Chapter 12, Problem 68AE

(a)

Interpretation Introduction

Interpretation:

Empirical formula of compound that contains $80.0\text{ \% C}$ and $20.0\text{ \% H}$ has to be determined.

Concept Introduction:

The steps required for calculation of empirical formula is as follows:

1 Convert the data provided of each element in grams if not given in grams. If data is provided in percentage than total mass of compound is taken as $100\text{ g}$.

2 Mass of each element is changed into moles by division of given mass of each element with its atomic mass respectively.

3 The value of mole of each element is divided by the smallest mole value and round off the resultant value to positive integer.

4 If resultant value is in fraction then multiply with smallest number to make it integer.

5 The final value of mole is used as a subscript for empirical formula.

Answer to Problem 68AE

Empirical formula of compound is ${\text{CH}}_3$.

Explanation of Solution

Compound contains $80.0\text{ \% C}$ and $20.0\text{ \% H}$. Therefore the empirical formula is calculated as follows:

Step 1: The percentage $"\%"$ is replaced by $"\text{g"}$. Thus, $100\text{ g}$ compound contain $80.0\text{ g C}$ and $20.0\text{ g H}$.

Step 2: The expression to calculate moles of elements is as follows:

  $\text{ Moles}=\frac{\text{given mass}}{\text{atomic mass}}$

The formula to calculate moles of $\text{C}$ is as follows:

  $\text{ Moles}\left(\text{C}\right)=\frac{\text{mass of C}}{\text{atomic mass of C}}$        (1)

Substitute $80.0\text{ g}$ for mass of $\text{C}$ and $12.01\text{ g/mol}$ for atomic mass of $\text{C}$ in equation (1).

  $\begin{array}{c}\text{ Moles}\left(\text{Fe}\right)=\frac{80.0\text{ g}}{12.01\text{ g/mol}}\\ =6.7\text{ mol}\end{array}$

The formula to calculate moles of $\text{H}$ is as follows:

  $\text{ Moles}\left(\text{H}\right)=\frac{\text{mass of H}}{\text{atomic mass of H}}$        (2)

Substitute $20.0\text{ g}$ for mass of $\text{H}$ and $1.0\text{ g/mol}$ for atomic mass of $\text{H}$ in equation (2).

  $\begin{array}{c}\text{ Moles}\left(\text{H}\right)=\frac{20.0\text{ g}}{1.0\text{ g/mol}}\\ =20.0\text{ mol}\end{array}$

Step 3: The positive integer for $\text{C}$ is calculated as follows:

  $\text{Subscript for C}=\frac{\text{moles of C }}{\text{smallest mole value}}$        (3)

Substitute $6.7\text{ mol}$ for moles of $\text{C}$ and $6.7\text{ mol}$ for smallest mole value in equation (3).

  $\begin{array}{c}\text{Subscript for C}=\frac{6.7\text{ mol}}{6.7\text{ mol}}\\ =1\end{array}$

The positive integer for $\text{H}$ is calculated as follows:

  $\text{Subscript for H}=\frac{\text{moles of H}}{\text{smallest mole value}}$        (4)

Substitute $20.0\text{ mol}$ for moles of $\text{H}$ and $6.7\text{ mol}$ for smallest mole value in equation (4).

  $\begin{array}{c}\text{Subscript for H}=\frac{20.0\text{ mol}}{6.7\text{ mol}}\\ =3\end{array}$

Therefore, the empirical formula for compound is ${\text{CH}}_3$.

(b)

Interpretation Introduction

Interpretation:

Molecular formula of compound that contains $80.0\text{ \% C}$ and $20.0\text{ \% H}$ has to be determined.

Concept Introduction:

Molecular formula represents number of atoms present in 1 molecule. Atomic formula represents atomic symbol for element.

Answer to Problem 68AE

Molecular formula of compound is ${\text{C}}_2{\text{H}}_6$.

Explanation of Solution

Conversion factor for conversion of $\text{g/L}$ to $\text{g/mol}$ at STP is as follows:

  $\left(\frac{22.4\text{ L}}{1\text{ mol}}\right)$

Therefore, molar mass of gas with mass $2.01\text{ g}$ and $1500\text{ mL}$ volume can be calculated as follows:

  $\begin{array}{c}\text{Molar mass}\left(\text{g/mol}\right)=\left(\frac{2.01\text{ g}}{1500\text{ mL}}\right)\left(\frac{22.4\text{ L}}{1\text{ mol}}\right)\left(\frac{{10}^3\text{ mL}}{1\text{ L}}\right)\\ =30.0\text{ g/mol}\end{array}$

Expression to calculate mass of empirical formula is as follows:

  $\text{Mass of}{\text{CH}}_3=\left[1\left(\text{atomic mass of C}\right)+3\left(\text{atomic mass of H}\right)\right]$        (5)

Substitute $12.01\text{ g}$ for atomic mass of $\text{C}$ and $1.0\text{ g}$ for atomic mass of $\text{H}$ in equation (5).

  $\begin{array}{c}\text{Mass of}{\text{CH}}_3=\left[1\left(12.01\text{ g}\right)+3\left(1.0\text{ g}\right)\right]\\ =15.01\text{ g}\end{array}$

The expression to calculate the value of $n$ is as follows:

  $n=\frac{\text{molecular mass}}{\text{empirical formula mass}}$        (6)

Substitute $30.0\text{ g}$ for molecular mass and $15.01\text{ g}$ for empirical formula mass in equation (6).

  $\begin{array}{c}n=\frac{30.0\text{ g}}{15.01\text{ g}}\\ =2\end{array}$

Since value of $n$ is 2 thus subscript values of empirical formula is multiplied by 2. Hence molecular formula of compound is ${\text{C}}_2{\text{H}}_6$.

(c)

Interpretation Introduction

Interpretation:

Lewis structure of ${\text{C}}_2{\text{H}}_6$ has to be drawn.

Concept Introduction:

Lewis structure or representation show bonding between atoms. It also shows lone pairs of electron if it is present in molecule. Only valence electrons on atoms have to be considered in Lewis structure.

Explanation of Solution

The total number of valence electrons of ${\text{C}}_2{\text{H}}_6$ is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}\left(\text{TVE}\right)=\left[2\left(4\right)+6\left(1\right)\right]\\ =14\end{array}$

In ${\text{C}}_2{\text{H}}_6$, total 14 valence electrons are used in formation of 7 single bonds. One bond is between carbon atoms and other 6 bonds are formed between carbon and hydrogen atoms to complete octet of each carbon atoms.

Thus Lewis structure of ${\text{C}}_2{\text{H}}_6$ can be drawn as follows: