24.97 CP - Chemistry: Molecular Nature Of Matter And Change

Question

Archeologists removed some charcoal from a Native American campfire, burned it in, and bubbled the ${\text{CO}}_{\text{2}}$ formed into  solution (limewater). The ${\text{CaCO}}_{\text{3}}$  that precipitated was filtered and dried. If  4.58 g of the ${\text{CaCO}}_{\text{3}}$ had a radioactivity of 3.2 d/min, how long ago was the campfire?

Step-by-Step Solution

Answer

Time period of the campfire burning is $8.0\times {10}^3$ years.

Step 1: Radiocarbon dating

The process of radiocarbon dating helps to measure the age of a particular object. In this problem, the amount of  ${}^{\text{12}}\text{C}$ present is calculated through analytical methods. Then the time period of the campfire is estimated.

Step 2: Calculation of the mass of the carbon present in the charcoal

The following reaction takes place when carbon dioxide is bubbled through $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$  to form ${\text{CaCO}}_{\text{3}}$ .

  ${\text{CO}}_{\text{2}}\left(\text{s}\right){\text{ + Ca(OH)}}_{\text{2}}\left(\text{aq}\right)\to {\text{CaCO}}_{\text{3}}\left(\text{s}\right){\text{ + H}}_{\text{2}}\text{O}\left(\text{l}\right)$

From the balanced equation it is evident that one mole of  combines with one mole of    to form one mole of    .

  data-custom-editor="chemistry" $\text{Mole} \text{ratio} \text{ = }\frac{\text{1} \text{mol} {\text{CO}}_{\text{2}}}{\text{1} \text{mol} {\text{CaCO}}_{\text{3}}}$

Considering the given information:

 

Mass of  $CaC{O}_3=4.58 g$

 

Half - time of   

Activity of  $CaC{O}_3=3.2 d/\min$

Number of moles of    formed  $=\frac{4.58 g}{100.09 g/mol}=0.0457 mol$

Number of moles of     $=0.0457 mol CaC{O}_3\times \frac{1 mol C{O}_2}{1 mol CaC{O}_3}=0.0457 mol C{O}_2$

Mass of Carbon   $=0.0457 mol CaC{O}_3\times \frac{1 mol C{O}_2}{1 mol CaC{O}_3}=0.0457 mol C{O}_2$

Step 3: Calculation of the rate constant from half-life

The rate constant, k, is calculated as follows:

 $$\begin{gathered} t_{1/2}=\frac{\ln 2}{k} \\ k=\frac{\ln 2}{t_{1/2}} \\ =\frac{0.693}{5730yr} \\ =1.209\times {10}^{-4}y{r}^{-1} \end{gathered}$$

Activity of ${\text{CaCO}}_{\text{3}}$    in   is given as follows,

Step 4: Calculation of the time of the campfire

Activity of   ${\text{CaCO}}_{\text{3}}$ d/min.g in   is given as follows,

 

 $$\begin{gathered} A_t=\frac{3.2 d/\min }{0.548 g} \\ =5.8228 d/\min .g \end{gathered}$$

Because charcoal came from a living organism, ${\text{A}}_{\text{0}}$  equals 15.3 d/min.g  .

The following equation can be used to calculate the duration of the campfire burning:

 $$\begin{gathered} t=\frac{1}{k}\ln \frac{A_0}{A_t} \\ =\frac{1}{\left(1.209\times {10}^{-4}y{r}^{-1}\right)}\ln \frac{(15.3 d/\min .g)}{(5.8228 d/\min .g)} \\ =8.0\times {10}^{3}years \end{gathered}$$

Therefore, the required time period of the campfire burning is $8.0\times {10}^3$ years.