Glucose anaylsis of sports drinks

[Tiptronic]

Just realized wrong forum...sorry.

I need some urgent help with this please!

This is referring to my glucose analysis coursework, using Quantitative Benedicts solution (with potassium thiocyanate) and titrations to find out how much glucose is present in Lucozade and Powerade.

I was told to use the fact that my Quantitative Benedicts Solution contained 18g of CuSO4.5H20, copper sulphate-5-water, in order to justify the percentage concentration of the Lucozade drink I used during the titration.

So, this is what i did, but i think it is most probably wrong.

I found the Molar mass of CuSO4.5H20 = 279 g mol-1

number of moles = 18/249 = 0.072 mol

In order to callibrate the Quantitative Benedicts Solution, I first used 0.5g glucose disolved in 100cm3 water and did a titration with that.

So, the molar mass for glucose is 180 g mol-1

I used 0.5g here, so:

number of moles = 0.5/180 = 0.0028 mol

I now have two values, 0.072 and 0.0028 mol.

Through wishful thinking i did this:

0.072/0.0028 = 25.71

25.71 x 0.5 = 12.9

So I then somehow would say that justifies using 12.9% concentrations for my glucose drinks that I am testing (Lucozade and Powerade)?

Help please?

 

[nate808]

Hello,

First of all, don't worry about posting in the wrong forum. We are here to help you with your coursework.

Your calculation for the molar mass of CuSO4.5H20 is correct, but there are some errors in your subsequent calculations. Let's break it down step by step:

1. Finding the number of moles of CuSO4.5H20 in 18g:
Number of moles = mass/molar mass
= 18/249 = 0.072 mol

2. Finding the number of moles of glucose in 0.5g:
Number of moles = mass/molar mass
= 0.5/180 = 0.0028 mol

3. Now, to find the concentration of the glucose solution, we need to use the equation:
M1V1 = M2V2
where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume.

In this case, M1 is the concentration of the CuSO4.5H20 solution, which we know is 0.072 mol. V1 is the volume of the CuSO4.5H20 solution used for the titration, which you haven't mentioned. Let's assume it is 25 mL (0.025 L). M2 is the concentration of the glucose solution, which is what we are trying to find. V2 is the volume of the glucose solution used for the titration, which again, you haven't mentioned. Let's assume it is 5 mL (0.005 L).

Substituting these values into the equation, we get:
(0.072 mol)(0.025 L) = (M2)(0.005 L)
M2 = (0.072 mol)(0.025 L)/(0.005 L)
M2 = 0.36 mol/L

So, the concentration of the glucose solution is 0.36 mol/L or 36%. This means that your glucose solution contains 36 g of glucose in 1 L of solution.

I hope this helps you with your coursework. Let me know if you have any further questions or need clarification on any steps. Good luck!

 

[Blueshift5]

I would first like to commend you for your efforts in conducting a glucose analysis of sports drinks. However, I would like to point out that this is not the appropriate forum for seeking urgent help with coursework. It is important to seek guidance from your teacher or a qualified expert in the field to ensure accurate results and understanding of the concepts.

That being said, I can offer some guidance on your approach to justifying the percentage concentration of your glucose drinks. It is important to note that the concentration of a substance is typically expressed in terms of mass per unit volume, not in moles. Therefore, your calculation using moles is not applicable in this case.

Instead, you can use the fact that the Quantitative Benedicts Solution contains 18g of CuSO4.5H20 in 100cm3 of solution. This means that the concentration of the solution is 18g/100cm3 or 0.18g/cm3. You can then use this value to calculate the concentration of glucose in your sports drinks.

For example, if you performed a titration using 10cm3 of the Quantitative Benedicts Solution and found that it required 20cm3 of the diluted glucose solution to reach the endpoint, you can calculate the concentration of glucose as follows:

0.18g/cm3 x 10cm3 = 1.8g of CuSO4.5H20 used
1.8g x (1 mol CuSO4.5H20/279g) = 0.00645 mol of CuSO4.5H20 used
0.00645 mol x (1 mol glucose/1 mol CuSO4.5H20) = 0.00645 mol of glucose in 20cm3 of solution
0.00645 mol x (180g/mol) = 1.161g of glucose in 20cm3 of solution
1.161g/20cm3 = 0.05805g/cm3 or 5.805% concentration of glucose in the diluted solution.

I hope this helps you to better understand the process and to justify the concentration of your sports drinks. Remember to always seek guidance from your teacher or a qualified expert to ensure accurate results and understanding. Good luck with your coursework!