Question

Write the formula of each compound, and determine its molecular (formula) mass:

a) Sodium dichromate;

b) Ammonium Perchlorate;

c) Magnesium nitrate trihydrate.

Short answer

(a) Theformula of the Sodium dichromate is${\mathbf{Na}}_{\mathbf{2}}{\mathbf{Cr}}_{\mathbf{2}}{\mathbf{o}}_{\mathbf{7}}$andMolecular mass is261.98 amu.

(b) the formula of the Ammonium Perchlorate is${\mathbf{NH}}_{\mathbf{4}}{\mathbf{CIO}}_{\mathbf{4}}$and molecular mass is 117.49 amu.

(c) the formula of the Magnesium nitrate trihydrate is $\mathbf{Mg}{\left({\mathrm{NO}}_2\right)}_{\mathbf{2}}\mathbf{.}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$and molecular mass is 170.37 amu.

Step-by-step solution

To Write the formula and calculate the molecular mass for sodium dichromate

Sodium is a group 1 element with a charge of +1. Dichromate is a polyatomic ion with the formula ${\mathrm{Cr}}_2{\mathrm{O}}_7^2$. The charges of each ion will just interchange as subscripts to give its molecular formula.

${\mathbf{Na}}^{\mathbf{+}}{\mathbf{Cr}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{7}}^{\mathbf{2}\mathbf{-}}\mathbf{\to }{\mathbf{Na}}_{\mathbf{2}}{\mathbf{Cr}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{7}}$

List the atomic mass of each of the elements present,

$\mathbf{Na}\mathbf{=}\mathbf{22}\mathbf{.}\mathbf{99}\mathbf{}\mathbf{u}\mathbf{,}\mathbf{}\mathbf{Cr}\mathbf{=}\mathbf{52}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{u}\mathbf{,}\mathbf{}\mathbf{O}\mathbf{=}\mathbf{16}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{u}$

Then add the atomic masses of the constituent atoms to find out the molecular formula.

Molecular mass

$$\begin{gathered} \mathbf{=}\mathbf{2}\mathbf{\times }\mathbf{Atomic}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{Na}\mathbf{+}\mathbf{2}\mathbf{\times }\mathbf{Atomic}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{Cr}\mathbf{+}\mathbf{7}\mathbf{\times }\mathbf{Atomic}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{O} \\ \mathbf{=}\mathbf{2}\left(22.99\right)\mathbf{+}\mathbf{2}\left(52.00\right)\mathbf{+}\mathbf{7}\left(16.00\right) \end{gathered}$$

Molecular massamu$\mathbf{=}\mathbf{261}\mathbf{.}\mathbf{98}\mathbf{}\mathbf{amu}\mathbf{}\mathbf{amu}$.

To Write the formula and calculate the molecular formula for ammonium Perchlorate

Ammonium is a polyatomic cation with the formula${\mathbf{NH}}_{\mathbf{4}}^{\mathbf{+}}$.

Perchlorate is a polyatomic anion with the formula${\mathbf{CIO}}_{\mathbf{4}}^{\mathbf{-}}$. The charges of each ion will just interchange as subscripts of its molecular formula.

${\mathbf{NH}}_{\mathbf{4}}^{\mathbf{+}}{\mathbf{CIO}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{\to }{\mathbf{NH}}_{\mathbf{4}}{\mathbf{CIO}}_{\mathbf{4}}$

List the atomic mass of each of the elements present.

$\mathbf{N}\mathbf{=}\mathbf{14}\mathbf{.}\mathbf{01}\mathbf{u}\mathbf{}\mathbf{}\mathbf{H}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{u}\mathbf{}\mathbf{}\mathbf{CI}\mathbf{=}\mathbf{35}\mathbf{.}\mathbf{45}\mathbf{}\mathbf{}\mathbf{O}\mathbf{=}\mathbf{16}\mathbf{.}\mathbf{00}\mathbf{u}$

Then add the atomic masses of the constituent atoms.

Molecular mass

$$\begin{gathered} \mathbf{Atomic}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{N}\mathbf{+}\mathbf{4}\mathbf{\times }\mathbf{Atomic}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{H}\mathbf{+}\mathbf{Atomic}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{CI}\mathbf{+}\mathbf{4}\mathbf{\times }\mathbf{Atomic}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{O} \\ \mathbf{=}\mathbf{14}\mathbf{.}\mathbf{01}\mathbf{+}\mathbf{4}\left(1.01\right)\mathbf{+}\mathbf{35}\mathbf{.}\mathbf{45}\mathbf{+}\mathbf{4}\left(16.00\right) \\ \mathrm{Molecular}\mathrm{mass}\mathbf{=}\mathbf{118}\mathbf{.}\mathbf{49}\mathbf{}\mathbf{amu} \end{gathered}$$

Molecular mass.

To Write the formula and calculate the molecular formula for Magnesium nitrate trihydrate

Magnesium is a group II element with a charge of +2. Nitrate is a polyatomic anion with the formula${\mathbf{NO}}_{\mathbf{2}}^{\mathbf{-}}$.

The Charges of each ion will just interchange as subscripts of its molecular formula.

${\mathbf{Mg}}^{\mathbf{2}\mathbf{+}}{\mathbf{NO}}_{\mathbf{2}}^{\mathbf{-}}\mathbf{\to }\mathbf{Mg}{\left({\mathrm{NO}}_2\right)}_{\mathbf{2}}$

The substance also has three hydrates from the prefix tri, which we can just attach$3{\mathrm{H}}_2\mathrm{O}$at the end of the formula.

So we have the molecular formula as$\mathbf{Mg}{\left({\mathrm{NO}}_2\right)}_{\mathbf{2}}\mathbf{.}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$

List the atomic mass of each of the elements present.

$\mathbf{Mg}\mathbf{=}\mathbf{24}\mathbf{.}\mathbf{31}\mathbf{u}\mathbf{}\mathbf{N}\mathbf{=}\mathbf{14}\mathbf{.}\mathbf{00}\mathbf{u}\mathbf{}\mathbf{O}\mathbf{=}\mathbf{16}\mathbf{.}\mathbf{00}\mathbf{u}\mathbf{}\mathbf{H}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{u}$

Then add the atomic masses of the constituent atoms.

Molecular mass

localid="1656652877966" $$\begin{gathered} \mathbf{=}\mathbf{Atomic}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{mg}\mathbf{+}\mathbf{2}\mathbf{\times }\mathbf{Atomic}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{N}\mathbf{+}\mathbf{7}\mathbf{\times }\mathbf{Atomic}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{O}\mathbf{+}\mathbf{6}\mathbf{\times }\mathbf{Atomic}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{H} \\ \mathbf{=}\mathbf{24}\mathbf{.}\mathbf{31}\mathbf{+}\mathbf{2}\mathbf{\times }\mathbf{14}\mathbf{.}\mathbf{00}\mathbf{+}\mathbf{7}\mathbf{\times }\mathbf{16}\mathbf{.}\mathbf{00}\mathbf{+}\mathbf{6}\mathbf{\times }\mathbf{1}\mathbf{.}\mathbf{01} \end{gathered}$$

Molecular mass$\mathbf{=}\mathbf{170}\mathbf{.}\mathbf{37}\mathbf{}\mathbf{amu}$.