Chapter 2, Problem 2.156QP

Interpretation Introduction

Interpretation:

Why the terms “molecules” and “formula units” are used for the two substances has to be explained and the other questions regarding the mixture needs to be answered.

Explanation of Solution

Molecule is the term that is used for compounds which contain only covalent bonding between the atoms present in it.  Formula unit is the term used for compounds which have ionic bonding which holds the anion and cation together.

In this case, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ is a molecule because there is only covalent bonding between the atoms of non-metals to give a unique molecular structure.  ${\text{Pb}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{4}}·6{\text{H}}_{\text{2}}\text{O}$ is a formula unit because this has ionic bonding where the lead(IV) cations and phosphate anions are held together.  When the dehydrated form is considered, it can be represented in a infinite number of ways.  Dehydrated form is ${\text{Pb}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{4}}$.  It can be represented as, ${\text{Pb}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{4}}$, ${{\text{(Pb}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{4}})}_2$, ${{\text{(Pb}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{4}})}_8$, ${{\text{(Pb}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{4}})}_{50,000}$, etc.  This is because, the crystal is viewed as an extended array of ions and not as distinct molecules.  The ionic compounds are represented by their simplest or empirical formula.  This is known as formula unit.  Hence, option (a) is the correct one.  The other options (b) and (c) are incorrect options.

The total atoms that is present in the mixture can be found by counting the atoms in a single molecule and formula unit.  The total number of atoms that is present in ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ is 12 atoms.  The total number of atoms present in ${\text{Pb}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{4}}·6{\text{H}}_{\text{2}}\text{O}$ is 41 atoms.  It is given that the mixture contains 4 molecules of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ and 5 formula units of ${\text{Pb}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{4}}·6{\text{H}}_{\text{2}}\text{O}$.  Hence, the total number of atoms in mixture can be calculated as shown,

$\begin{array}{l}\text{4}{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}\text{molecules × }\frac{\text{12 atoms}}{\text{1 molecule}}+5{\text{Pb}}_{\text{3}}{{\text{(PO}}_{\text{4}}}_{\text{)}}·6{\text{H}}_{\text{2}}\text{O}\text{f}\text{.u}\times \frac{\text{41 atoms}}{\text{1 f}\text{.u}}\\ \text{= 253 atoms}\end{array}$

The total number of atoms in the given mixture was found as 253 atoms.

The total number of protons that is present in the mixture can be found identifying the protons that is present in each atom.  Carbon has 6 protons, hydrogen has 1 proton, oxygen contains 8 protons, Lead contains 82 protons, phosphorous contains 15 protons.  The total number of atoms present in mixture is 12 Carbon atoms, 92 Hydrogen atoms, 8 Oxygen atoms, 15 Lead atoms, 20 Phosphorous atoms.  Hence, the total number of protons in the mixture is calculated and found to be 2606 protons.

The total number of electrons that is present in the mixture can be found as shown similar to the total number of protons.  As the substances are electrically neutral, the total number of protons and electrons must be equal.  Hence, the mixture contains 2606 electrons.

The theoretical maximum number of ${\text{O}}_{\text{3}}$ molecules that can be formed from the mixture can be found by calculating the total number of oxygen atoms that is present in the mixture and dividing the same by 3.

$\begin{array}{l}\text{4}{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}\text{molecules × }\frac{\text{1 O atom}}{\text{1 molecule}}+5{\text{Pb}}_{\text{3}}{{\text{(PO}}_{\text{4}}}_{\text{)}}·6{\text{H}}_{\text{2}}\text{O}\text{f}\text{.u}\times \frac{\text{22 O atoms}}{\text{1 f}\text{.u}}\\ \text{= 114 O atoms}\end{array}$

The total number of oxygen atoms that is present in the mixture was found as 114.  Each ${\text{O}}_{\text{3}}$ molecule contains three oxygen atoms.  Hence, 114 oxygen atoms would give 38 ${\text{O}}_{\text{3}}$ molecules.

$\text{114 O atoms }\times \frac{{\text{1 O}}_{\text{3}}\text{ molecule}}{\text{3 O atoms}}{\text{ = 38 O}}_{\text{3}}\text{ molecules}$.

The symbol for potassium is $\text{K}$.  The total number of potassium atoms that is present in the mixture is zero.  This is because, both the substance that is present in the mixture does not contain potassium atom.

The total number of lead atoms in the mixture can be found by considering how many formula units are present in the mixture.  Each formula unit contains three lead atoms.  As the mixture contains five formula units, the total number of lead atoms, that is present in the mixture is 15.

The name of the formula unit can be given considering the cation and anion.  The hydrate is named by considering how many times that water molecule is present in the formula unit.  The cation is named showing the oxidation state of the element in Roman letter followed by the anion and hydrate.  Hence, the name of ${\text{Pb}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{4}}·6{\text{H}}_{\text{2}}\text{O}$ can be given as Lead(IV) phosphate hexahydrate.

Oxoanion that is present in ${\text{Pb}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{4}}·6{\text{H}}_{\text{2}}\text{O}$ is the phosphate ion.  The phosphate ion has “-3” charge on it.  Hence, the oxoacid from which it is formed can be found by adding three hydrogen atoms.  The formula of the oxoacid was found as ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$.  The name of oxoacid can be given by replacing “-ate” with “-ic” followed by the word acid.  Hence, the name of oxoacid can be given as Phosphoric acid.