Chapter 25, Problem 100AP

Interpretation Introduction

Interpretation:

The amounts, in milligrams, of $\text{C}$, $\text{H}$, and $\text{O}$ that were present in the original sample of $\text{Y}$ are to be calculated, the empirical formula of $\text{Y}$ is to be derived, and a plausible structure for $\text{Y}$ is to be suggested in case the empirical formula is the same as the molecular formula.

Concept Introduction:

In an empirical formula, the mass of each element changes into moles by taking the molar mass from the periodic table.

Each mole value is divided by the smallest number of calculated moles.

Answer to Problem 100AP

Solution:

(a) The amounts of $\text{C,}$ $\text{H,}$ and $\text{O}$ are calculated as follows:

$15.81\text{mg C,}$ $1.32\text{mg H,}$ and $3.49\text{mg O}$

(b) $\text{Empirical formula}={\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\text{O}$

(c)

Explanation of Solution

Given information: The mass of compound Y = $20.63\text{mg}\text{.}$

The mass of ${\text{CO}}_{\text{2}}$ formed = $57.94\text{mg}\text{.}$

The mass of ${\text{H}}_{\text{2}}\text{O}$ formed = $11.85\text{mg}\text{.}$

a)The mass of $\text{C,}\text{H,}\text{and}\text{O}$ were present in the original sample of $\text{Y}$

The mass of $\text{C}$ in $57.94\text{mg}$ of ${\text{CO}}_{\text{2}}$ is calculated as follows:

$\begin{array}{c}\text{Mass}\text{of C}=57.94\times {10}^{-3}\text{g}{\text{of CO}}_2\times \frac{1\cancel{{\text{mol CO}}_2}}{44.01\cancel{{\text{g CO}}_2}}\times \frac{1\cancel{\text{mol C}}}{1\cancel{{\text{mol CO}}_2}}\times \frac{12.01\text{g C}}{1\cancel{\text{mol C}}}\\ =15.81\text{mg}\text{.}\end{array}$

The mass of $H$ in $11.85\text{mg}$ of ${\text{H}}_{\text{2}}\text{O}$ is calculated as follows:

$\begin{array}{c}\text{Mass}\text{of H}=\left(11.85\times {10}^{-3}\cancel{\text{g}{\text{of H}}_2\text{O}}\right)\left(\frac{1\cancel{{\text{mol H}}_2\text{O}}}{18.02\cancel{{\text{g H}}_2\text{O}}}\right)\left(\frac{2\cancel{\text{mol H}}}{1\cancel{{\text{mol H}}_2\text{O}}}\right)\left(\frac{1.008\text{g H}}{1\cancel{\text{mol H}}}\right)\\ =1.32\text{mg}\text{.}\end{array}$

The mass of oxygen is calculated as follows:

$\begin{array}{c}\text{Mass}\text{of O}=\text{Mass}\text{of sample}-\left(\text{Mass}\text{of C}+\text{Mass}\text{of H}\right)\\ =20.63\text{mg}-\left(15.81\text{mg}+1.32\text{mg}\right)\\ =3.49\text{mg}\text{.}\end{array}$

b)Derive the empirical formula of $\text{Y}\text{.}$

The calculated value of compound $\text{Y}$ is given as follows:

$15.81\text{mg C}$ or $15.81\times {10}^{-3}g\text{ C}$;

$1.32\text{mg H}$ or $1.32\times {10}^{-3}\text{ g H}$;

$3.49\text{mg O}$ or $3.49\times {10}^{-3}\text{ g O}$.

The mass of each element changes into moles by taking the molar mass from the periodic table as follows:

$\left(15.81\times {10}^{-3}\cancel{\text{g C}}\times \frac{1\text{mol C}}{12.0\cancel{\text{g C}}}\right)=1.31\times {10}^{-3}\text{mol C}$

$\left(1.32\times {10}^{-3}\cancel{\text{g H}}\times \frac{1\text{mol H}}{1.01\cancel{\text{g H}}}\right)=1.31\times {10}^{-3}\text{mol H}$

$\left(3.49\times {10}^{-3}\cancel{\text{g O}}\times \frac{1\text{mol O}}{16.0\cancel{\text{g O}}}\right)=0.21\times {10}^{-3}\text{mol O}\text{.}$

Each mole is divided by the smallest number of calculated moles and taken round to the nearest whole number.

$\begin{array}{l}\frac{1.31\times {10}^{-3}}{0.21\times {10}^{-3}}\text{mol C}\\ \Rightarrow \text{6}\text{.27}\approx \text{6}\end{array}$

$\begin{array}{l}\frac{1.31\times {10}^{-3}}{0.21\times {10}^{-3}}\text{mol H}\\ \Rightarrow \text{6}\text{.21}\approx \text{6}\end{array}$

$\frac{0.21\times {10}^{-3}}{0.21\times {10}^{-3}}\text{mol O}\Rightarrow \text{1}\text{.}$

The value of each element is considered as the mole ratio of the elements, which is represented by subscripts in the empirical formula, as follows:

$\text{Empirical formula}={\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\text{O}$

c) Plausible structure for $\text{Y}$ if the empirical formula is the same as the molecular formula.

The compound $\text{Y}$ contains the molecular formula ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\text{O}$, which means two structures can have different functional groups.

The structures are represented below.