# Homework Help: Number of moles = volume/molar mass?

1. Oct 30, 2006

### Tiptronic

Question:
0.2ml glucose solution.
Molar mass of glucose = 180 g/mol

How many moles?

If it was a mass in grams it would be easy:

n = mass/molar mass

But what do I do now its a volume?
Can I still use the same relationship, since grams is equivalent to ml isnt it?
If yes, then what does the unit of 'n' become...ml/mol?

2. Oct 30, 2006

### geoffjb

Grams are not equal to mL. $cm^{3}$ are equal to mL.

Are you given the concentration (measured as M or mol/L)?

3. Oct 30, 2006

### Gokul43201

Staff Emeritus
The question is either incomplete or wrong. Please post the COMPLETE question, EXACTLY as it is given to you.

Correct.

No, you can't.

The number of moles only has one unit : "moles". Anything that is dimensionally different from this can not be a unit for the number of moles.

4. Oct 30, 2006

### Tiptronic

Thanks for the quick replies guys.

Glucose + 3 Phenylhydrazine hydrochloride --> Osazone + Phenylamine + Ammonia + Water

These are what Im using in this reaction:

0.2ml of a Lucozade drink (not diluted, so its 100% concentrated as you'd buy it off the shelf)
0.4g Phenylhydrazine hydrochloride
0.6g crystallized sodium acetate
4ml distilled water.

What i need to do is justify the use of 0.2ml of the glucose drink by using the ratio in the equation (1:3) and by working out the no. of moles of Phenylhydrazine Hydrochloride... n = 0.4/142.5 = 2.8 x 10^23

Now I need to work out the number of moles of the Glucose drink.

Any ideas?

Last edited: Oct 30, 2006
5. Oct 30, 2006

### Tiptronic

Just thought of something:

Could i use this:

n = concentration x volume

Though i doubt "100%" is a concentration that can be used in the above relationship???

6. Oct 30, 2006

### geoffjb

No, 100% is not a concentration (since n = concentration x volume, with n in units of moles and volume in units of L, then concentration must have units of mol/L).

7. Oct 30, 2006

### Tiptronic

So, there is no way I can work out the concentration from the information I have?

So could you tell me how I can justify the use of 0.2ml of the Lucozade Glucose drink?

8. Oct 30, 2006

### Tiptronic

Sorry Im an idiot. http://www.tamscc.org/forum/images/smilies/bang_head.gif [Broken]

The whole point of this investigation was to find the glucose amount: I did an experiment and found that in one bottle (380ml) there is 47g of glucose.

Thus, if 380ml contained 47g, then 0.2ml would contain: (0.2ml/380ml) x 47g = 0.025g

So now I know this:

0.2ml of 100% concentrated glucose containing 0.025g glucose.
Glucose molar mass = 180 g/mol

So can I now find 'n'?

n = mass/Molar mass
n = 0.025/180
n = 0.000139 mol

Please tell me this works now? And does this get me any closer to justifying why Ive used 0.2ml in this Osazone test?

The Glucose is 0.000139 mol and the Phenylhydrazine hydrochloride is 0.0028 mol.

According to the equation the ratio is 1:3.

0.0028/0.000139 = 20 so its not as if there is three times as many molecules in the Phenylhydrazine hydrochloride than the glucose...

Help...

Last edited by a moderator: May 2, 2017
9. Oct 30, 2006

### geoffjb

Get rid of "100% concentrated glucose"; it is wrong (you have a solution of glucose, meaning dissolved solute [glucose] in a solvent [likely water]).

Assuming your molar mass is exactly 180 g/mol, then I get 1.37E-4 (not 1.39E-4, as you get, although it might not matter much in the end, depending on sig figs).

Based on what you said you needed to do, you can take this amount and determine whether or not it is a reasonable amount of glucose needed in the reaction. If it is, you have justified yourself.

Last edited by a moderator: May 2, 2017
10. Oct 30, 2006

### Tiptronic

Thank you for the reply geoffjb.

Would this be a good explanation?

The Glucose is 0.000139 mol and the Phenylhydrazine hydrochloride is 0.0028 mol.

According to the equation the ratio is 1:3.

