Q15P - Chemistry: Molecular Nature Of Matter And Change

Question

3.15 Calculate each of the following quantities:
 (a) Total number of ions in 38.1 g of ${\mathbf{SrF}}_{\mathbf{2}}$
 (b) Mass in kilograms of 3.58 mol of ${\mathbf{CuCl}}_{\mathbf{2}\mathbf{ }}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$
 (c) Mass in milligrams of $\mathbf{2}\mathbf{.}\mathbf{88}\mathbf{\times }{\mathbf{10}}^{\mathbf{22}}$formula unit $\mathbf{Bi}{\mathbf{\left(}{\mathbf{NO}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{3}}$$\mathbf{5}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$.

Step-by-Step Solution

Answer

1. The total number of ions in 38.1 g is $\mathbf{5}\mathbf{.}\mathbf{49}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{ions}$
2. Mass in kilograms of 3.58 mol of ${\mathbf{CuCl}}_{\mathbf{2}}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$  is $\mathbf{5}\mathbf{.}\mathbf{49}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{ }\mathbf{ions}$.
3. Mass in milligrams of $\mathbf{2}\mathbf{.}\mathbf{88}\mathbf{\times }{\mathbf{10}}^{\mathbf{22}}$ formula unit is$2.32\times {10}^4 \mathrm{mg} \mathrm{Bi}{\left({\mathrm{NO}}_3\right)}_3.5{\mathrm{H}}_2\mathrm{O}$.

Step 1: Introduction to the Concept

The mass of a substance made up of an equal number of fundamental units is defined as a mole.

Step 2: Solution Explanation

a)

The following molar mass${\mathbf{SrF}}_{\mathbf{2}}$,

$$\begin{gathered} {\mathrm{M}}_{\mathrm{\omega }} \mathrm{of} \mathrm{SrF}={\mathrm{M}}_{\mathrm{\omega }} \mathrm{of} \mathrm{Sr}+2\times ({\mathrm{M}}_{\mathrm{\omega }} \mathrm{of} \mathrm{F}) \\ =87.62\frac{\mathrm{g}}{\mathrm{mol}}+2\times \left(19.00\frac{\mathrm{g}}{\mathrm{mol}}\right) \\ =125.62\frac{\mathrm{g}}{\mathrm{mol}} \end{gathered}$$

The number of formula units (FU) of is calculated by multiplying the given mass of ${\mathbf{SrF}}_{\mathbf{2}}$ by the reciprocal of its molar mass and the Avogadro's number.

$$\begin{gathered} \mathbf{Number}\mathbf{ }\mathbf{FU}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{SrF}}_{\mathbf{2}}\mathbf{=}\mathbf{38}\mathbf{.}\mathbf{1}\mathbf{g}\mathbf{ }\mathbf{SrF}\mathbf{\times }\left(\frac{\mathrm{mol} \mathrm{SrF}}{125.62 g \mathrm{SrF}}\right)\mathbf{\times }\left(\frac{6.022\times {10}^{23}\mathrm{FU} {\mathrm{SrF}}_2}{\mathrm{mol} {\mathrm{SrF}}_2}\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{83}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{ }\mathbf{FU}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{SrF}}_{\mathbf{2}} \end{gathered}$$

Because each FU of ${\mathbf{SrF}}_{\mathbf{2}}$ contains three ions, the total number of ions in 38.1 g of ${\mathbf{SrF}}_{\mathbf{2}}$is as follows.

$$\begin{gathered} \mathrm{Total} \mathrm{numbers} \mathrm{of} \mathrm{ions} \mathrm{of} {\mathrm{SrF}}_2=1.83\times {10}^{23}\mathrm{FU} {\mathrm{SrF}}_2\left(\frac{3 \mathrm{ions}}{\mathrm{FU} {\mathrm{SrF}}_2}\right) \\ =\boxed{5.49\times {10}^{23} \mathrm{ions}} \end{gathered}$$

Step 3: Solution Explanation

b)

The following molar mass ${\mathbf{CuCl}}_{\mathbf{2}}\mathbf{ }\mathbf{.}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$

$$\begin{gathered} {\mathbf{M}}_{\mathbf{\omega }}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{CuCl}\mathbf{ }\mathbf{\times }\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{=}{\mathbf{M}}_{\mathbf{\omega }}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Cu}\mathbf{ }\mathbf{+}\mathbf{2}\mathbf{\times }\left(M_{\omega } \mathrm{of} \mathrm{Cl}\right)\mathbf{+}\mathbf{4}\mathbf{\times }\left(M_{\omega } \mathrm{of} H\right)\mathbf{+}\mathbf{2}\mathbf{\times }\mathbf{(}{\mathbf{M}}_{\mathbf{\omega }}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{O}\mathbf{)} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{63}\mathbf{.}\mathbf{55}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{+}\mathbf{2}\mathbf{\times }\left(35.45\frac{g}{\mathrm{mol}}\right)\mathbf{+}\mathbf{4}\mathbf{\times }\left(1.01\frac{g}{\mathrm{mol}}\right)\mathbf{+}\mathbf{2}\mathbf{\times }\left(16.00\frac{g}{\mathrm{mol}}\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{170}\mathbf{.}\mathbf{49}\frac{\mathbf{g}}{\mathbf{mol}} \end{gathered}$$

