Q105P - Chemistry: Molecular Nature Of Matter And Change

Question

Give the number of atoms of the specified element in a formula unit of each of the following compounds, and calculate the molecular(formula) mass:

$$\begin{gathered} \mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{ }\mathbf{Hydrogen}\mathbf{ }\mathbf{in}\mathbf{ }\mathbf{ammonium}\mathbf{ }\mathbf{banzoate}\mathbf{,}\mathbf{ }{\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{5}}{\mathbf{COONH}}_{\mathbf{4}} \\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{ }\mathbf{Nitrogen}\mathbf{ }\mathbf{in}\mathbf{ }\mathbf{hydrazinium}\mathbf{ }\mathbf{sulfate}\mathbf{,}\mathbf{ }{\mathbf{N}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{6}}{\mathbf{O}}_{\mathbf{4}} \\ \mathbf{ }\mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{ }\mathbf{Oxygen}\mathbf{ }\mathbf{in}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{mineral}\mathbf{ }\mathbf{leadhillite}\mathbf{,}\mathbf{ }{\mathbf{Pb}}_{\mathbf{4}}{\mathbf{SO}}_{\mathbf{4}}{\left({\mathrm{CO}}_3\right)}_{\mathbf{2}}{\left(\mathrm{OH}\right)}_{\mathbf{2}} \end{gathered}$$

Step-by-Step Solution

Answer

(a) The Number of atoms is 9H atoms, and the Molecular mass is 139.14 amu.

(b) The Number of atoms is 2Natoms, and the Molecular mass is 130.09 amu.

(c) The Number of atoms is 12O atoms, and the Molecular mass is 1078.79 amu

Step 1:Solving an i.e., ammonium benzoate, C 6 H 5 COONH 4

It is clear from the molecular formula that there are 9 H-atoms in ammonium benzoate, ${\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{5}}{\mathbf{COONH}}_{\mathbf{4}}\mathbf{.}$

List the atomic mass of each of the elements present :

$\mathbf{C}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{ }\mathbf{u}\mathbf{,}\mathbf{H}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{ }\mathbf{u}\mathbf{,}\mathbf{O}\mathbf{=}\mathbf{15}\mathbf{.}\mathbf{99}\mathbf{ }\mathbf{u}\mathbf{,}\mathbf{ }\mathbf{N}\mathbf{=}\mathbf{14}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{u}$

Then add the atomic masses of the constituent atoms.

Molecular mass can be calculated as:

$$\begin{gathered} \mathbf{=}\mathbf{7}\mathbf{\times }\mathbf{atomic}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{C}\mathbf{+}\mathbf{9}\mathbf{\times }\mathbf{atomic}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{H}\mathbf{+}\mathbf{2}\mathbf{\times }\mathbf{atomic}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{O} \\ \mathbf{+}\mathbf{atomic}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{N} \\ \mathbf{=}\mathbf{7}\mathbf{\times }\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{+}\mathbf{9}\mathbf{\times }\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{+}\mathbf{2}\mathbf{\times }\mathbf{15}\mathbf{.}\mathbf{99}\mathbf{+}\mathbf{14}\mathbf{.}\mathbf{00} \\ Molecular mass\mathbf{=}\mathbf{139}\mathbf{.}\mathbf{14}\mathbf{ }\mathbf{amu} \end{gathered}$$

Step 2: Solving b i.e., hydrazinium sulfate, N 2 H 6 SO 4

There are 2 N-atoms in ${\mathbf{N}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{6}}{\mathbf{SO}}_{\mathbf{4}}$.

List the atomic mass of each of the elements present.

$\mathbf{N}\mathbf{=}\mathbf{14}\mathbf{.}\mathbf{00}\mathbf{u}\mathbf{,}\mathbf{ }\mathbf{H}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{u}\mathbf{,}\mathbf{ }\mathbf{S}\mathbf{=}\mathbf{32}\mathbf{.}\mathbf{07}\mathbf{u}\mathbf{,}\mathbf{ }\mathbf{O}\mathbf{=}\mathbf{15}\mathbf{.}\mathbf{99}\mathbf{u}$

Then add the atomic masses of the constituent atoms to get the molecular mass.

Molecular mass

$\mathbf{=}\mathbf{2}\mathbf{\times }\mathbf{Atomic}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{N}\mathbf{+}\mathbf{6}\mathbf{\times }\mathbf{Atomic}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{H}\mathbf{+}\mathbf{ }\mathbf{Atomic}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{S}\mathbf{+}\mathbf{4}\mathbf{\times }\mathbf{ }\mathbf{Atomic}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{O}$

Molecular mass $\mathbf{=}\mathbf{130}\mathbf{.}\mathbf{09}\mathbf{ }\mathbf{amu}\mathbf{.}$

Step 3: Solving c i.e., mineral leadhillite, Pb 4 SO 4 ( CO 3 ) 2 ( OH ) 2

There are 4 oxygen atoms in Sulfate and there are also 3 oxygen atoms in each carbonate and there is 2 carbonate. There is also  1oxygen atom in each hydroxide ion and there are 2 hydroxide ions in the formula.

Overall, we have in $\mathbf{4}\mathbf{+2\times 3+2\times 1=12 }\mathbf{O}\mathbf{-}\mathbf{atoms}\mathbf{ }{\mathbf{Pb}}_{\mathbf{4}}{\mathbf{SO}}_{\mathbf{4}}{\left({\mathrm{CO}}_3\right)}_{\mathbf{2}}{\left(\mathrm{OH}\right)}_{\mathbf{2}}\mathbf{.}$

List the atomic mass of each of the elements present.

$\mathbf{Pb}\mathbf{=}\mathbf{207}\mathbf{.}\mathbf{20}\mathbf{u}\mathbf{,}\mathbf{ }\mathbf{S}\mathbf{=}\mathbf{32}\mathbf{.}\mathbf{07}\mathbf{u}\mathbf{,}\mathbf{ }\mathbf{O}\mathbf{=}\mathbf{15}\mathbf{.}\mathbf{99}\mathbf{u}\mathbf{,}\mathbf{ }\mathbf{C}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{u}\mathbf{,}\mathbf{ }\mathbf{H}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{ }\mathbf{u}$

Then add the atomic masses of the constituent atoms.

Molecular mass

$$\begin{gathered} \mathbf{=}\mathbf{4}\mathbf{ }\mathbf{Atomic}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Pb}\mathbf{+}\mathbf{ }\mathbf{Atomic}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{S}\mathbf{+}\mathbf{12}\mathbf{ }\mathbf{Atomic}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{O} \\ \mathbf{+}\mathbf{2}\mathbf{ }\mathbf{Atomic}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{C}\mathbf{+}\mathbf{2}\mathbf{Atomic}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{H} \\ \mathbf{=}\mathbf{4}\mathbf{(}\mathbf{207}\mathbf{.}\mathbf{20}\mathbf{)}\mathbf{+}\mathbf{32}\mathbf{.}\mathbf{07}\mathbf{+}\mathbf{12}\mathbf{(}\mathbf{15}\mathbf{.}\mathbf{99}\mathbf{)}\mathbf{+}\mathbf{2}\mathbf{(}\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{)}\mathbf{+}\mathbf{2}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{)} \end{gathered}$$

Molecular mass $\mathbf{=}\mathbf{1078}\mathbf{.}\mathbf{79}\mathbf{ }\mathbf{amu}\mathbf{.}$