Question

3.10 Calculate the molar mass of each of the following:

(a)$\mathbf{SnO}$ (b) ${\mathbf{BaF}}_{\mathbf{2}}$ (c) ${\mathbf{AI}}_{\mathbf{2}}{\left({\mathrm{SO}}_4\right)}_{\mathbf{3}}$ (d)${\mathbf{MnCI}}_{\mathbf{2}}$

Short answer

The molar mass of each of the following:

a) $\mathbf{SnO}\mathbf{=}\mathbf{134}\mathbf{.}\mathbf{71}\frac{\mathbf{g}}{\mathbf{mol}}$

b) ${\mathbf{BaF}}_{\mathbf{2}}\mathbf{=}\mathbf{175}\mathbf{.}\mathbf{33}\frac{\mathbf{g}}{\mathbf{mol}}$

c) ${\mathbf{AI}}_{\mathbf{2}}\mathbf{}{\left({\mathrm{SO}}_4\right)}_{\mathbf{3}}\mathbf{=}\mathbf{342}\mathbf{.}\mathbf{17}\frac{\mathbf{g}}{\mathbf{mol}}$

d) ${\mathbf{MnCI}}_{\mathbf{2}}\mathbf{=}\mathbf{125}\mathbf{.}\mathbf{84}\frac{\mathbf{g}}{\mathbf{mol}}$

Step-by-step solution

Relation between mass and number of moles

The molar mass of a material is defined as the mass of the substance contained in one mole of the substance, and it is equal to the total mass of the component atoms. It may be stated mathematically as follows:

$\mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{=}\frac{\mathbf{Weightgrams}}{\mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{moles}}$

Step 2: Calulate the molar mass of SnO

a)

A compound's molar mass (M) is equal to the sum of the molar masses of the atoms in the formula.

The molar mass of are $\mathbf{118}\mathbf{.}\mathbf{71}\frac{\mathbf{g}}{\mathbf{mol}}$and $\mathbf{16}\mathbf{.}\mathbf{00}\frac{\mathbf{g}}{\mathbf{mol}}$respectively.

role="math" localid="1657106505360" $$\begin{gathered} \mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{SnO}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{Sn}\mathbf{+}\mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{O} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{118}\mathbf{.}\mathbf{71}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{+}\mathbf{16}\mathbf{.}\mathbf{00}\frac{\mathbf{g}}{\mathbf{mol}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\boxed{\mathbf{134}\mathbf{.}\mathbf{71}\frac{\mathbf{g}}{\mathbf{mol}}} \end{gathered}$$

Step 3: Calculate the molar mass of BaF2

b)

Here themolar massofBa and F is given as $\mathbf{137}\mathbf{.}\mathbf{33}\frac{\mathbf{g}}{\mathbf{mol}}$and $\mathbf{19}\mathbf{.}\mathbf{00}\frac{\mathbf{g}}{\mathbf{mol}}$.

role="math" localid="1657106836978" $$\begin{gathered} \mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{BaF}}_{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{Ba}\mathbf{+}\mathbf{2}\mathbf{\times }\mathbf{(}\mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{F}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{137}\mathbf{.}\mathbf{33}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{+}\mathbf{2}\mathbf{(}\mathbf{19}\mathbf{.}\mathbf{00}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\boxed{\mathbf{175}\mathbf{.}\mathbf{33}\frac{\mathbf{g}}{\mathbf{mol}}} \end{gathered}$$

Step 4: Determine the molar mass of AI2(SO4)3

c)

Here themolar mass ofAl and S is given as $\mathbf{26}\mathbf{.}\mathbf{98}\frac{\mathbf{g}}{\mathbf{mol}}$and $\mathbf{32}\mathbf{.}\mathbf{07}\frac{\mathbf{g}}{\mathbf{mol}}$

$$\begin{gathered} \mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{SnO}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{2}\mathbf{\times }\mathbf{}\left(\mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{AI}\right)\mathbf{+}\mathbf{3}\mathbf{\times }\left(\mathrm{Molar}\mathrm{mass}\mathrm{of}S\right)\mathbf{+}\mathbf{12}\mathbf{\times }\mathbf{(}\mathbf{Molar}\mathbf{}\mathbf{mass}\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{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{2}\mathbf{\times }\left(26.98\frac{g}{\mathrm{mol}}\right)\mathbf{+}\mathbf{3}\mathbf{\times }\mathbf{(}\mathbf{32}\mathbf{.}\mathbf{07}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{)}\mathbf{+}\mathbf{12}\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{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\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{342}\mathbf{.}\mathbf{17}\frac{\mathbf{g}}{\mathbf{mol}}} \end{gathered}$$

Step 5: Determine the molar mass of MnCI2

d)

Here themolar massofMn and Cl is given as $\mathbf{54}\mathbf{.}\mathbf{94}\frac{\mathbf{g}}{\mathbf{mol}}$and $\mathbf{35}\mathbf{.}\mathbf{45}\frac{\mathbf{g}}{\mathbf{mol}}$

$$\begin{gathered} \mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{MnCI}}_{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{Mn}\mathbf{+}\mathbf{2}\mathbf{\times }\mathbf{(}\mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{CI}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{54}\mathbf{.}\mathbf{94}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{+}\mathbf{2}\mathbf{\times }\mathbf{(}\mathbf{35}\mathbf{.}\mathbf{45}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\boxed{\mathbf{125}\mathbf{.}\mathbf{84}\frac{\mathbf{g}}{\mathbf{mol}}} \end{gathered}$$