Question

A sample of LSD (D-lysergic acid diethylamide,${\mathbf{C}}_{\mathbf{24}}{\mathbf{H}}_{\mathbf{30}}{\mathbf{N}}_{\mathbf{3}}\mathbf{O}$)is added to some table salt (sodiumchloride) to form a mixture. Given that $\mathbf{1}\mathbf{.}\mathbf{00}\mathit{g}$asample of the mixture undergoes combustion toproduce$\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{g}$of${\mathbf{CO}}_{\mathbf{2}}$, what is the mass percentage of LSD in the mixture?

Short answer

The mass percentage of LSD ${\mathrm{C}}_{24}{\mathrm{H}}_{30}{\mathrm{N}}_3\mathrm{O}$ in the mixture is$42.86\mathrm{percent}$

Step-by-step solution

Definition

The molar mass of a materialis defined as the mass of the substance contained in one mole of the substance, and it is equal to the total mass of the component atoms. It may be stated mathematically as follows:

$\mathrm{Molar}\mathrm{mass}=\frac{\mathrm{Weight}\mathrm{in}\mathrm{g}}{\mathrm{number}\mathrm{of}\mathrm{moles}}$

Determination of number of moles of CO2

Write the following as the LSD (D-lysergic acid diethylamide) combustion reaction:

${\mathrm{C}}_{24}{\mathrm{H}}_{30}{\mathrm{N}}_3\mathrm{O}+34{\mathrm{O}}_2\to 24{\mathrm{CO}}_2+15{\mathrm{H}}_2\mathrm{O}+3{\mathrm{NO}}_2$

One mole of LSD${\mathrm{C}}_{24}{\mathrm{H}}_{30}{\mathrm{N}}_3\mathrm{O}$ creates two moles of carbon dioxide, according to the stoichiometric equationabove.

When a 1.0 g sample of mixture is burned, 1.20 g of is produced.

So, to find the number of moles in 1.20 g of, do the following:

$\begin{array}{rcl}\mathrm{Number}\mathrm{of}\mathrm{moles}& =& \frac{\mathrm{Mass}}{\mathrm{Molecular}\mathrm{weight}}\\ & =& \frac{1.20\mathrm{g}}{44.0{\displaystyle \raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mole}$}\right.}}\\ & =& 0.0273\mathrm{mol}\end{array}$

Here the number of moles of ${\mathrm{CO}}_2$are $0.0273\mathrm{mol}$

Determination of number of moles of C24H30N3O

Fromthe stoichiometric equation,

$\begin{array}{rcl}1\mathrm{mol}\mathrm{of}{\mathrm{C}}_{24}{\mathrm{H}}_{30}{\mathrm{N}}_3\mathrm{O}& =& 24\mathrm{mol}\mathrm{of}{\mathrm{CO}}_2\\ {\mathrm{C}}_{24}{\mathrm{H}}_{30}{\mathrm{N}}_3\mathrm{O}& =& 0.0273\mathrm{mol}{\mathrm{CO}}_2\times \frac{1\mathrm{mole}\mathrm{of}{\mathrm{C}}_{24}{\mathrm{H}}_{30}{\mathrm{N}}_3\mathrm{O}}{24\mathrm{mol}\mathrm{of}{\mathrm{CO}}_2}\\ & =& 0.00114\mathrm{mol}\end{array}$

Therefore the number of moles ${\mathrm{C}}_{24}{\mathrm{H}}_{30}{\mathrm{N}}_3\mathrm{O}$of needed to produce $0.0273\mathrm{mol}\mathrm{of}{\mathrm{CO}}_2$ are$0.00114\mathrm{mol}$.

Determination of mass of C24H30N3O

$\begin{array}{rcl}\mathrm{Molar}\mathrm{mass}\left({\mathrm{MM}}_{{\mathrm{C}}_{24}{\mathrm{H}}_{30}{\mathrm{N}}_3\mathrm{O}}\right)& =& 24\left({\mathrm{MM}}_{\mathrm{C}}\right)+30\left({\mathrm{MM}}_{\mathrm{H}}\right)+3\left({\mathrm{MM}}_{\mathrm{N}}\right)+1\left({\mathrm{MM}}_{\mathrm{O}}\right)\\ & =& 24\times 12\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.+30\times 1\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.+3\times 14\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.+1\times 16\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.\\ & =& 376\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.\end{array}$

Let’s compute mass by converting $0.00114mol$of${\mathrm{C}}_{24}{\mathrm{H}}_3{\mathrm{N}}_3\mathrm{O}$.
role="math" localid="1655206369821" $\begin{array}{rcl}\mathrm{Mass}\mathrm{of}{\mathrm{C}}_{24}{\mathrm{H}}_{30}{\mathrm{N}}_3\mathrm{O}& =& 0.00114\mathrm{mol}\times 376\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.\\ & =& 0.4286\mathrm{g}\end{array}$

Thus, Mass of${\mathrm{C}}_{24}{\mathrm{H}}_{30}{\mathrm{N}}_3\mathrm{O}$ is$0.4286\mathrm{g}$.

Determination of mass percentage of C24H30N3O

Let’s calculate mass percentage,

$\begin{array}{rcl}\mathrm{Mass}\mathrm{of}{\mathrm{C}}_{24}{\mathrm{H}}_{30}{\mathrm{N}}_3\mathrm{O}& =& \left(\frac{\mathrm{Mass}\mathrm{of}{\mathrm{C}}_{24}{\mathrm{H}}_{30}{\mathrm{N}}_3\mathrm{O}}{\mathrm{Total}\mathrm{mass}\mathrm{of}\mathrm{sample}}\right)\times 100\\ & =& \left(\frac{0.4286\mathrm{g}}{1.00\mathrm{g}}\right)\times 100\\ & =& 42.86\mathrm{percent}\end{array}$

Therefore the mass percentage is$42.86\mathrm{Percent}$.