Q. 7.16 - An Introduction to General, Organic, and Biological Chemistry

Question

Calculate the molar mass for each of the following:

a. ${\mathrm{FeSO}}_4$, iron supplement

b. ${\mathrm{Al}}_2{\mathrm{O}}_3$, absorbent and abrasive

c. ${\mathrm{C}}_7{\mathrm{H}}_5{\mathrm{NO}}_3\mathrm{S}$, saccharin, artificial sweetener

Step-by-Step Solution

Answer

a. The molar mass of ${\mathrm{FeSO}}_4$ is $151.86 \mathrm{g}$

b. The molar mass of ${\mathrm{Al}}_2{\mathrm{O}}_3$ is $101.93 \mathrm{g}$$101.93g$

c. The molar mass of ${\mathrm{C}}_7{\mathrm{H}}_5{\mathrm{NO}}_3\mathrm{S}$ is $183.10 \mathrm{g}$

Part (a) Step 1 : Given information

We have to find the molar mass of ${\mathrm{FeSO}}_4$

Part (a) Step 2 : Simplification

We know that,

Molar mass of one $\mathrm{Fe}$ atom = $55.84 \mathrm{g}$

Molar mass of one $\mathrm{S}$ atom = $32.06 \mathrm{g}$

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

So, the molar mass of ${\mathrm{FeSO}}_4$ = $55.84 + 32.06 + 4\left(15.99\right)$ = $151.86 \mathrm{g}$

Part (b) Step 1 : Given information

We have to find the molar mass of ${\mathrm{Al}}_2{\mathrm{O}}_3$

Part (b) Step 2 : Simplification

We know that,

Molar mass of one $\mathrm{Al}$ atom = $26.98 \mathrm{g}$

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

So, the molar mass of ${\mathrm{Al}}_2{\mathrm{O}}_3$ = $2\left(26.98\right) + 3\left(15.99\right)$ = $101.93 \mathrm{g}$

Part (c) Step 1 : Given information

We have to find the molar mass of ${\mathrm{C}}_7{\mathrm{H}}_5{\mathrm{NO}}_3\mathrm{S}$

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 $\mathrm{N}$ atom = $14.00 \mathrm{g}$

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

Molar mass of one $\mathrm{S}$ atom = $32.06 \mathrm{g}$

So, the molar mass of ${\mathrm{C}}_7{\mathrm{H}}_5{\mathrm{NO}}_3\mathrm{S}$= $7\left(12.01\right) + 5\left(1.00\right) + 14.00 + 3\left(15.99\right) + 32.06$ = $183.10 \mathrm{g}$