To calculate the mass in kilograms of 3.58 mol of ${\mathbf{CuCl}}_{\mathbf{2}}\mathbf{.}\mathbf{ }\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ multiply the provided number of moles by its molar mass and divide by 1000.

$$\begin{gathered} \mathbf{moles}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{CuCl}}_{\mathbf{2}}\mathbf{\times }\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{58}\mathbf{ }\mathbf{mol}\mathbf{ }{\mathbf{CuCl}}_{\mathbf{2}}\mathbf{\times }\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\times }\left(\frac{170.49 g {\mathrm{CuCl}}_2\times 2H_{2}O}{\mathrm{mol} {\mathrm{CuCl}}_2 \times 2H_{2}O}\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\boxed{\mathbf{0}\mathbf{.}\mathbf{610}\mathbf{ }\mathbf{kg}\mathbf{ }{\mathbf{CuCl}}_{\mathbf{2}}\mathbf{ }\mathbf{\times }\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}} \end{gathered}$$

Step 4: Solution Explanation

c)

The following molar mass $\mathbf{Bi}{\left({\mathrm{NO}}_3\right)}_{\mathbf{3}}\mathbf{ }\mathbf{.}\mathbf{5}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{,}$

$$\begin{gathered} {\mathbf{M}}_{\mathbf{\omega }}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Bi}{\left({\mathrm{NO}}_3\right)}_{\mathbf{3}}\mathbf{\times }\mathbf{5}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{=}{\mathbf{M}}_{\mathbf{\omega }}\mathbf{ }\mathbf{of}\mathbf{ }\left(M_{\omega } \mathrm{of} N\right)\mathbf{+}\mathbf{14}\mathbf{\times }\left(M_{\omega } \mathrm{of} O\right)\mathbf{+}\mathbf{10}\mathbf{\times }\left(M_{\omega } \mathrm{of} H\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{208}\mathbf{.}\mathbf{98}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{+}\mathbf{3}\mathbf{\times }\left(14.01\frac{g}{\mathrm{mol}}\right)\mathbf{+}\mathbf{14}\mathbf{\times }\left(16.00\frac{g}{\mathrm{mol}}\right)\mathbf{+}\mathbf{10}\mathbf{\times }\left(1.01\frac{g}{\mathrm{mol}}\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{485}\mathbf{.}\mathbf{11}\frac{\mathbf{g}}{\mathbf{mol}} \end{gathered}$$

The number of formula units (FU) of $\mathbf{Bi}{\left({\mathrm{NO}}_3\right)}_{\mathbf{3}}\mathbf{.}\mathbf{ }\mathbf{5}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ is calculated by multiplying the given mass of $\mathbf{Bi}{\left({\mathrm{NO}}_3\right)}_{\mathbf{3}}\mathbf{.}\mathbf{5}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ by the reciprocal of its molar mass and the Avogadro's number.

$$\begin{gathered} \mathbf{moles}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Bi}{\left({\mathrm{NO}}_3\right)}_{\mathbf{3}}\mathbf{\times }\mathbf{5}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{88}\mathbf{\times }{\mathbf{10}}^{\mathbf{22}}\mathbf{FU}\mathbf{\times }\left(\frac{\mathrm{mol} \mathrm{Bi}{\left({\mathrm{NO}}_3\right)}_3\times 5H_{2}O}{6.022\times {10}^{23}\mathrm{FU}}\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{78}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{mol}\mathbf{ }\mathbf{Bi}{\left({\mathrm{NO}}_3\right)}_{\mathbf{3}}\mathbf{\times }\mathbf{5}{\mathbf{H}}_{\mathbf{2}}\mathbf{O} \\ \mathrm{Molar} \mathrm{masses} \mathrm{of} \mathbf{Bi}{\left({\mathrm{NO}}_3\right)}_{\mathbf{3}}\mathbf{5}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{,} \\ \mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Bi}\mathbf{ }{\mathbf{\left(}{\mathbf{NO}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{3}}\mathbf{\times }\mathbf{5}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{78}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Bi}\mathbf{ }{\mathbf{\left(}{\mathbf{NO}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{3}}\mathbf{\times }\mathbf{5}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\times }\left(\frac{485.11g \mathrm{Bi}{\left({\mathrm{NO}}_3\right)}_3\times 5H_{2}O}{\mathrm{mol} \mathrm{Bi}{\left({\mathrm{NO}}_3\right)}_3\times 5H_{2}O}\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\boxed{\mathbf{2}\mathbf{.}\mathbf{32}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{ }\mathbf{mg}\mathbf{ }\mathbf{Bi}{\left({\mathrm{NO}}_3\right)}_{\mathbf{3}}\mathbf{\times }\mathbf{5}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}} \end{gathered}$$