Q12P - Chemistry: Molecular Nature Of Matter And Change

Question

Calculate each of the following quantities:
 (a) Mass in grams of 0.68 mol of  ${\mathbf{KMnO}}_{\mathbf{4}}$
 (b) Moles of O atoms in 8.18 g of $\mathbf{Ba}{\left({\mathrm{NO}}_3\right)}_{\mathbf{2}}$ 
 (c) Number of O atoms in $\mathbf{7}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{g}$ of ${\mathbf{CaSO}}_{\mathbf{4}\mathbf{ }}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$

Step-by-Step Solution

Answer

1. Mass in grams of 0.68 mol is $\mathbf{107}\mathbf{.}\mathbf{47}\mathbf{ }\mathbf{g}\mathbf{ }{\mathbf{KMnO}}_{\mathbf{4}}$ .
2. Moles of O atoms in 8.18 g is 0.188 mol
3. The number of O atoms in $\mathbf{7}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}$  is  $\mathbf{1}\mathbf{.}\mathbf{53}\mathbf{\times }{\mathbf{10}}^{\mathbf{20}}$ $\mathbf{O}\mathbf{ }\mathbf{atoms}$

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: Determination of Mass in grams of 0.68 mol of KMnO 4

a)

Let’s compute for molar mass of ${\mathbf{KMnO}}_{\mathbf{4}}$,

$$\begin{gathered} {\mathbf{M}}_{\mathbf{\omega }}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{KMnO}}_{\mathbf{4}}\mathbf{=}{\mathbf{M}}_{\mathbf{\omega }}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{K}\mathbf{+}{\mathbf{M}}_{\mathbf{\omega }}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Mn}\mathbf{+}\mathbf{4}\mathbf{\times }\left(M_{\psi }\mathrm{of} O\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{39}\mathbf{.}\mathbf{10}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{+}\mathbf{54}\mathbf{.}\mathbf{94}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{+}\mathbf{4}\mathbf{\times }\left(16.00\frac{g}{\mathrm{mol}}\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{158}\mathbf{.}\mathbf{04}\frac{\mathbf{g}}{\mathbf{mol}} \end{gathered}$$

We multiply the supplied number of moles of ${\mathbf{KMnO}}_{\mathbf{4}}$  by the mass in grams to get the mass in grams.

$$\begin{gathered} \mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{KMnO}}_{\mathbf{4}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{68}\mathbf{ }\mathbf{mol}\mathbf{ }{\mathbf{KMnO}}_{\mathbf{4}}\mathbf{\left(}\frac{\mathbf{158}\mathbf{.}\mathbf{04}\mathbf{ }\mathbf{g}\mathbf{ }{\mathbf{KMnO}}_{\mathbf{4}}}{\mathbf{mol}\mathbf{ }{\mathbf{KMnO}}_{\mathbf{4}}}\mathbf{\right)} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\boxed{\mathbf{107}\mathbf{.}\mathbf{47}\mathbf{ }\mathbf{g}\mathbf{ }{\mathbf{KMnO}}_{\mathbf{4}}} \end{gathered}$$

Step 3: Determination of Moles of O atoms in 8.18 g of Ba ( NO 3 ) 2

b)

The following is the molar mass of  $\mathbf{Ba}{\mathbf{\left(}{\mathbf{NO}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{2}}$

$$\begin{gathered} {\mathbf{M}}_{\mathbf{\omega }\mathbf{ }}\mathbf{of}\mathbf{ }\mathbf{Ba}{\mathbf{\left(}{\mathbf{NO}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{2}}\mathbf{=}{\mathbf{M}}_{\omega }\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Ba}\mathbf{ }\mathbf{+}\mathbf{2}\mathbf{\times }\mathbf{(}{\mathbf{M}}_{\mathbf{\omega }\mathbf{ }}\mathbf{of}\mathbf{ }\mathbf{N}\mathbf{)}\mathbf{+}\mathbf{6}\mathbf{\times }\mathbf{(}{\mathbf{M}}_{\mathbf{\omega }}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{O}\mathbf{)}\mathbf{ } \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{137}\mathbf{.}\mathbf{33}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{+}\mathbf{2}\mathbf{\times }\left(14.01\frac{g}{\mathrm{mol}}\right)\mathbf{+}\mathbf{6}\mathbf{\times }\left(16.00\frac{g}{\mathrm{mol}}\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{261}\mathbf{.}\mathbf{35}\frac{\mathbf{g}}{\mathbf{mol}} \end{gathered}$$

The mass of $\mathbf{Ba}{\mathbf{\left(}{\mathbf{NO}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{2}}$  is multiplied by the reciprocal of its molar mass. We multiply 6 to it because there are 6 moles of O atoms per mole of  $\mathbf{Ba}{\mathbf{\left(}{\mathbf{NO}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{2}}$.

