Q. 18.93 - An Introduction to General, Organic, and Biological Chemistry

Question

One cell at work may break down a $2$ $\mathrm{million}$ $\left(2000000\right)\mathrm{ATP}$ molecules in one second. Some researchers estimate that the human body has about ${10}^{13}\mathrm{cells}$. $(18.1)$

a. How much energy, in Kilocalories, would be used by the cells in the body in one day?

b. If ATP has a molar mass of $507 \raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mole}$}\right.$, how many grams of ATP is hydrolyzed in one day?

Step-by-Step Solution

Answer

a. $21 \mathrm{kcal}$ energy, would be used by the cells in the body in one day.

b. $1454 \mathrm{g} \mathrm{ATP}$is hydrolyzed in one day.

Part (a) step 1: Given Information

We need to find the cells in the body that would be used how much energy in one day.

Part (a) step 2: Simplify

Now,

One mole of ATP generates $7.3 \mathrm{kcal}$ of energy, therefore the amount of energy for one mole of ATP is

$\frac{7.3 \mathrm{kcal}}{\mathrm{mol}}\left(\frac{1 \mathrm{mol}}{6.023\times {10}^{23} \mathrm{molecules}}\right)=\frac{7.3 \mathrm{kcal}}{6.023\times {10}^{23} \mathrm{molecules}}$

On the Avogadro's number. Also, in one day, there are $86400 \mathrm{seconds}$.

So, 

$$\begin{gathered} \mathrm{E}=\frac{\left(2\times {{10}^6}^{{\displaystyle \frac{\mathrm{molecules}}{8}}}\right)\left(86400 \mathrm{s}\right)\left(7.3 \mathrm{kcal}\right)}{6.023\times {10}^{23} \mathrm{molecules}} \\ =21 \mathrm{kcal} \end{gathered}$$

Part (b) step 1: Given Information

We need to find the grams of ATP is hydrolyzed in one day.

Part (b) step 2: Simplify

The human body has an estimated ${10}^{13}$ cells. Based on the Avogadro's numbers, so

${10}^{13 }\mathrm{cells}\left(\frac{1 \mathrm{mol}}{6.023\times {10}^{23} \mathrm{cells}}\right)=1.66\times {10}^{-11} \mathrm{mol}$

In one day, there are $86400 \mathrm{seconds}$. So,

$$\begin{gathered} \mathrm{mol} \mathrm{ATP}=\left(2\times {10}^6\frac{\mathrm{cells}}{\mathrm{s}}\right)\left(86400 \mathrm{s}\right)\left(1.66\times {10}^{-11} \mathrm{mol}\right) \\ =2.86848 \mathrm{mol} \mathrm{ATP} \end{gathered}$$

The molar mass of ATP $\left(507 \raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mole}$}\right.\right)$,

$$\begin{gathered} \mathrm{g} \mathrm{ATP} =2.86848 \mathrm{mol}\left(\frac{507 \mathrm{g}}{\mathrm{mol}}\right) \\ =1454 \mathrm{g} \mathrm{ATP} \end{gathered}$$