Problem 46 - General Chemistry: Principles and Modern Applications

Question

A particular gaseous hydrocarbon that is \(82.7 \%\) C and \(17.3 \%\) H by mass has a density of \(2.33 \mathrm{g} / \mathrm{L}\) at \(23^{\circ} \mathrm{C}\) and \(746 \mathrm{mm} \mathrm{Hg} .\) What is the molecular formula of this hydrocarbon?

Step-by-Step Solution

Answer

The molecular formula of the given hydrocarbon is \( C_2H_5 \).

Step 1: Calculate molar mass and ratio of elements

First, convert the percentages to grams (since the total molar mass is 100g). 82.7% C is \( 82.7g \) of C and 17.3% H is \( 17.3g \) of H. The number of moles of C and H can then be calculated using atomic masses (C is \(12.01 \, g/mol \) and H is \( 1.008 \, g/mol \)) which yields: \( n_C = \frac{82.7g}{12.01 \, g/mol}= 6.88 mol \) and \ \( n_H=\frac{17.3g}{1.008 \, g/mol}= 17.16 mol \).

Step 2: Determine the empirical formula

The ratio of number of moles calculated in step 1 gives \(\frac{17.16 mol}{6.88 mol}= 2.49 \) which rounds approximately to \( 2.5 \). Given that molecular formulas must be in whole numbers, this suggests the empirical formula is \( C_2 H_5 \) .

Step 3: Apply the Ideal Gas Law

To calculate the molar mass of the gas, the Ideal Gas Law: \( PV=nRT \) can be rearranged to: \( \frac{PM}{RT} = d \), where the molar mass M can be expressed as: \( M=\frac{dRT}{P} \). With temperatures in Kelvin, pressure in atm, R as \(\frac{0.0821 \, L \cdot atm}{mol \cdot K} \) and converting the given condition of pressure and temperature, you can find the molar mass. Replacing values, you get: \( M=\frac{2.33g/L \cdot 0.0821 \, L \cdot atm / (mol \cdot K) \cdot (23+273) \, K}{0.9803 \, atm}=30.07 \, g/mol \)

Step 4: Determine the Molecular Formula

Next, determine how many empirical formula units are in the molecular formula. The molecular formula is found by simply multiplying the subscripts of the empirical formula by the factor. By: \( n=\frac{M_{molar}}{M_{empirical}}=\frac{30.07 \, g/mol}{29.07 \, g/mol}=1.03 \) which rounded off gives \(1\) . Hence, the molecular formula is \( C_2H_5 \) which is the same as the empirical formula because the factor is close to \(1\) .

Key Concepts

Empirical Formula

When determining the empirical formula of a compound, we are looking for the simplest whole-number ratio of elements in the compound. For example, if a compound is made up of 82.7% carbon (C) and 17.3% hydrogen (H) by mass, we first convert these percentages to grams, assuming a 100g sample size.
This means we deal with 82.7g of C and 17.3g of H. To find the number of moles, divide each element's mass by its respective atomic mass: for carbon, it's approximately 12.01 g/mol, and for hydrogen, it's about 1.008 g/mol.

Calculating gives us:

• Carbon: \( n_C = \frac{82.7g}{12.01 \, g/mol} = 6.88 \) moles
• Hydrogen: \( n_H = \frac{17.3g}{1.008 \, g/mol} = 17.16 \) moles

The ratio of moles is found by dividing each by the smallest number of moles. This yields a ratio of approximately 1:2.5, which suggests an empirical formula of \( C_2H_5 \) when expressed in whole numbers.

Ideal Gas Law

The Ideal Gas Law is a vital tool in determining various properties of gases, including their molar mass. The formula is expressed as \( PV = nRT \), where:

• \( P \) is the pressure
• \( V \) is the volume
• \( n \) is the number of moles
• \( R \) is the ideal gas constant
• \( T \) is the temperature in Kelvin

Rearranging for molar mass \( M \) and using the density \( d \), we get \( M = \frac{dRT}{P} \).
For example, with a gas having a density of 2.33 g/L, temperature at 23°C (convert it to 296 K), and pressure of 746 mmHg (0.9803 atm), we substitute these into the rearranged equation alongside \( R = 0.0821 \, L \cdot atm / (mol \cdot K)\).
This yields a molar mass of approximately 30.07 g/mol, aiding further in molecular formula determination.

Molar Mass Calculation

Understanding molar mass is crucial for molecular formula determination. Once you have determined the empirical formula, which for our example was \( C_2H_5 \), compute its empirical molar mass. Calculate by adding together the atomic masses of all atoms present in the empirical formula. This means for \( C_2H_5 \):

• 2 carbon atoms = \( 2 \times 12.01 \, g/mol = 24.02 \, g/mol \)
• 5 hydrogen atoms = \( 5 \times 1.008 \, g/mol = 5.04 \, g/mol \)

Summing gives an empirical molar mass of approximately 29.07 g/mol.
To find the molecular formula, divide the molar mass calculated using the Ideal Gas Law (about 30.07 g/mol) by the empirical molar mass (29.07 g/mol). The result of around 1 suggests that the molecular formula is also \( C_2H_5 \), as the calculated factor approximates an integer.