Question

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.

Short answer

The empirical formula of the compound is \(\text{C}_{2}\text{H}_{3}\text{N}\) and the molecular formula is also \(\text{C}_{2}\text{H}_{3}\text{N}\).

Step-by-step solution

Calculate the mass percentages for each element in the 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: $$ \% N = 100\% - 58.51\% - 7.37\% = 34.12\% $$

Convert mass percentages to mole ratios

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: $$ \frac{34.12\,\text{g}}{14.01\,\text{g/mol}} = 2.43\,\text{moles} $$

Determine the empirical formula

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 \(\text{C}_{2}\text{H}_{3}\text{N}\).

Determine the molecular formula using effusion rates

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 = \(\text{C}_{2}\text{H}_{3}\text{N}\)

Key concepts

Mass Percentage

Mass percentage is a way of expressing the concentration of an element within a compound as a percentage of the total mass of the compound. It gives students an understanding of the proportion each element contributes to the overall mass of the compound. In chemistry, calculating mass percentages is crucial for determining the empirical formula of a compound.
To calculate mass percentage for any particular element in a compound, use the formula:

• Mass percentage = \( \frac{\text{mass of the element}}{\text{total mass of the compound}} \times 100\% \)

For instance, with the elements Carbon (C) and Hydrogen (H) in the provided example:

• We already know the mass percentages of Carbon (\(58.51\%\)) and Hydrogen (\(7.37\%\)).
• To determine the mass percentage of Nitrogen (N), subtract the sum of known mass percentages from \(100\%\).
  After subtraction, \(\% \text{N} = 34.12\%\).

Understanding mass percentage helps in converting these values into mole ratios, which is the next step in solving for the empirical formula.

Mole Ratio

The mole ratio is an expression of the quantities of elements in a compound in terms of moles. Unlike mass percentage, mole ratios don't depend on the weight of an element but on the number of atoms of each element. This makes mole ratios particularly useful when trying to find the empirical formula of a compound.
To calculate the mole ratio, you follow these steps:

• Divide the mass percentage of each element by its respective molar mass (found on the periodic table).
• This gives you the number of moles of each element present per \(100\) grams of the compound.

For example, using the exercise:

• Carbon: \( \frac{58.51}{12.01} = 4.87 \text{ moles} \)
• Hydrogen: \( \frac{7.37}{1.01} = 7.30 \text{ moles} \)
• Nitrogen: \( \frac{34.12}{14.01} = 2.43 \text{ moles} \)

The final step involves dividing each of these calculated mole values by the smallest one among them to simplify them. This provides a whole number ratio of each element, leading us to the empirical formula, \(\text{C}_2\text{H}_3\text{N}\).
Remember, understanding mole ratios is fundamental to finding empirical formulas because they show the smallest whole number ratio of atoms in a compound.

Graham's Law of Effusion

Graham's Law of Effusion describes the rate at which gas molecules move through a small opening or pore from one compartment to another. It's a useful concept for understanding molecular movement and molecular weight relationships.
The law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass:

• \[\frac{\text{Rate of A}}{\text{Rate of B}} = \sqrt{\frac{\text{Molar mass of B}}{\text{Molar mass of A}}}\]

In the context of the exercise, helium effuses \(3.20\) times faster than the compound. By using Graham's Law, we calculated:

• \[3.20 = \sqrt{\frac{\text{Molar mass of Compound}}{4.00}}\]
• Solving the equation yields the molar mass of the compound to be \(40.96\, \text{g/mol}\).

This calculated molar mass allows us to compare it with the empirical formula's molar mass to determine if the empirical and molecular formula differ or are the same. Using these tools, you gain a deeper understanding of molecular structures and how they relate to physical properties.