Q3.6P - Chemistry: Molecular Nature Of Matter And Change

Question

3.6 You need to calculate the number of ${\mathbf{P}}_{\mathbf{4}}$ molecules that can form from 2.5 g of ${\mathbf{Ca}}_{\mathbf{3}}{\left({\mathrm{PO}}_4\right)}_{\mathbf{2}}$ Explain how you would proceed. (That is, write a solution "Plan," without actually doing any calculations.)

Step-by-Step Solution

Answer

The number of ${\mathbf{P}}_{\mathbf{4}}$ molecules that can form from 2.5 g of ${\mathbf{Ca}}_{\mathbf{3}}{\mathbf{\left(}{\mathbf{PO}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{2}}$is calculated as follows,

$$\begin{gathered} \left(2.5 g C_{a_3}{\left({\mathrm{PO}}_4\right)}_2\right)\mathbf{\times }\left(\frac{\mathrm{mol} {\mathrm{Ca}}_3 {\left({\mathrm{PO}}_4\right)}_2}{\mathrm{molar} \mathrm{mass} \mathrm{in} g {\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2}\right)\mathbf{\times }\left(\frac{2 \mathrm{mol} P \mathrm{atoms}}{1 \mathrm{mol} {\mathrm{Ca}}_3 {\left({\mathrm{PO}}_4\right)}_2}\right)\mathbf{\times }\left(\frac{1 \mathrm{mol} P_4 \mathrm{molecules}}{4 \mathrm{mol} P \mathrm{atoms}}\right) \\ \mathbf{=}\mathbf{\times }\left(\frac{\mathrm{Avogadro}\text{'}s \mathrm{number} P_4 \mathrm{molecules}}{\mathrm{mol} P_4 \mathrm{molecules}}\right) \end{gathered}$$

Step 1: Relation between mass and mole of the substance

The mass of a substance made up of an equal number of fundamental units is defined as a mole.

Step 2: Solution Explanation

To find the number of moles of  divide the supplied mass by the compound's molar mass. We might simply multiply the provided mass by the reciprocal of the molar mass for a simple demonstration.

$\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{ }\mathbf{g}\mathbf{ }{\mathbf{C}}_{{\mathbf{a}}_{\mathbf{3}}}{\mathbf{\left(}{\mathbf{PO}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{2}}\mathbf{)}\mathbf{\times }\mathbf{\left(}\frac{\mathbf{mol}\mathbf{ }{\mathbf{Ca}}_{\mathbf{3}}\mathbf{ }{\mathbf{\left(}{\mathbf{PO}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{2}}}{\mathbf{molar}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{in}\mathbf{ }\mathbf{g}\mathbf{ }{\mathbf{Ca}}_{\mathbf{3}}{\mathbf{\left(}{\mathbf{PO}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{2}}}\mathbf{\right)}$

By chemical formula, 2 moles of P atoms in a mole of  ${\mathbf{Ca}}_{\mathbf{3}}{\mathbf{\left(}{\mathbf{PO}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{2}}$

$\left(\mathrm{mol} {\mathrm{Ca}}_3 {\left({\mathrm{PO}}_4\right)}_2\right)$$\mathbf{\times }\mathbf{\left(}\frac{\mathbf{2}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{P}\mathbf{ }\mathbf{atoms}}{\mathbf{1}\mathbf{ }\mathbf{mol}\mathbf{ }{\mathbf{Ca}}_{\mathbf{3}}\mathbf{ }{\mathbf{\left(}{\mathbf{PO}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{2}}}\mathbf{\right)}$

As there are 4 moles of P atoms in a atoms ${\mathbf{P}}_{\mathbf{4}}$ atoms

$\left(\mathrm{mol} P \mathrm{atoms}\right)$ $\mathbf{\times }\mathbf{\left(}\frac{\mathbf{1}\mathbf{ }\mathbf{mol}\mathbf{ }{\mathbf{P}}_{\mathbf{4}}\mathbf{ }\mathbf{molecules}}{\mathbf{4}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{P}\mathbf{ }\mathbf{atoms}}\mathbf{\right)}$

Finally, we calculate the number of ${\mathit{P}}_{\mathit{4}}$ molecules by multiplying the number of moles of  ${\mathit{P}}_{\mathit{4}}$ molecules by Avogadro's number.

width="111" height="20" style="max-width: none;" $\mathbf{(}\mathbf{mol}\mathbf{ }\mathbf{P}\mathbf{ }\mathbf{atoms}\mathbf{)}$ $\mathbf{\times }\mathbf{\left(}\frac{\mathbf{Avogadro}\mathbf{\text{'}}\mathbf{s}\mathbf{ }\mathbf{number}\mathbf{ }{\mathbf{P}}_{\mathbf{4}}\mathbf{ }\mathbf{molecules}}{\mathbf{mol}\mathbf{ }{\mathbf{P}}_{\mathbf{4}}\mathbf{ }\mathbf{molecules}}\mathbf{\right)}$

The plan's summary is presented below,

$$\begin{gathered} \mathbf{(}\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{ }\mathbf{g}\mathbf{ }{\mathbf{C}}_{{\mathbf{a}}_{\mathbf{3}}}{\mathbf{\left(}{\mathbf{PO}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{2}}\mathbf{)}\mathbf{\times }\mathbf{\left(}\frac{\mathbf{mol}\mathbf{ }{\mathbf{Ca}}_{\mathbf{3}}\mathbf{ }{\mathbf{\left(}{\mathbf{PO}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{2}}}{\mathbf{molar}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{in}\mathbf{ }\mathbf{g}\mathbf{ }{\mathbf{Ca}}_{\mathbf{3}}{\mathbf{\left(}{\mathbf{PO}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{2}}}\mathbf{\right)}\mathbf{\times }\mathbf{\left(}\frac{\mathbf{2}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{P}\mathbf{ }\mathbf{atoms}}{\mathbf{1}\mathbf{ }\mathbf{mol}\mathbf{ }{\mathbf{Ca}}_{\mathbf{3}}\mathbf{ }{\mathbf{\left(}{\mathbf{PO}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{2}}}\mathbf{\right)}\mathbf{\times }\mathbf{\left(}\frac{\mathbf{1}\mathbf{ }\mathbf{mol}\mathbf{ }{\mathbf{P}}_{\mathbf{4}}\mathbf{ }\mathbf{molecules}}{\mathbf{4}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{P}\mathbf{ }\mathbf{atoms}}\mathbf{\right)} \\ \mathbf{=}\mathbf{\times }\mathbf{\left(}\frac{\mathbf{Avogadro}\mathbf{\text{'}}\mathbf{s}\mathbf{ }\mathbf{number}\mathbf{ }{\mathbf{P}}_{\mathbf{4}}\mathbf{ }\mathbf{molecules}}{\mathbf{mol}\mathbf{ }{\mathbf{P}}_{\mathbf{4}}\mathbf{ }\mathbf{molecules}}\mathbf{\right)} \end{gathered}$$