Q4.81P - Chemistry: Molecular Nature Of Matter And Change

Question

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

 

 $16H^{+}\left(aq\right)+2C{r}_2{O}_7^{2-}\left(aq\right)+C_2{H}_{5}OH\left(aq\right)\to 4C{r}^{3+}\left(aq\right)+2C{O}_2\left(g\right)+11H_{2}O\left(l\right)$

 

If 35.46mL of 0.05961M Cr₂O₇^2- is required to titrate 28.00 g of plasma, what is the mass percent of alcohol in the blood?

Step-by-Step Solution

Answer

You need to find the mass percent of alcohol in the blood.

Step 1: Calculation of required moles of Cr 2 O 7 2-

As you know,

 $n_{C{r}_2{O}_7^{2-}}=V_{C{r}_2{O}_7^{2-}}\times M_{C{r}_2{O}_7^{2-}}$

Where, 

 

$n_{C{r}_2{O}_7^{2-}}$= moles of Cr₂O₇^2-

 

$V_{C{r}_2{O}_7^{2-}}$= volume of Cr₂O₇^2-

 

$M_{C{r}_2{O}_7^{2-}}$= Molarity of Cr₂O₇^2-

 

According to the question;

 

 $V_{C{r}_2{O}_7^{2-}}$ = 35.46mL = 0.03546L

$M_{C{r}_2{O}_7^{2-}}$  = 0.05961 mol/L

Now, you can write

$$\begin{gathered} n_{C{r}_2{O}_7^{2-}}=V_{C{r}_2{O}_7^{2-}}\times M_{C{r}_2{O}_7^{2-}} \\ {n}_{C{r}_2{O}_7^{2-}}=0.03546\times 0.05961\left(mol\right) \\ {n}_{C{r}_2{O}_7^{2-}}=0.0021\left(mol\right) \end{gathered}$$

 

 

Step 2: Conclusion

0.0021 mol of Cr₂O₇^2- were required to titrate 28g of plasma.

Step 3: Calculation of required moles of C 2 H 5 OH

According to the balanced equation

$16H^{+}\left(aq\right)+2C{r}_2{O}_7^{2-}\left(aq\right)+C_2{H}_{5}OH\left(aq\right)\to 4C{r}^{3+}\left(aq\right)+2C{O}_2\left(g\right)+11H_{2}O\left(l\right)$

2 moles of Cr₂O₇^2- reacts with 1 moles of C₂H₅OH

 

Hence, moles of C₂H₅OH () can be calculated as:

$$\begin{gathered} n_{C_2{H}_{5}OH}=n_{C{r}_2{O}_7^{2-}}\times \frac{1}{2} \\ {n}_{C_2{H}_{5}OH}=0.0021\times \frac{1}{2}\left(mol\right) \\ {n}_{C_2{H}_{5}OH}=0.00105\left(mol\right) \end{gathered}$$

Step 4: Conclusion

0.00105 moles of C₂H₅OH were present in the 28g sample of plasma.

Step 5: Calculation of grams of C 2 H 5 OH in the blood

As you know,

 $m_{C_2{H}_{5}OH}=n_{C_2{H}_{5}OH}\times M_{C_2{H}_{5}OH}$

 

Where,

$m_{C_2{H}_{5}OH}$= mass of C₂H₅OH

$n_{C_2{H}_{5}OH}$= moles of C₂H₅OH = 0.00105 mol

$M_{C_2{H}_{5}OH}$= molar mass of C₂H₅OH = 46.07 g/mol

Now you can write,

$$\begin{gathered} m_{C_2{H}_{5}OH}=n_{C_2{H}_{5}OH}\times M_{C_2{H}_{5}OH} \\ {m}_{C_2{H}_{5}OH}=0.00105\times 46.07\left(g\right) \\ {m}_{C_2{H}_{5}OH}=0.04837\left(g\right) \end{gathered}$$

Step 6: Conclusion

0.04837 grams of C₂H₅OH were in the blood.

Step 7: Calculation of mass percent of C 2 H 5 OH in the blood

As you know,

Mass percent of C₂H₅OH 

 $=\frac{m_{C_2{H}_{5}OH}}{m_S}\times 100\%$

Where, m_S = mass of sample.

Now you can write

Mass percent of C₂H₅OH 

 $$\begin{gathered} =\frac{m_{C_2{H}_{5}OH}}{m_S}\times 100\% \\ =\frac{0.04837}{28}\times 100\% \\ =0.173\% \\ \backslash \mathrm{endgathered} \end{gathered}$$

Step 8: Conclusion

The mass percentage of alcohol in the blood is 0.173%.