As shown below, the number of moles of O atoms in 8.18 g of  $\mathbf{Ba}{\mathbf{\left(}{\mathbf{NO}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{2}}$ has been computed.

$\mathbf{Moles}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{atoms}\mathbf{ }\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{18}\mathbf{ }\mathbf{g}\mathbf{ }\mathbf{Ba}{\mathbf{\left(}{\mathbf{NO}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\times }\left(\frac{\mathrm{mol} \mathrm{Ba} (\mathrm{NO}}{}\right)$₃)₂261.35 g Ba(NO₃)₂×(6 mol O atoms1 mol Ba NO32)                              =0.188 mol O atoms

Step 4: Determination of Number of O atoms in 7 . 3 × 10 - 3   g of CaSO 4 . 2 H 2 O

c)

The following is the molar mass of ${\mathbf{CaSO}}_{\mathbf{4}}\mathbf{ }\mathbf{.}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{ }\mathbf{.}$.

$$\begin{gathered} {\mathbf{M}}_{\mathbf{\omega }\mathbf{ }}\mathbf{of}\mathbf{ }{\mathbf{CaSO}}_{\mathbf{4}\mathbf{ }}\mathbf{.}\mathbf{ }\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{=}{\mathbf{M}}_{\mathbf{\omega }\mathbf{ }}\mathbf{of}\mathbf{ }\mathbf{Ca}\mathbf{+}{\mathbf{M}}_{\mathbf{\omega }\mathbf{ }}\mathbf{of}\mathbf{ }\mathbf{S}\mathbf{ }\mathbf{+}\mathbf{6}\mathbf{\times }\mathbf{(}{\mathbf{M}}_{\mathbf{\omega }\mathbf{ }}\mathbf{of}\mathbf{ }\mathbf{ }\mathbf{O}\mathbf{)}\mathbf{+}\mathbf{4}\mathbf{\times }\mathbf{(}{\mathbf{M}}_{\mathbf{\omega }\mathbf{ }\mathbf{ }}\mathbf{of}\mathbf{ }\mathbf{H}\mathbf{)} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{40}\mathbf{.}\mathbf{08}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{+}\mathbf{32}\mathbf{.}\mathbf{07}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{+}\mathbf{6}\mathbf{\times }\left(16.00\frac{g}{\mathrm{mol}}\right)\mathbf{+}\mathbf{4}\mathbf{\times }\left(1.01\frac{g}{\mathrm{mol}}\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{172}\mathbf{.}\mathbf{19}\frac{\mathbf{g}}{\mathbf{mol}} \end{gathered}$$

The mass of   is multiplied by the reciprocal of its molar mass, the number of moles of O atoms per mole of   and Avogadro's number.

$$\begin{gathered} \mathbf{Moles}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{atoms}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{g}\mathbf{ }{\mathbf{CaSO}}_{\mathbf{4}}\mathbf{ }\mathbf{.}\mathbf{ }\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\times }\left(\frac{\mathrm{mol} g {\mathrm{CaSO}}_4 . 2H_{2}O}{172.19 g {\mathrm{CaSO}}_4 . 2H_{2}O}\right)\mathbf{\times }\left(\frac{6 \mathrm{mol} O \mathrm{atoms}}{1 \mathrm{mol} {\mathrm{CaSO}}_4 . 2H_{2}O}\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{\times }\left(\frac{6.022\times {10}^{23} O \mathrm{atoms}}{\mathrm{mol} O \mathrm{atoms}}\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\boxed{\mathbf{1}\mathbf{.}\mathbf{53}\mathbf{\times }{\mathbf{10}}^{\mathbf{20}}\mathbf{ }\mathbf{O}\mathbf{ }\mathbf{atoms}} \end{gathered}$$