Question

A sample of a hydrocarbon (a compound consisting of only carbon and hydrogen) contains $\mathbf{2}\mathbf{.}\mathbf{59}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}$atoms of hydrogen and is $\mathbf{17}\mathbf{.}\mathbf{3}\mathbf{}\mathbf{Percent}$ hydrogen by mass. If the molar mass of the hydrocarbon is between$\mathbf{55}$

and$\mathbf{65}\mathbf{}\raisebox{1ex}{$\mathbf{g}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{mol}$}\right.$, how many moles of compound are present, and what is the mass of the sample?

Short answer

The moles of hydrocarbon present are $4.3\times {10}^{-2}$and the mass of the hydrocarbon is $29$grams.

Step-by-step solution

Definition

Themolar mass of a material is defined asthe mass of the substance contained inone mole of the substance, and it isequal to the total mass of the component atoms. It may be stated mathematically as follows:
$\mathrm{Molar}\mathrm{mass}=\frac{\mathrm{Weight}\mathrm{in}\mathrm{g}}{\mathrm{Number}\mathrm{of}\mathrm{moles}}$

Determining the mass of hydrogen in hydrocarbon

Number of moles of ‘M’ atoms atoms $=2.59\times {10}^{23}\mathrm{atoms}$

$$\begin{gathered} =\left(2.59\times {10}^{23}\mathrm{atoms}\right)\left(\frac{1\mathrm{mole}}{6.023\times {10}^{23}\mathrm{atoms}}\right) \\ =0.4300\mathrm{Moles} \end{gathered}$$

Here,

$1\mathrm{mole}\mathrm{of}\mathrm{hydrogen}=1.0078\mathrm{grams}$

Thus, the mass of hydrogen in hydrocarbon

$$\begin{gathered} =\left(0.4300\right)\left(1.0078\right) \\ =0.4333\mathrm{g} \end{gathered}$$

Determining the total mass of hydrocarbon

As, the hydrogen mass in hydrocarbon $=17.3$

$\begin{array}{rcl}\mathrm{Carbon}\mathrm{mass}& =& 100-17.3\\ & =& 82.7\mathrm{Percent}\end{array}$

Lets determine the total mass of hydrocarbon x,

$\begin{array}{rcl}\left(17.3\mathrm{Percent}\right)\left(\mathrm{x}\right)& =& 0.4333\\ \mathrm{x}& =& \left(0.4333\right)\left(\frac{100}{17.3}\right)\\ & =& 2.5046\mathrm{g}\end{array}$

Number of moles of compound $=\frac{\mathrm{Mass}}{\mathrm{Molar}}$

$=\left(82.7\mathrm{Percent}\right)\left(2.5046\right)$

The total mass of the hydrocarbon sample $=2.0713g$

Determining the molar mass of hydrocarbon

Moles of $\begin{array}{rcl}\mathrm{C}& =& \frac{2.0713}{12}\\ & =& 0.1726\mathrm{mol}\end{array}$

To derive an empirical formula for a compound, divide the number of moles by the lowest number.

$$\begin{gathered} \mathrm{Therefore},\frac{0.4300}{0.1726} \\ =\frac{5}{2}\times 2\mathrm{Atoms}\mathrm{of}\mathrm{M} \\ \frac{0.4300}{0.1726}=1 \\ =1\times 2\mathrm{Atoms}\mathrm{of}\mathrm{C} \end{gathered}$$

The molar mass of hydrocarbon

$$\begin{gathered} =\left(12\right)+5\left(1\right) \\ =29 \end{gathered}$$

Determining the moles of hydrocarbon

Here the range of given molar mass is in the middle of $55-65\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.$.

Thus, $\left(29\right)\times 2=58\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.$could be themolar mass of the hydrocarbon

Moles of hydrocarbon

$$\begin{gathered} =\frac{2.5046}{58} \\ =0.431 \\ =4.3\times {10}^{-2}\mathrm{mol} \end{gathered}$$

Therefore, $4.3\times {10}^{-2}$moles present and the mass are $29$grams.