Chapter 10, Problem 135A

Interpretation Introduction

Interpretation:

The mass of $\text{3}\text{.50}\times {\text{10}}^{-2}{\text{ mole C}}_{\text{14}}{\text{H}}_{\text{10}}{\text{O}}_{\text{4}}$ should be calculated.

Concept introduction:

Number of moles of a compound is defined as the ratio of given mass of the compound to the molar or molecular mass of the compound.

The mathematical expression is given by:

Number of moles = $\frac{\text{mass of the compound}}{\text{molar mass of the compound}}$

The one mole of any substance is equal to the Avogadro number i.e. $6.022\times {10}^{23}$ atoms.

Answer to Problem 135A

Mass of the ${\text{C}}_{\text{14}}{\text{H}}_{\text{10}}{\text{O}}_{\text{4}}$ = $8.47\text{8 g}$

Explanation of Solution

Molar mass of ${\text{C}}_{\text{14}}{\text{H}}_{\text{10}}{\text{O}}_{\text{4}}$ = $14\left(12.01\text{ g/mole}\right)+10\left(1.0078\text{ g/mole}\right)+4\left(15.999\text{ g/mole}\right)$

= 242.23 g/mole

Number of moles of ${\text{C}}_{\text{14}}{\text{H}}_{\text{10}}{\text{O}}_{\text{4}}$ = $\text{3}\text{.50}\times {\text{10}}^{-2}\text{ mole}$

Number of moles = $\frac{\text{mass of the compound}}{\text{molar mass of the compound}}$

Put the values,

$\text{3}\text{.50}\times {\text{10}}^{-2}\text{ mole}$ = $\frac{\text{mass of the compound}}{242.23\text{ g/mole}}$

Mass of the ${\text{C}}_{\text{14}}{\text{H}}_{\text{10}}{\text{O}}_{\text{4}}$ = $\text{3}\text{.50}\times {\text{10}}^{-2}\text{ mole}\times 242.23\text{ g/mole}$

= $8.47\text{8 g}$