Q13P - Chemistry: Molecular Nature Of Matter And Change

Question

3.13 Calculate each of the following quantities:
(a) Mass in kilograms of $\mathbf{4}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{21}}$molecules of ${\mathbf{NO}}_{\mathbf{2}}$
(b) Moles of Cl atoms in 0.0615 g of ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}{\mathbf{Cl}}_{\mathbf{2}}$
(c) Number of ions in 5.82 g of ${\mathbf{SrH}}_{\mathbf{2}}$

Step-by-Step Solution

Answer

1. Mass in kilograms of $4.6\times {10}^{21}$molecules of ${\mathrm{NO}}_2$is $3.5\times {10}^{-4} \mathrm{kg}$.
2. Moles of Cl atoms in 0.0615 g of ${\mathrm{C}}_2{\mathrm{H}}_4{\mathrm{Cl}}_2$is $1.24\times {10}^{-3} \mathrm{mol} \mathrm{Cl} \mathrm{atoms}$.
3. Number of ions in 5.82 g of ${\mathrm{SrH}}_2$ is $7.82\times {10}^{22} {\mathrm{H}}^{-} \mathrm{ions}$ .

Step 1: Relation between mass and mole of the substance

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

Step 2: Determination of Mass in kilograms of 4 . 6 × 10 21 molecules of N O 2

a)

Let’s compute for molar mass of ${\mathbf{NO}}_{\mathbf{2}}$

$$\begin{gathered} {\mathbf{M}}_{\mathbf{\omega }}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{NO}}_{\mathbf{2}}\mathbf{=}{\mathbf{M}}_{\mathbf{\omega }}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{N}\mathbf{+}\mathbf{2}\mathbf{\times }\mathbf{(}{\mathbf{M}}_{\mathbf{\omega }}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{O}\mathbf{)} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{14}\mathbf{.}\mathbf{01}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{+}\mathbf{2}\mathbf{\times }\left(16.00\frac{g}{\mathrm{mol}}\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{46}\mathbf{.}\mathbf{01}\mathbf{ }\frac{\mathbf{g}}{\mathbf{mol}} \end{gathered}$$

We start by calculating the mass in grams. Multiply the provided quantity of ${\mathrm{NO}}_2$molecules by the reciprocal of Avogadro's number and the molar mass of ${\mathrm{NO}}_2$ To get the mass in kilograms, multiply the value by 1000.

$$\begin{gathered} \mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{NO}}_{\mathbf{2}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{21}}\mathbf{ }\mathbf{molecules}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{NO}}_{\mathbf{2}}\mathbf{ }\mathbf{\times }\left(\frac{\mathrm{mol} {\mathrm{NO}}_2}{6.022\times {10}^{23} \mathrm{molecules} {\mathrm{NO}}_2}\right)\mathbf{\times }\left(\frac{46.01 g {\mathrm{NO}}_2}{\mathrm{mol} {\mathrm{NO}}_2}\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\boxed{\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{ }\mathbf{kg}\mathbf{ }{\mathbf{NO}}_{\mathbf{2}}} \end{gathered}$$

Step 3: Determination of Moles of Cl atoms in 0.0615 g of C 2 H 4 Cl 2

b)

The following is the molar mass of ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}{\mathbf{Cl}}_{\mathbf{2}}$

$$\begin{gathered} {\mathbf{M}}_{\mathbf{\omega }}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}{\mathbf{Cl}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{\times }\mathbf{(}{\mathbf{M}}_{\mathbf{\omega }}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{C}\mathbf{)}\mathbf{+}\mathbf{4}\mathbf{\times }\mathbf{(}{\mathbf{M}}_{\mathbf{\omega }}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{H}\mathbf{)}\mathbf{+}\mathbf{2}\mathbf{\times }\mathbf{(}{\mathbf{M}}_{\mathbf{\omega }}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Cl}\mathbf{)} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{2}\mathbf{\times }\left(12.01\frac{g}{\mathrm{mol}}\right)\mathbf{ }\mathbf{+}\mathbf{4}\mathbf{\times }\left(1.01\frac{g}{\mathrm{mol}}\right)\mathbf{+}\mathbf{2}\mathbf{\times }\left(35.45\frac{g}{\mathrm{mol}}\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{98}\mathbf{.}\mathbf{94}\mathbf{ }\frac{\mathbf{g}}{\mathbf{mol}} \end{gathered}$$

