A person's blood alcohol (C2H5OH) level can be determined by titrating a sample of blood plasma with a potassium dichromate solution. The balanced equation is

Question

16H⁺(aq) + 2Cr₂O₇^2-(aq) + C₂H₅OH(aq) > 4Cr³⁺(aq) + 2CO₂(g) + 11H₂O(l)

If 35.46 mL of 0.05961 M Cr₂O₇^2- is required to titrate 27.90 g of plasma, what is the mass percent of alcohol in the blood?

got 0.7452 percent, but it is wrong.

Answer 1

Hello Dawson,

Let me help you out on this problem.

We are going to start out with what we know. We know we have 35.46mL used of a 0.05961M (mol/L) Cr2O7

What are we trying to look for? How many grams of C₂H₅OH(aq)

So now we start some FUN math!

35.46mL always change to L = 34.46 * 1L/1000mL = 0.03546L

0.05961mol/L of Cr2O7 * 0.03546L = 0.002114 mol of Cr2O7

Now we have how many moles were used of Cr2O7 but we are asked about C₂H₅OH(aq). So we need to use the balanced equation. 1 mol of C₂H₅OH(aq) is equal to how many Cr2O7? Answer:2. So now we get:

0.002114 mol Cr2O7 * 1 mol C₂H₅OH(aq) / 2 mol Cr2O7 = 0.001057 mol C₂H₅OH(aq)

Now how do we get from moles to grams? We use the molar mass of the compound. So for C₂H₅OH(aq) we get 46grams

0.001057mol C₂H₅OH(aq) * 46 g C₂H₅OH(aq)/ 1mol C₂H₅OH(aq) = 0.04862 gC₂H₅OH(aq)

Now to get the percentage of mass we need to use the formula (mass it took to titrate / mass of plasma)*100

(0.04862 g C₂H₅OH(aq) / 27.90g ) * 100% = 0.1743%

Please let me know if you need me to explain it in another way!