Answer to Question #269720 in General Chemistry for nedo

Question

. A person’s blood alcohol (C2H5OH) level can be determined by titrating a sample of blood plasma with a potassium dichromate solution (reaction below). If 35.46 mL of 0.05961 M Cr2O7 2- is required to titrate 28.00 g of plasma, what is the mass percent of alcohol in the blood? Cr2O7 2− (aq) + C2H5OH (aq) → Cr3+ (aq) + CO2 (g) (Not balanced, acidic conditions)

Answer 1

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%