clearer understanding of this subject now. I'll try to implement that formula in all of my future excercises.

And thank you to both of you for correcting my mistakes! Without your help I'd still be stuck on that problem 

Glad to help.

That's a major function of this site.

🙂

Ask again if you need more help.