Chapter 3, Problem 3.155QE

Interpretation Introduction

Interpretation:

The following table has to be filled.

    $\begin{array}{l}\text{Name Empiricalformula Molar mass Molecular formula}\\ \text{ (g/mol)}\\ {\text{Dimethyl C}}_{\text{2}}{\text{H}}_{\text{6}}\text{SO 78 -----------}\\ \text{ sulfoxide}\\ \\ {\text{Cyclopropane ------------ ----------- C}}_{\text{3}}{\text{H}}_{\text{6}}\\ \\ {\text{Tryptamine C}}_{\text{5}}{\text{H}}_{\text{6}}\text{N 160 --------}\\ \\ {\text{Lactose ---------- -------- C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}\end{array}$

Explanation of Solution

The empirical formula of dimethyl sulfoxide is ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{SO}$.

Molar mass of dimethyl sulfoxide is $\text{78 g/mol}$

Empirical formula mass is

  $\begin{array}{l}\text{(2}\times \text{12}\text{.008)g/mol C + (6}\times \text{1}\text{.008)g/mol H + (31}\text{.998)g/mol S + (15}\text{.999)g/mol O}\\ \text{= 78}\text{.061g/mol}\end{array}$.

    $\text{n = }\frac{\text{molar mass}}{\text{empirical formula mass}}\text{=}\frac{\text{78}\text{.061g/mol}}{\text{78g/mol}}=1$

Molecular formula is ${{\text{(C}}_{\text{2}}{\text{H}}_{\text{6}}\text{SO)}}_{\text{n}}={\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{SO}$.

Molecular formula of cyclopropane is ${\text{C}}_3{\text{H}}_{\text{6}}$.

Molar mass of cyclopropane is $\text{(3×12}\text{.008)g/mol C + (6×1}\text{.008)g/mol H = 42}\text{.072g/mol}$.

The molecular formula can be written as ${\left(\text{empirical formula}\right)}_{\text{n}\text{.}}$.

Therefore empirical formula is ${{\text{(CH}}_{\text{2}}\text{)}}_{\text{3}}$.

The empirical formula of tryptamine is ${\text{C}}_{\text{5}}{\text{H}}_{\text{6}}\text{N}$.

Molar mass of tryptamine is $\text{160 g/mol}$.

Empirical formula mass is $\begin{array}{l}\text{(5}\times \text{12}\text{.008)g/mol C + (6}\times \text{1}\text{.008)g/mol H + (14}\text{.008)g/mol N}\\ \text{= 80}\text{.096g/mol}\end{array}$.

    $\text{n = }\frac{\text{molar mass}}{\text{empirical formula mass}}\text{=}\frac{\text{160g/mol}}{\text{80 g/mol}}=2$

Molecular formula is ${{\text{(C}}_5{\text{H}}_6\text{N)}}_{\text{n}}={\text{C}}_{10}{\text{H}}_{12}{\text{N}}_2$.

Molecular formula of lactose is ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$

Molar mass of lactose is

  $\text{(12×12}\text{.008)g/mol C + (22×1}\text{.008)g/mol H + (11}\times \text{15}\text{.99)g/mol O = 342}\text{.261g/mol}$

The molecular formula can be written as ${\left(\text{empirical formula}\right)}_{\text{n}\text{.}}$.

Here the number of moles is not divisible by a common factor.

Therefore empirical formula is ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$.

    $\begin{array}{l}\text{Name Empiricalformula Molar mass Molecular formula}\\ \text{ (g/mol)}\\ {\text{Dimethyl C}}_{\text{2}}{\text{H}}_{\text{6}}{\text{SO 78 C}}_{\text{2}}{\text{H}}_{\text{6}}\text{SO}\\ \text{ sulfoxide}\\ \\ {\text{Cyclopropane CH}}_2\text{ 42}{\text{.072 C}}_{\text{3}}{\text{H}}_{\text{6}}\\ \\ {\text{Tryptamine C}}_{\text{5}}{\text{H}}_{\text{6}}{\text{N 160 C}}_{10}{\text{H}}_{12}{\text{N}}_2\\ \\ {\text{Lactose C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}\text{ 342}{\text{.26 C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}\end{array}$