Question

Calculate the molar mass of each of the following:

(a)${\left({\mathrm{NH}}_4\right)}_{\mathbf{3}}\mathbf{}{\mathbf{PO}}_{\mathbf{4}}$ (b)${\mathbf{CH}}_{\mathbf{2}}{\mathbf{CI}}_{\mathbf{2}}$ (c)${\mathbf{CuSO}}_{\mathbf{4}}\mathbf{.}\mathbf{5}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ (d)${\mathbf{BrF}}_{\mathbf{3}}$

Short answer

The molar massof each of the following:

a)${\left({\mathrm{NH}}_4\right)}_{\mathbf{3}}{\mathbf{PO}}_{\mathbf{4}}\mathbf{=}\mathbf{149}\mathbf{.}\mathbf{12}\frac{\mathbf{g}}{\mathbf{mol}}$

b)${\mathbf{CH}}_{\mathbf{2}}{\mathbf{CI}}_{\mathbf{2}}\mathbf{=}\mathbf{84}\mathbf{.}\mathbf{93}\frac{\mathbf{g}}{\mathbf{mol}}$

c)${\mathbf{CuSO}}_{\mathbf{4}}\mathbf{.}\mathbf{5}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{=}\mathbf{249}\mathbf{.}\mathbf{72}\frac{\mathbf{g}}{\mathbf{mol}}$

d)${\mathbf{BrF}}_{\mathbf{3}}\mathbf{=}\mathbf{136}\mathbf{.}\mathbf{90}\frac{\mathbf{g}}{\mathbf{mol}}$

Step-by-step solution

Step 1: Relation between mass and number of moles

Themolar mass of a materialis defined as the mass of the substance contained inone mole of the substance, and it isequal to the total massof the component atoms. It may be stated mathematically as follows:

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

Step 2: Calculate the molar mass of (NH4)3 PO4

a)

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

N, H, P, and O have molar masses of$\mathbf{14}\mathbf{.}\mathbf{01}\mathbf{,}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{,}\mathbf{30}\mathbf{.}\mathbf{97}\mathbf{}\mathbf{and}\mathbf{}\mathbf{16}\mathbf{.}\mathbf{00}\frac{\mathbf{g}}{\mathbf{mol}}$ respectively.
role="math" localid="1657109135688" $$\begin{gathered} \mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{\left(}\mathbf{N}{\mathbf{H}}_{\mathbf{1}}\mathbf{\right)}}_{\mathbf{3}}\mathit{P}{\mathit{O}}_{\mathbf{4}}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{3}\mathbf{\times }\mathbf{}\mathbf{(}\mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{N}\mathbf{)}\mathbf{+}\mathbf{12}\mathbf{\times }\mathbf{(}\mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{H}\mathbf{)}\mathbf{+}\mathbf{(}\mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{P}\mathbf{)}\mathbf{+}\mathbf{4}\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{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{3}\mathbf{\times }\mathbf{(}\mathbf{}\mathbf{14}\mathbf{.}\mathbf{01}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{)}\mathbf{+}\mathbf{12}\mathbf{\times }\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{01}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{)}\mathbf{+}\mathbf{(}\mathbf{30}\mathbf{.}\mathbf{97}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{)}\mathbf{+}\mathbf{4}\mathbf{\times }\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{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\boxed{\mathbf{149}\mathbf{.}\mathbf{12}\frac{\mathbf{g}}{\mathbf{mol}}} \end{gathered}$$

Step 3: Calculate the molar mass of CH2CI2

b)

Here themolar mass ofC and Cl is given as $\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{}\mathbf{and}\mathbf{}\mathbf{35}\mathbf{.}\mathbf{45}\frac{\mathbf{g}}{\mathbf{mol}}$

role="math" localid="1657109106142" $$\begin{gathered} \mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{CH}}_{\mathbf{2}}{\mathbf{CI}}_{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{C}\mathbf{+}\mathbf{2}\mathbf{\times }\mathbf{(}\mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{H}\mathbf{)}\mathbf{+}\mathbf{2}\mathbf{\times }\mathbf{(}\mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathit{C}\mathit{I}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{12}\mathbf{.}\mathbf{01}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{+}\mathbf{2}\mathbf{\times }\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{01}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{)}\mathbf{+}\mathbf{2}\mathbf{\times }\mathbf{(}\mathbf{35}\mathbf{.}\mathbf{45}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\boxed{\mathbf{84}\mathbf{.}\mathbf{93}\frac{\mathbf{g}}{\mathbf{mol}}} \end{gathered}$$

Step 4: Calculate the molar mass of CuSO4.5H2O

c)

Cu and S have molar masses of $\mathbf{63}\mathbf{.}\mathbf{955}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\mathbf{32}\mathbf{.}\mathbf{07}\frac{\mathbf{g}}{\mathbf{mol}}$respectively.

role="math" localid="1657109605476" $$\begin{gathered} \mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathit{C}\mathit{u}\mathit{S}{\mathit{O}}_{\mathbf{4}}\mathbf{.}\mathbf{5}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathit{C}\mathit{u}\mathbf{+}\mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathit{S}\mathbf{+}\mathbf{9}\mathbf{\times }\mathbf{(}\mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{O}\mathbf{)}\mathbf{+}\mathbf{(}\mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathit{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{}\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{32}\mathbf{.}\mathbf{07}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{+}\mathbf{9}\mathbf{\times }\mathbf{(}\mathbf{16}\mathbf{.}\mathbf{00}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{)}\mathbf{+}\mathbf{(}\mathbf{16}\mathbf{.}\mathbf{00}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{)}\mathbf{+}\mathbf{10}\mathbf{\times }\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{01}\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{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\boxed{\mathbf{249}\mathbf{.}\mathbf{72}\frac{\mathbf{g}}{\mathbf{mol}}} \end{gathered}$$

Step 5: Determine the molar mass of BrF3

d)

Here the molar massofBr and F is given as$\mathbf{79}\mathbf{.}\mathbf{90}\mathbf{}\mathbf{and}\mathbf{}\mathbf{19}\mathbf{.}\mathbf{00}\frac{\mathbf{g}}{\mathbf{mol}}$

$$\begin{gathered} \mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{BrF}}_{\mathbf{3}}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{Br}\mathbf{+}\mathbf{3}\mathbf{\times }\left(\mathrm{Molar}\mathrm{mass}\mathrm{of}F\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{79}\mathbf{.}\mathbf{90}\frac{\mathbf{g}}{\mathbf{mol}}\mathbf{+}\mathbf{3}\mathbf{\times }\left(19.00\frac{g}{\mathrm{mol}}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\boxed{\mathbf{136}\mathbf{.}\mathbf{90}\frac{\mathbf{g}}{\mathbf{mol}}} \end{gathered}$$