Q. 7.15 - An Introduction to General, Organic, and Biological Chemistry

Question

Calculate the molar mass for each of the following:

a. $\mathrm{KCl}$, salt substitute

b. ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NO}}_2$, niacin (vitamin $\mathrm{B}3$) needed for cell metabolism

c. ${\mathrm{C}}_{19}{\mathrm{H}}_{20}{\mathrm{FNO}}_3$, Paxil, antidepressant

Step-by-Step Solution

Answer

a. The molar mass of  $\mathrm{KCl}$ is $74.54 \mathrm{g}$

b. The molar mass of ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NO}}_2$  is $123.04 \mathrm{g}$

c. The molar mass of ${\mathrm{C}}_{19}{\mathrm{H}}_{20}{\mathrm{FNO}}_3$  is $329.15 \mathrm{g}$

Part (a) Step 1 : Given information

We have to find the molar mass of $\mathrm{KCl}$

Part (a) Step 2 : Simplification

We know that,

Molar mass of one $\mathrm{K}$ atom = $39.09 \mathrm{g}$

Molar mass of one $\mathrm{Cl}$ atom = $35.45 \mathrm{g}$

So, the molar mass of $\mathrm{KCl}$ = $39.09 + 35.45$ = $74.54 \mathrm{g}$

Part (b) Step 1 : Given information

We have to find the molar mass of ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NO}}_2$

Part (b) Step 2 : Simplification

We know that,

Molar mass of one $\mathrm{C}$ atom = $12.01 \mathrm{g}$

Molar mass of one $\mathrm{H}$ atom = $1.00 \mathrm{g}$

Molar mass of one $\mathrm{N}$ atom = $14.00 \mathrm{g}$

Molar mass of one $\mathrm{O}$ atom = $15.99 \mathrm{g}$

So, the molar mass of ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NO}}_2$ = $6\left(12.01\right) + 5\left(1.00\right) + 14.00 + 2\left(15.99\right)$ = $123.04 \mathrm{g}$

Part (c) Step 1 : Given information

We have to find the molar mass of ${\mathrm{C}}_{19}{\mathrm{H}}_{20}{\mathrm{FNO}}_3$

Part (c) Step 2 : Simplification

We know that,

Molar mass of one $\mathrm{C}$ atom = $12.01 \mathrm{g}$

Molar mass of one $\mathrm{H}$ atom = $1.00 \mathrm{g}$

Molar mass of one $\mathrm{F}$ atom = $18.99 \mathrm{g}$

Molar mass of one $\mathrm{N}$ atom = $14.00 \mathrm{g}$

Molar mass of one $\mathrm{O}$ atom = $15.99 \mathrm{g}$

So, the molar mass of ${\mathrm{C}}_{19}{\mathrm{H}}_{20}{\mathrm{FNO}}_3$ = $19\left(12.01\right) + 20\left(1.00\right) + 18.99 + 14.00 + 3\left(15.99\right)$

                                                               = $329.15 \mathrm{g}$