A compound contains only \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{N}\). It is \(58.51 \% \mathrm{C}\) and \(7.37 \% \mathrm{H}\) by mass. Helium effuses through a porous frit \(3.20\) times as fast as the compound does. Determine the empirical and molecular formulas of this compound.

Since we know the mass percentages of carbon and hydrogen, we can calculate the mass percentage of nitrogen by subtracting them from 100%. $$ \% N = 100\% - \% C - \% H $$ So, using the given values: $$ \% N = 100\% - 58.51\% - 7.37\% = 34.12\% $$

To determine the empirical formula, we need to convert the mass percentages to mole ratios. We'll do this by dividing the mass percentage of each element by its molar mass: Mole Ratio of C: $$ \frac{58.51\,\text{g}}{12.01\,\text{g/mol}} = 4.87\,\text{moles} $$ Mole Ratio of H: $$ \frac{7.37\,\text{g}}{1.01\,\text{g/mol}} = 7.30\,\text{moles} $$ Mole Ratio of N: $$ \frac{34.12\,\text{g}}{14.01\,\text{g/mol}} = 2.43\,\text{moles} $$

To find the empirical formula, we need to divide all the mole ratios by the smallest mole ratio and then round off the results to the nearest whole number: Divide by the smallest mole ratio (2.43): C: $$ \frac{4.87}{2.43} \approx 2 $$ H: $$ \frac{7.30}{2.43} \approx 3 $$ N: $$ \frac{2.43}{2.43} \approx 1 $$ Therefore, the empirical formula is \(\text{C}_{2}\text{H}_{3}\text{N}\).

We're given the information that helium effuses 3.20 times as fast as the compound. Using Graham's law of effusion, we can write the relationship between effusion rates and molar masses: $$ \frac{\text{Rate of He}}{\text{Rate of Compound}} = \sqrt{\frac{\text{Molar mass of Compound}}{\text{Molar mass of He}}} $$ Since the rate of He is 3.20 times faster, we can plug in the values and solve for the molar mass of the compound: $$ 3.20 = \sqrt{\frac{\text{Molar mass of Compound}}{4.00\,\text{g/mol}}} $$ Square both sides of the equation: $$ 10.24 = \frac{\text{Molar mass of Compound}}{4.00\,\text{g/mol}} $$ Multiply both sides by the molar mass of He (4.00 g/mol): $$ \text{Molar mass of Compound} = 40.96\,\text{g/mol} $$ Finally, since we now know the molar mass of the compound and we have the empirical formula, we can find the molecular formula. Divide the molar mass of the compound by the molar mass of the empirical formula: $$ \frac{40.96\,\text{g/mol}}{(2 \times 12.01\,\text{g/mol}) + (3 \times 1.01\,\text{g/mol}) + (1 \times 14.01\,\text{g/mol})} = \frac{40.96\,\text{g/mol}}{41.07\,\text{g/mol}} \approx 1 $$ Since the ratio is approximately 1, the molecular formula is the same as the empirical formula: Molecular Formula = \(\text{C}_{2}\text{H}_{3}\text{N}\)