0.0028/0.000139 = 20 (2sf)

This means that the phenylhydrazine hydrochloride is more than 3 times the amount of glucose (in terms of mols) thus satisfying the ratio of the equation.

Looking at the equation:

Glucose + 3 Phenylhydrazine hydrochloride --> Osazone + Phenylamine + Ammonia + Water

The Phenylhydrazine hydrochloride should be in excess so that all the glucose is turned into the osazone. I have shown that the Phenylhydrazine hydrochloride is definately in excess, satisfying the ratio of the equation, and thus this justifies using a small quantity of glucose drink (0.2ml) for the osazone test.

How does that sound?

11. Oct 30, 2006

### geoffjb

If the theoretical ratio is 1:3 and the actual ratio is 1:20, are you justified in using only 0.2 mL of glucose?

If this is a scientific lab report, don't use "I".

12. Oct 30, 2006

### Tiptronic

Firstly, do I not need to multiply the 0.0028 mol by 3, if the equation says that 3 mols reacts with 1 mol of glucose?

So you're saying 0.2ml is too little? Doesnt 1:20 means that the phenylhydrazine hydrochloride is in excess? Then isnt that ok, since there is enough to react with all of the glucose? I dont quite understand.

13. Oct 30, 2006

### geoffjb

Based on your equation, three moles of phenylhydrazine hydrochloride are required to react with one mole of glucose. Your multiplication factors must account for this.

You're right; it's definitely enough to react with all the glucose. However, since I don't know the original wording of the question, I don't know if such excess is "justified", as you were initially asking. Only you can determine this.

14. Oct 30, 2006

### Tiptronic

Sorry, I didnt understand. So do I multiply the 0.0028 mol of Phenylhydrazine hydrochloride by 3, and then form a ratio from that?

Its part of my write-up of my investigation. So Im ok in saying that such a small amount is justified, as the Phenylhydrazine hydrochloride should be in excess anyway for best results of this Osazone test?

15. Oct 30, 2006

### geoffjb

Yes, in determining limiting reagents, you use a multiplication factor.

Okay.

16. Oct 30, 2006

### Tiptronic

Thanks for that, really appreciate it.

Btw, you wouldnt happen to know of any websites with information on the osazone, why its solid at room temperature and why it precipitates out? Ive looked a lot but to no avail.

Thanks.

17. Oct 30, 2006

### geoffjb

The physical state of osazone at room temperature would be dependent on the intermolecular forces between the molecules.

It would precipitate out simply because it is insoluble in the solvent. Remember that "like dissolves like".

More information on either question requires more in-depth analysis of the molecular structure of osazone.

http://jchemed.chem.wisc.edu/jcesoft/cca/cca5/MAIN/1ORGANIC/ORG18/TRAM18/C/THUMBS.HTM [Broken] might help.

Last edited by a moderator: May 2, 2017
18. Oct 30, 2006

### Tiptronic

Thanks again. I hope I havent taken up too much of your time.

19. Oct 30, 2006

### geoffjb

It was my pleasure.

20. Oct 31, 2006

### Tiptronic

Hi again. Ive stayed up all night and Ive nearly lost the will to live.

Is the science of this ok?

The osazone that is precipitated in this reaction is a solid at room temperature. It has a high boiling point due to its structure and the intermolecular forces acting on it. There are a number of reasons for its high boiling point. Firstly, the size of this osazone molecule is quite big (its molecular mass is 358 gmol-1), which means that it has more electrons and more nuclei that create more Van der Waals forces. Molecular shape is another factor – molecules linear in shape have higher boiling points than their isomers with spherical shapes because the former has a greater contact surface area and thus relatively greater Van der Waals attractive forces.

This osazone is not soluble in water, because “like dissolves like”, and since water is a polar solvent and osazone is a non-polar molecule, the osazone will stay as a precipitate.

Can i add any more? This osazone is not polar is it - so it cant have permanant dipoles? And it doesnt do hydrogen bonding does it?

http://jchemed.chem.wisc.edu/jcesoft/cca/cca5/GRAPHTRAM/TRAM18/18C.GIF [Broken]

Thanks once again.

Last edited by a moderator: May 2, 2017