Chapter 5, Problem 71P

What is the mass in grams of each quantity of lactic acid ${\text{(C}}_{\text{3}}{\text{H}}_{\text{6}}{\text{O}}_{\text{3}}\text{, molar mass 90}\text{.08 g/mol),}$ the compound responsible for the aching feeling of tired muscles during vigorous exercise?

1. 3.60 mol
2. 0.580 mol
3. $7.3\times {10}^{24}\text{molecules}$
4. $6.56\times {10}^{22}\text{molecules}$

Interpretation Introduction

(a)

Interpretation:

The mass (in grams) of lactic acid ${\text{(C}}_3{\text{H}}_6{\text{O}}_3)$ present in $3.60\text{ mole}$ of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$ is to be determined.

Concept Introduction:

Following is the formula that will be used to calculate the mass of the substance:

  $numberofmoles=\frac{mass}{Molarmass}$

Or,

  $mass=numberofmoles\times Molarmass$

Answer to Problem 71P

The mass of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$ present in $3.60$ mol of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$ is $342.3g$.

Explanation of Solution

Number of moles of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$

  $\text{= 3}\text{.60 mol}$

Molar mass of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$

  $\text{= 90}\text{.01 g/mol}$

The formula will be used to calculate the mass of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$ −

  $numberofmoles=\frac{mass}{Molarmass}$

Or,

  $mass=numberofmoles\times Molarmass$

  $mass=3.60mol\times 90.01g/mol$

  $mass=324.3g$

Thus, the mass of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$ present in $3.60$ mol of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$ is $324.3g$.

Interpretation Introduction

(b)

Interpretation:

The mass (in grams) of lactic acid ${\text{(C}}_3{\text{H}}_6{\text{O}}_3)$ present in $0.580\text{ mole}$ of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$ is to be determined.

Concept Introduction:

Following is the formula that will be used to calculate the mass of the substance −

  $numberofmoles=\frac{mass}{Molarmass}$

Or,

  $mass=numberofmoles\times Molarmass$

Answer to Problem 71P

The mass of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$ present in $0.580$ mol of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$ is $52.25g$.

Explanation of Solution

Number of moles of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$

  $\text{= 0}\text{.580 mol}$

Molar mass of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$

  $\text{= 90}\text{.01 g/mol}$

The formula will be used to calculate the mass of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$ −

  $numberofmoles=\frac{mass}{Molarmass}$

Or,

  $mass=numberofmoles\times Molarmass$

  $mass=0.580mol\times 90.01g/mol$

Thus, the mass of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$ present in $0.580$ mol of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$ is $52.25g$.

Interpretation Introduction

(c)

Interpretation:

The mass (in grams) of lactic acid ${\text{(C}}_3{\text{H}}_6{\text{O}}_3)$ present in $\text{7}\text{.3}\times {\text{10}}^{24}$ molecules of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$ is to be determined.

Concept Introduction:

In order to convert the number of molecules into grams, divide the number of molecules by Avogadro's number. This will give the value of the number of moles of that substance. Avogadro's number is $6.023\times {10}^{23}$ molecules. Then, the molar mass of the substance is multiplied by the number of moles. This will give the value of amount of substance in grams.

Answer to Problem 71P

The mass of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$ present in $\text{7}\text{.3}\times {\text{10}}^{24}$ molecules of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$ is $\text{1}\text{.1}\times {\text{10}}^{3}g$.

Explanation of Solution

Molar mass of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$

  $\text{= 90}\text{.01 g/mol}$

Molecules of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$ = $\text{7}\text{.3}\times {\text{10}}^{24}$ molecules

Avogadro's number of $\text{=}$

  $6.023\times {10}^{23}$ molecules

Since,

  $\begin{array}{l}numberofmoles=\frac{numberofmolecules}{Avogadro\text{'}snumber}\\ massingrams=numberofmoles\times molarmass\\ or\\ massingrams=\frac{numberofmolecules}{Avogadro\text{'}snumber}\times molarmass\end{array}$

So,

  $mass=\frac{\text{7}\text{.3}\times {\text{10}}^{24}}{6.023\times {10}^{23}}\times 90.08g$

  $\text{mass =1}\text{.1}\times {\text{10}}^{3}g$

Thus, the mass of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$ present in $\text{7}\text{.3}\times {\text{10}}^{24}$ molecules of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$ is $\text{1}\text{.1}\times {\text{10}}^{3}g$.

Interpretation Introduction

(d)

Interpretation:

The mass (in grams)of lactic acid ${\text{(C}}_3{\text{H}}_6{\text{O}}_3)$ present in $\text{6}\text{.56}\times {\text{10}}^{22}$ molecules of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$ is to be determined.

Concept Introduction:

In order to convert the number of molecules into grams, divide the number of molecules by Avogadro's number. This will give the value of the number of moles of that substance. Avogadro's number is $6.023\times {10}^{23}$ molecules. Then, the molar mass of the substance is multiplied by the number of moles. This will give the value of amount of substance in grams.

Answer to Problem 71P

The mass of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$ present in $\text{6}\text{.56}\times {\text{10}}^{22}$ molecules of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$ is $9.81g$.

Explanation of Solution

Molar mass of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$

  $\text{= 90}\text{.01 g/mol}$

Molecules of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$ = $\text{6}\text{.56}\times {\text{10}}^{22}$ molecules

Avogadro's number of $\text{=}$

  $6.023\times {10}^{23}$ molecules

Since,

  $\begin{array}{l}numberofmoles=\frac{numberofmolecules}{Avogadro\text{'}snumber}\\ massingrams=numberofmoles\times molarmass\\ or\\ massingrams=\frac{numberofmolecules}{Avogadro\text{'}snumber}\times molarmass\end{array}$

So,

  $mass=\frac{\text{6}\text{.56}\times {\text{10}}^{22}}{6.023\times {10}^{23}}\times 90.08g$

  $\text{mass = 9}\text{.81}g$

The mass of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$ present in $\text{6}\text{.56}\times {\text{10}}^{22}$ molecules of ${\text{C}}_3{\text{H}}_6{\text{O}}_3$ is $9.81g$.