Chapter 10.2, Problem 16SP

Interpretation Introduction

Interpretation: The mass in grams for $4.52\times {10}^{-3}\text{ mol}$ of ${\text{C}}_{\text{20}}{\text{H}}_{\text{42}}$ should be calculated.

Concept Introduction: The mass in grams of one mole of the compound is said to be its molar mass.

Answer to Problem 16SP

The mass in grams for $4.52\times {10}^{-3}\text{ mol}$ of ${\text{C}}_{\text{20}}{\text{H}}_{\text{42}}$ is $\text{1}\text{.27 g}$ .

Explanation of Solution

To determine the mass of $4.52\times {10}^{-3}\text{ mol}$ of ${\text{C}}_{\text{20}}{\text{H}}_{\text{42}}$ the molar mass of ${\text{C}}_{\text{20}}{\text{H}}_{\text{42}}$ is calculated first by adding up the sum of the masses of C and H elements.

The moles of C and H are converted to grams of C and H as:

  $\begin{array}{l}\text{20 mol C }\times \frac{12.0\text{ g C}}{1\text{ mol C}}\text{ = 240}\text{.0 g C}\\ \text{42 mol H }\times \frac{\text{1}\text{.0 g H}}{1\text{ mol H}}\text{ = 42}\text{.0 g H}\end{array}$

The mass of 1 mol of ${\text{C}}_{\text{20}}{\text{H}}_{\text{42}}$ is:

  $\begin{array}{l}{\text{mass of 1 mol of C}}_{\text{20}}{\text{H}}_{\text{42}}\text{ = 240}\text{.0 g C + 42}\text{.0 g H}=\text{ 282}\text{.0 g}\\ {\text{molar mass of C}}_{\text{20}}{\text{H}}_{\text{42}}=\text{282}\text{.0 g/mol}\end{array}$

Thus, the molar mass of ${\text{C}}_{\text{20}}{\text{H}}_{\text{42}}$ is $\text{282}\text{.0 g/mol}$ .

Now, the mass in grams for $4.52\times {10}^{-3}\text{ mol}$ of ${\text{C}}_{\text{20}}{\text{H}}_{\text{42}}$ is:

  $4.52\times {10}^{-3}\text{ mol}\times \frac{\text{282}\text{.0 g}}{1\text{ mol}}\text{ = 1}\text{.27 g}$

Hence, the mass in grams for $4.52\times {10}^{-3}\text{ mol}$ of ${\text{C}}_{\text{20}}{\text{H}}_{\text{42}}$ is $\text{1}\text{.27 g}$ .