The mass of ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}{\mathbf{Cl}}_{\mathbf{2}}$is multiplied by the reciprocal of its molar mass. We multiply 2 to it because there are 2 moles of Cl atoms per mole of ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}{\mathbf{Cl}}_{\mathbf{2}}$

As shown below, the number of moles of Cl atoms in 0.0615 g of  ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}{\mathbf{Cl}}_{\mathbf{2}}$has been computed.

$$\begin{gathered} \mathbf{Moles}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Cl}\mathbf{ }\mathbf{atoms}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{0615}\mathbf{ }\mathbf{g}\mathbf{ }{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}{\mathbf{Cl}}_{\mathbf{2}}\mathbf{\times }\mathbf{\left(}\frac{\mathbf{mol}\mathbf{ }{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}{\mathbf{Cl}}_{\mathbf{2}}}{\mathbf{98}\mathbf{.}\mathbf{94}\mathbf{ }\mathbf{g}\mathbf{ }{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}{\mathbf{Cl}}_{\mathbf{2}}}\mathbf{\right)}\mathbf{\times }\mathbf{\left(}\frac{\mathbf{2}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{Cl}\mathbf{ }\mathbf{atoms}}{\mathbf{1}\mathbf{ }\mathbf{mol}\mathbf{ }{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}{\mathbf{Cl}}_{\mathbf{2}}}\mathbf{\right)} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\boxed{\mathbf{1}\mathbf{.}\mathbf{24}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{Cl}\mathbf{ }\mathbf{atoms}} \end{gathered}$$

Step 4: Determination of Number of H ions in 5.82 g of SrH 2

c)

The following is the molar mass of ${\mathbf{SrH}}_{\mathbf{2}}$ 

$$\begin{gathered} {\mathbf{M}}_{\mathbf{\omega }}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{SrH}}_{\mathbf{2}}\mathbf{=}{\mathbf{M}}_{\mathbf{\omega }}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Sr}\mathbf{ }\mathbf{+}\mathbf{2}\mathbf{\times }\mathbf{(}{\mathbf{M}}_{\mathbf{\omega }}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{H}\mathbf{)} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{87}\mathbf{.}\mathbf{62}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{+}\mathbf{2}\mathbf{\times }\left(1.01\frac{g}{\mathrm{mol}}\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{89}\mathbf{.}\mathbf{64}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{ } \end{gathered}$$

The mass of  ${\mathbf{SrH}}_{\mathbf{2}}$is multiplied by the reciprocal of its molar mass, the number of moles of  ${\mathbf{H}}^{\mathbf{-}}$ions per mole of ${\mathbf{SrH}}_{\mathbf{2}}$and Avogadro's number. 

$$\begin{gathered} \mathbf{Moles}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{H}}^{\mathbf{-}}\mathbf{ }\mathbf{ions}\mathbf{ }\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{82}\mathbf{ }\mathbf{g}\mathbf{ }{\mathbf{SrH}}_{\mathbf{2}}\mathbf{\times }\left(\frac{\mathrm{mol} g {\mathrm{SrH}}_2}{89.64 g {\mathrm{SrH}}_2}\right)\mathbf{\times }\left(\frac{2 \mathrm{mol} H^{-} \mathrm{ions}}{\mathrm{mol} {\mathrm{SrH}}_2}\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{\times }\left(\frac{6.022\times {10}^{23}{H}^{-} \mathrm{ions}}{\mathrm{mol} H^{-} \mathrm{ions}}\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\boxed{\mathbf{7}\mathbf{.}\mathbf{82}\mathbf{\times }{\mathbf{10}}^{\mathbf{22}}\mathbf{ }{\mathbf{H}}^{\mathbf{-}}\mathbf{ }\mathbf{ions}} \end{gathered}$$