. 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)
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%
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