Chapter 10, Problem 141A

Interpretation Introduction

Interpretation:

The mass of $\text{3}\text{.62}\times {\text{10}}^{24}{\text{ molecules C}}_6{\text{H}}_{\text{12}}{\text{O}}_6$ should be calculated.

Concept introduction:

The number of moles of a compound is defined as the ratio of the 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 141A

Mass of the ${\text{C}}_6{\text{H}}_{\text{12}}{\text{O}}_6$ = $1.08\times {10}^3\text{ g}$

Explanation of Solution

Molar mass of ${\text{C}}_6{\text{H}}_{\text{12}}{\text{O}}_6$ = $6\left(12.01\text{ g/mole}\right)+12\left(1.0078\text{ g/mole}\right)+6\left(15.999\text{ g/mole}\right)$

= 180.2 g/mole

Number of molecules of ${\text{C}}_6{\text{H}}_{\text{12}}{\text{O}}_6$ = $\text{3}\text{.62}\times {\text{10}}^{24}\text{ molecules}$

Mass of ${\text{C}}_6{\text{H}}_{\text{12}}{\text{O}}_6$ = $\text{3}\text{.62}\times {\text{10}}^{24}\text{ molecules}\times \frac{1\text{ mole}}{6.022\times {10}^{23}\text{ molecules}}\times 180.2\text{ g/mole}$

Mass of the ${\text{C}}_6{\text{H}}_{\text{12}}{\text{O}}_6$ = $1.08\times {10}^3\text{ g}$