Chapter 2, Problem 2.92QE

(a)

Interpretation Introduction

Interpretation:

Formula mass of $\text{NaOH}$ has to be calculated.

Concept Introduction:

Formula mass is similar to molecular mass of compound.  Formula mass is the sum of atomic mass of all the atoms present in empirical formula of ionic compound.  In case of ionic compounds, the relative number is only taken and not the exact formula.  Hence, formula mass is used.

Explanation of Solution

Given compound is $\text{NaOH}$.  This contains one sodium cation and a hydroxide anion.  One formula unit of $\text{NaOH}$ contains one sodium atom, one hydrogen atom, and one oxygen atom.  The formula mass can be calculated as shown below.

  $\begin{array}{c}\text{Formula}\text{mass}\text{of}\text{NaOH}=1(\text{at}\text{.}\text{mass}\text{of}\text{Na})+1(\text{at}\text{.}\text{mass}\text{of}\text{O)+4(at}\text{.}\text{mass}\text{of}\text{H})\\ =1(22.99u)+1(16.00\text{u)+1(101}\text{u)}\\ \text{=22}\text{.99 u + 16}\text{.00 u + 1}\text{.01 u}\\ \text{=40}\text{.00u}\end{array}$

Therefore, formula mass of $\text{NaOH}$ is $\text{40}\text{.00}\text{u}$.

(b)

Interpretation Introduction

Interpretation:

Formula mass of ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ has to be calculated.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given compound is ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$.  This contains two potassium cation and a carbonate anion.  One formula unit of ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ contains two potassium atoms, one carbon atom, and three oxygen atoms.  The formula mass can be calculated as shown below.

  $\begin{array}{c}\text{Formula}\text{mass}\text{of}{\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}=2(\text{at}\text{.}\text{mass}\text{of}\text{K})+1(\text{at}\text{.}\text{mass}\text{of}\text{C)+3(at}\text{.}\text{mass}\text{of}\text{O})\\ =2(39.10u)+1(12.01\text{u)+3(16}\text{.00}\text{u)}\\ \text{=78}\text{.20 u + 12}\text{.01 u + 48}\text{.00 u}\\ \text{=138}\text{.21}\text{u}\end{array}$

Therefore, formula mass of ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ is $\text{138}\text{.21}\text{u}$.

(c)

Interpretation Introduction

Interpretation:

Formula mass of ${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ has to be calculated.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given compound is ${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$.  This contains two phosphate anion and three calcium cation.  One formula unit of ${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ contains two calcium atom, two phosphorus atom and eight oxygen atoms.  The formula mass can be calculated as shown below.

  $\begin{array}{c}\text{Formula}\text{mass}\text{of}{\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}}_{\text{)}}=3(\text{at}\text{.}\text{mass}\text{of}\text{Ca})+1(\text{at}\text{.}\text{mass}\text{of}\text{P)+8(at}\text{.}\text{mass}\text{of}\text{O})\\ =3(40.08u)+2(30.97\text{u)+8(16}\text{.00}\text{u)}\\ \text{=120}\text{.24 u + 61}\text{.94 u + 128}\text{.00 u}\\ \text{=310}\text{.18}\text{u}\end{array}$

Therefore, formula mass of ${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ is $\text{310}\text{.18}\text{u}$.