Question

An anesthetic gas contains \(64.81 \%\) carbon, \(13.60 \%\) hydrogen, and \(21.59 \%\) oxygen, by mass. If \(2.00 \mathrm{~L}\) of the gas at \(25^{\circ} \mathrm{C}\) and \(0.420 \mathrm{~atm}\) weighs \(2.57 \mathrm{~g},\) what is the molecular formula of the anesthetic?

Short answer

The molecular formula of the anesthetic gas is \( C_{4}H_{10}O \).

Step-by-step solution

Calculate the moles of the gas

Use the ideal gas law equation, \( PV = nRT \), to find the number of moles (\( n \)) of the gas in the container. Here, \( P = 0.420 \, \text{atm} \), \( V = 2.00 \, \text{L} \), \( R = 0.0821 \, \text{L} \, \text{atm} \, \text{K}^{-1} \, \text{mol}^{-1} \), and \( T = 298 \, \text{K} \). \[ n = \frac{PV}{RT} \] Substituting the given values: \[ n = \frac{0.420 \, \text{atm} \times 2.00 \, \text{L}}{0.0821 \, \text{L} \, \text{atm} \, \text{K}^{-1} \, \text{mol}^{-1} \times 298 \, \text{K}} \approx 0.0345 \, \text{mol} \]

Determine the molar mass of the gas

The molar mass (\( M \)) of the gas can be found using the mass of the gas and the number of moles calculated in Step 1. \[ M = \frac{\text{mass}}{n} \] Substituting the values: \[ M = \frac{2.57 \, \text{g}}{0.0345 \, \text{mol}} \approx 74.49 \, \text{g/mol} \]

Convert the percent composition to grams

Assume 100 grams of the gas to simplify calculations. This gives: - Carbon: \( 64.81 \, \text{g} \) - Hydrogen: \( 13.60 \, \text{g} \) - Oxygen: \( 21.59 \, \text{g} \)

Calculate the moles of each element

Use the atomic masses of each element to convert grams to moles. - Carbon: \( \frac{64.81 \, \text{g}}{12.01 \, \text{g/mol}} \approx 5.40 \, \text{mol} \) - Hydrogen: \( \frac{13.60 \, \text{g}}{1.008 \, \text{g/mol}} \approx 13.49 \, \text{mol} \) - Oxygen: \( \frac{21.59 \, \text{g}}{16.00 \, \text{g/mol}} \approx 1.35 \, \text{mol} \)

Find the simplest mole ratio

Divide the number of moles of each element by the smallest number of moles: - Carbon: \( \frac{5.40}{1.35} = 4 \) - Hydrogen: \( \frac{13.49}{1.35} \approx 10 \) - Oxygen: \( \frac{1.35}{1.35} = 1 \) Thus, the empirical formula is \( C_{4}H_{10}O \)

Determine the molecular formula

Compare the empirical formula mass with the molar mass calculated in Step 2. \[ \text{Empirical formula mass} = 4(12.01) + 10(1.008) + 16.00 = 74.12 \, \text{g/mol} \] Since the empirical formula mass (\( 74.12 \, \text{g/mol} \)) is nearly equal to the molar mass (\( 74.49 \, \text{g/mol} \)), the molecular formula is the same as the empirical formula, \( C_{4}H_{10}O \).

Key concepts

Ideal Gas Law

The Ideal Gas Law equation, represented as \( PV = nRT \), is a fundamental concept in chemistry. It links pressure (\( P \)), volume (\( V \)), temperature (\( T \)), and the number of moles (\( n \)) of a gas. The constant \( R \) is known as the gas constant and its value depends on the units used.
For instance, in this exercise, we've utilized \( R = 0.0821 \, \text{L} \, \text{atm} \, \text{K}^{-1} \, \text{mol}^{-1} \). By inserting the given values into the equation, we calculated the number of moles of the gas.
Steps to follow:

• Identify the known variables: pressure, volume, temperature.
• Make sure units are consistent (e.g., temperature in Kelvin).
• Solve for \( n \) using the equation \( n = \frac{PV}{RT} \).

This approach connects macroscopic observations of gases to their microscopic properties.

Percent Composition

Percent composition lets us understand the makeup of a compound. It shows the percentage of each element by mass in a substance.
In this exercise, we were given:

• Carbon: \( 64.81 \% \)
• Hydrogen: \( 13.60 \% \)
• Oxygen: \( 21.59 \% \)

To use this information, assume a 100-gram sample. This converts percentage directly to grams:

• Carbon: \(64.81 \, \text{g}\)
• Hydrogen: \(13.60 \, \text{g}\)
• Oxygen: \(21.59 \, \text{g}\)

From here, we can easily convert these masses to moles. Percent composition is essential for both empirical and molecular formula determination.

Molar Mass Calculation

Molar mass is a key concept in figuring out the substance's total mass relative to one mole of its molecules. It's calculated as the mass of given substance divided by the number of moles.
In our example, we calculated the molar mass (\( M \)) using:

• Mass of gas (2.57g)
• Moles (0.0345 mol)

Calculated by:
\( M = \frac{2.57 \, \text{g}}{0.0345 \, \text{mol}} \approx 74.49 \, \text{g/mol} \)
This value helps confirm the molecular formula and connects the macroscopic world with molecular dimension. When the empirical formula's mass equals or closely matches the molar mass, we establish our molecular formula.

Empirical Formula

Empirical formulas show the simplest whole-number ratio of elements in a compound. To find it from percent composition (by mass), convert the grams of each element to moles.
Using our example:

• Carbon: \( \frac{64.81 \, \text{g}}{12.01 \, \text{g/mol}} = 5.40 \, \text{mol} \)
• Hydrogen: \( \frac{13.60 \, \text{g}}{1.008 \, \text{g/mol}} = 13.49 \, \text{mol} \)
• Oxygen: \( \frac{21.59 \, \text{g}}{16.00 \, \text{g/mol}} = 1.35 \, \text{mol} \)

Next, find the smallest number of moles among them and divide all moles by this number:

• Carbon: \( \frac{5.40}{1.35} = 4 \)
• Hydrogen: \( \frac{13.49}{1.35} \approx 10 \)
• Oxygen: \( \frac{1.35}{1.35} = 1 \)

Thus, the empirical formula is \( C_{4}H_{10}O \). Comparing the empirical formula mass to the molar mass from the molar mass calculation confirms that our empirical formula equals the molecular formula.