Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: SPASH on September 05, 2005, 10:00:23 AM
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You are working in a forensic lab and are given a blood sample from a suspected drunk driver and asked to determine the person's blood alcohol level. The legal limit is 0.1% by mass in most states. You use the following reaction in your determination, assuming only the ethanol (C2H5OH) in the blood sample will react with the dichromate.
16H+(aq) + 2Cr2O72-(aq) + C2H5OH(l)4Cr3+(aq) + 11H2O(l) + 2CO2(g)
If 14.6 mL of 0.050 M K2Cr2O7 are needed to react completely with 28 g of blood, what is the blood alcohol level of the driver?
1) 0.23%
2) 0.07%
3) 0.03%
4) 0.12%
5) 0.034%
I used the volume of concentration and volume of dichromate ion to find the number of moles of dichromate, and then used the stoichometric factor to find moles of ethanol. Then i ued the molecular weight of ethanol to find the mass of ethanol. THen I divded by the mass of blood and multiplied by 100 to get the weight percent.
Doing so gave me an answer of .12%. However, the correct answer is .07%. Can anyone explain to me why this is and how they arrived at the correct answer?
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Hello I am not sure what you actually did, but how about this:
14.6 ml K2Cr2O7 soln * (1L/1000ml) * (0.050 mol K2Cr2O7(aq)/1L K2Cr2O7 soln) *
(1 mol Cr2O7^-2/1mol K2Cr2O7(aq)) * (1 mol C2H5OH(l)/2 mol Cr2O7^-2(aq))*
(46 g C2H5OH/1 mol C2H5OH) = 0.01679 g approx. 0.02g
(0.02g C2H5OH/28g blood)*100% = 0.07%
Please correct me if I am wrong.
:)
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SPASH - check stoichiometry!