Chapter 12, Problem 4E

Calculate (to the nearest 0.1 u) the formula mass of these compounds:

1. (a) glucose, C₆H₁₂O₆
2. (b) sucrose, C₁₂H₂₂O₁₁
3. (c) azithromycin, C₃₈H₇₂N₂O₁₂

To determine

The formula mass of glucose $\left({\text{C}}_6{\text{H}}_{12}{\text{O}}_6\right)$.

Answer to Problem 4E

The formula mass of glucose $\left({\text{C}}_6{\text{H}}_{12}{\text{O}}_6\right)$ is $180.0\text{ u}$.

Explanation of Solution

Formula mass of any compound is given as the sum of the atomic masses of the constituent atoms given in the chemical formula of a substance. Formula mass and atomic mass are abbreviated as FM and AM respectively.

The formula mass of glucose $\left({\text{C}}_6{\text{H}}_{12}{\text{O}}_6\right)$ is given as,

$\text{FM of }\left({\text{C}}_6{\text{H}}_{12}{\text{O}}_6\right)\text{ = }\left[\begin{array}{c}\left(\text{6}\times \text{AM of carbon atom}\right)+\left(\text{12}\times \text{AM of hydrogen atom}\right)+\\ \left(6\times \text{AM of oxygen atom}\right)\end{array}\right]$ (1)

Atomic mass (AM) of carbon atom is $12.0\text{ u}$.

Atomic mass (AM) of hydrogen atom is $1.0\text{ u}$.

Atomic mass (AM) of oxygen atom is $16.0\text{ u}$.

Substitute $12.0\text{ u}$ for $\text{AM of carbon atom}$, $1.0\text{ u}$ for $\text{AM of hydrogen atom}$ and $16.0\text{ u}$ for $\text{AM of oxygen atom}$ in equation (1).

$\begin{array}{c}{\text{FM of C}}_6{\text{H}}_{12}{\text{O}}_6\text{ = }\left(6\times 12.0\text{ u}\right)+\left(12\times 1.0\text{ u}\right)+\left(6\times 16.0\text{ u}\right)\\ =180.0\text{ u}\end{array}$

Conclusion:

Therefore, the formula mass of glucose $\left({\text{C}}_6{\text{H}}_{12}{\text{O}}_6\right)$ is $180.0\text{ u}$.

To determine

The formula mass of sucrose $\left({\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{\text{11}}\right)$.

Answer to Problem 4E

The formula mass of sucrose $\left({\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{\text{11}}\right)$ is $342.0\text{ u}$.

Explanation of Solution

Formula mass of any compound is given as the sum of the atomic masses of the constituent atoms given in the chemical formula of a substance. Formula mass and atomic mass are abbreviated as FM and AM respectively.

The formula mass of sucrose $\left({\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{\text{11}}\right)$ is given as,

$\text{FM of }\left({\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{\text{11}}\right)\text{ =}\left[\begin{array}{c}\left(\text{12}\times \text{AM of carbon atom}\right)+\\ \left(\text{22}\times \text{AM of hydrogen atom}\right)+\\ \left(11\times \text{AM of oxygen atom}\right)\end{array}\right]$ (2)

Atomic mass (AM) of carbon atom is $12.0\text{ u}$.

Atomic mass (AM) of hydrogen atom is $1.0\text{ u}$.

Atomic mass (AM) of oxygen atom is $16.0\text{ u}$.

Substitute $12.0\text{ u}$ for $\text{AM of carbon atom}$, $1.0\text{ u}$ for $\text{AM of hydrogen atom}$ and $16.0\text{ u}$ for $\text{AM of oxygen atom}$ in equation (2).

$\begin{array}{c}{\text{FM of C}}_{12}{\text{H}}_{22}{\text{O}}_{11}\text{ = }\left(12\times 12.0\text{ u}\right)+\left(22\times 1.0\text{ u}\right)+\left(11\times 16.0\text{ u}\right)\\ =342.0\text{ u}\end{array}$

Conclusion:

Therefore, the formula mass of sucrose $\left({\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{\text{11}}\right)$ is $342.0\text{ u}$.

To determine

The formula mass of azithromycin $\left({\text{C}}_{38}{\text{H}}_{72}{\text{N}}_2{\text{O}}_{\text{12}}\right)$.

Answer to Problem 4E

The formula mass of azithromycin $\left({\text{C}}_{38}{\text{H}}_{72}{\text{N}}_2{\text{O}}_{\text{12}}\right)$ is $748.0\text{ u}$.

Explanation of Solution

Formula mass of any compound is given as the sum of the atomic masses of the constituent atoms given in the chemical formula of a substance. Formula mass and atomic mass are abbreviated as FM and AM respectively.

The formula mass of azithromycin $\left({\text{C}}_{38}{\text{H}}_{72}{\text{N}}_2{\text{O}}_{\text{12}}\right)$ is given as,

$\text{FM of }\left({\text{C}}_{38}{\text{H}}_{72}{\text{N}}_2{\text{O}}_{\text{12}}\right)\text{ =}\left[\begin{array}{c}\left(\text{38}\times \text{AM of carbon atom}\right)+\\ \left(\text{72}\times \text{AM of hydrogen atom}\right)+\\ \left(2\times \text{AM of nitrogen atom}\right)+\\ \left(12\times \text{AM of oxygen atom}\right)\end{array}\right]$ (3)

Atomic mass (AM) of carbon atom is $12.0\text{ u}$.

Atomic mass (AM) of hydrogen atom is $1.0\text{ u}$.

Atomic mass (AM) of oxygen atom is $16.0\text{ u}$.

Atomic mass (AM) of nitrogen atom is $14.0\text{ u}$.

Substitute $12.0\text{ u}$ for $\text{AM of carbon atom}$, $1.0\text{ u}$ for $\text{AM of hydrogen atom}$, $14.0\text{ u}$ for $\text{AM of nitrogen atom}$ and $16.0\text{ u}$ for $\text{AM of oxygen atom}$ in equation (3).

$\begin{array}{c}{\text{FM of C}}_{38}{\text{H}}_{72}{\text{N}}_2{\text{O}}_{\text{12}}\text{ = }\left(38\times 12.0\text{ u}\right)+\left(72\times 1.0\text{ u}\right)+\left(2\times 14.0\text{ u}\right)+\left(12\times 16.0\text{ u}\right)\\ =748.0\text{ u}\end{array}$

Conclusion:

Therefore, the formula mass of azithromycin $\left({\text{C}}_{38}{\text{H}}_{72}{\text{N}}_2{\text{O}}_{\text{12}}\right)$ is $748.0\text{ u}$.