Question

Complete the following table

Mass of sample	Moles of sample	Molecules in sample	Total atoms in a sample
$\mathbf{4}\mathbf{.}\mathbf{24}{\mathbf{gC}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{6}}$
	$\mathbf{0}\mathbf{.}\mathbf{224}\mathbf{mol}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$
		$\mathbf{2}\mathbf{.}\mathbf{71}\mathbf{\times }{\mathbf{10}}^{\mathbf{22}}\mathbf{}\mathbf{molecules}\mathbf{}{\mathbf{CO}}_{\mathbf{2}}$
			$\mathbf{3}\mathbf{.}\mathbf{35}\mathbf{\times }{\mathbf{10}}^{\mathbf{22}}\mathbf{total}\mathbf{}\mathbf{atom}\mathbf{}\mathbf{in}\mathbf{}{\mathbf{CH}}_{\mathbf{3}}\mathbf{OH}\mathbf{}\mathbf{sample}$

Short answer

The table is given as:

Mass of sample	Moles of sample	Molecules in sample	Total atoms in sample
$4.24{\mathrm{gC}}_6{\mathrm{H}}_6$	$0.0543\mathrm{moles}$	$3.27\times {10}^{22}\mathrm{molecules}$	$3.92\times {10}^{23}\mathrm{atoms}$
$4.04\mathrm{g}{\mathrm{H}}_2\mathrm{O}$	$0.224\mathrm{moles}$	$1.35\times {10}^{23}\mathrm{molecules}$	$4.05\times {10}^{23}\mathrm{atoms}$
$1.98\mathrm{g}{\mathrm{CO}}_2$	$4.50\times {10}^{-2}\mathrm{moles}$	$2.71\times {10}^{22}\mathrm{molecules}$	$8.13\times {10}^{22}\mathrm{atoms}$
$0.297\mathrm{g}{\mathrm{CH}}_3\mathrm{OH}$	$9.27\times {10}^{-3}\mathrm{moles}$	$5.58\times {10}^{21}\mathrm{molecules}$	$3.35\times {10}^{22}\mathrm{atoms}$

Step-by-step solution

Definition of Atomic number

Atoms of the same element with the same number of protons, but different neutrons are called isotopes. The atomic number of an element equals its number of protons.

Explanation of percent composition

% of abundance ­of ${}^{185}\mathrm{Re}$ +% of abundance of ${}^{187}\mathrm{Re}$equals 100.

So,

% of abundance of+62.60=100

% of abundance of ${}^{185}\mathrm{Re}$=100-62.60

% of abundance of${}^{185}\mathrm{Re}$ =37.40%

Calculation of the atomic mass of % of the abundance of  Re185

$$\begin{gathered} \mathrm{Average}\mathrm{atomic}\mathrm{mass}\mathrm{of}\mathrm{Re}=\left(37.40\%\right)\left(\mathrm{mass}\right)+\left(62.60\%\right)\left(186.956\right) \\ 186.207=\left(\frac{37.40}{100}\right)\left(\mathrm{mass}\right)+\left(\frac{62.60}{100}\right)\left(186.956\right) \\ \mathrm{Atomic}\mathrm{mass}\mathrm{of}{}^{185}\mathrm{Re}=\left(186.207-117.0344\right)\left(\frac{100}{37.40}\right) \\ \mathrm{Atomic}\mathrm{mass}\mathrm{of}{}^{185}\mathrm{Re}=\left(69.1726\right)\left(\frac{100}{37.40}\right) \\ \mathrm{Atomic}\mathrm{mass}\mathrm{of}{}^{185}\mathrm{Re}=184.953\mathrm{amu} \\ \mathrm{Atomic}\mathrm{mass}\mathrm{of}{}^{185}\mathrm{Re};185\mathrm{amu} \\ \mathrm{Therefore},\mathrm{the}\mathbf{atomic}\mathbf{}\mathbf{mass}\mathrm{of}{}^{185}\mathrm{Re}\mathrm{will}\mathrm{be}185\mathrm{amu} \end{gathered}$$

Completion of table

The table is given as:

Mass of sample	Moles of sample	Molecules in sample	Total atoms in sample
$4.24{\mathrm{gC}}_6{\mathrm{H}}_6$	$0.0543\mathrm{moles}$	$3.27\times {10}^{22}\mathrm{molecules}$	$3.92\times {10}^{23}\mathrm{atoms}$
$4.04\mathrm{g}{\mathrm{H}}_2\mathrm{O}$	$0.224\mathrm{moles}$	$1.35\times {10}^{23}\mathrm{molecules}$	$4.05\times {10}^{23}\mathrm{atoms}$
$1.98\mathrm{g}{\mathrm{CO}}_2$	$4.50\times {10}^{-2}\mathrm{moles}$	$2.71\times {10}^{22}\mathrm{molecules}$	$8.13\times {10}^{22}\mathrm{atoms}$
$0.297\mathrm{g}{\mathrm{CH}}_3\mathrm{OH}$	$9.27\times {10}^{-3}\mathrm{moles}$	$5.58\times {10}^{21}\mathrm{molecules}$	$3.35\times {10}^{22}\mathrm{atoms}$