Q. 7.96 - An Introduction to General, Organic, and Biological Chemistry

Question

How many moles are in $4.00$ g of each of the following compounds? ($7.3$)

a. ${\mathrm{NH}}_3$

b. $\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2$

c. ${\mathrm{SO}}_3$

Step-by-Step Solution

Answer

(a) The no. of moles of ${\mathrm{NH}}_3$ in style="max-width: none; vertical-align: -4px;" $4.00$ g are $0.235$ mole

(b) The no. of moles of $\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2$ in $4.00$ g are style="max-width: none; vertical-align: -4px;" $0.024$ mole

(c) The no. of moles of ${\mathrm{SO}}_3$ in $4.00$ g are $0.05$ mole

Part (a) Step 1: Given information

We need to find no. of moles in given compound i.e. $N{H}_3$

Part (a) Step 2: Explanation

We know that

Molar mass of ${\mathrm{NH}}_3$ is $17 \mathrm{g}/\mathrm{mole}$

And molar mass is equal to a mole

Therefore,

$$\begin{gathered} 17 \mathrm{g}=1 \mathrm{mole} \\ 4.00 \mathrm{g}=\frac{1 \mathrm{mole}}{17 \cancel{\mathrm{g}}}\times 4 \cancel{\mathrm{g}}=0.235 \mathrm{mole} \end{gathered}$$

Part (b) Step 1: Given information

We need to find no. of moles in given compound i.e. $\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2$

Part (b) Step 2: Explanation

We know that

Molar mass of $\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2$ is $164 \mathrm{g}/\mathrm{mole}$

And molar mass is equal to a mole

Therefore,

$$\begin{gathered} 164 \mathrm{g}=1 \mathrm{mole} \\ 4.00 \mathrm{g}=\frac{1 \mathrm{mole}}{164 \cancel{\mathrm{g}}}\times 4 \cancel{\mathrm{g}}=0.024 \mathrm{mole} \end{gathered}$$

Part (c) Step 1: Given information

We need to find no. of moles in given compound i.e. ${\mathrm{SO}}_3$

Part (c) Step 2: Explanation

We know that

Molar mass of ${\mathrm{SO}}_3$ is $80 \mathrm{g}/\mathrm{mole}$

And molar mass is equal to a mole

Therefore,

$$\begin{gathered} 80 \mathrm{g}=1 \mathrm{mole} \\ 4.00 \mathrm{g}=\frac{1 \mathrm{mole}}{80 \cancel{\mathrm{g}}}\times 4.00 \cancel{\mathrm{g}}=0.05 \mathrm{mole} \end{gathered}$$