Question

A 2.56-g sample of a colorless liquid was vaporized in a 250 -mL flask at \(121^{\circ} \mathrm{C}\) and \(786 \mathrm{mmHg}\). What is the molecular weight of this substance?

Short answer

The molecular weight of the substance is 320 g/mol.

Step-by-step solution

Convert Units

First, convert the given temperature from Celsius to Kelvin by using the formula: \[ K = ^{\circ}C + 273.15 \]So, the temperature in Kelvin is \[ K = 121 + 273.15 = 394.15 \ \text{K} \]Next, convert the pressure from mmHg to atm using:\[ 1 \ \text{atm} = 760 \ \text{mmHg} \] so \[ 786\ \text{mmHg} \times \frac{1\ \text{atm}}{760\ \text{mmHg}} \approx 1.034\ \text{atm} \].

Calculate Moles Using Ideal Gas Law

We use the ideal gas law to find the number of moles of the vaporized substance. The ideal gas law is given by:\[ PV = nRT \]Where \( P \) is the pressure (1.034 atm), \( V \) is volume (0.250 L), \( R \) is the ideal gas constant (0.0821 \( \text{L atm} \cdot \text{mol}^{-1} \cdot K^{-1} \)), and \( T \) is the temperature (394.15 K). Solve for \( n \) (number of moles): \[ n = \frac{PV}{RT} \]\[ n = \frac{1.034 \times 0.250}{0.0821 \times 394.15} \approx 0.008 \ \text{mol} \]

Determine Molecular Weight

The molecular weight \( M \) of the substance can be determined using the formula:\[ M = \frac{\text{mass of substance}}{\text{number of moles}} \]Here, the mass of the substance is given as 2.56 g and the number of moles we calculated is 0.008 mol. Therefore,\[ M = \frac{2.56}{0.008} = 320 \ \text{g/mol} \].

Key concepts

Molecular Weight Determination

Determining the molecular weight of a substance is crucial in chemistry. It helps understand the composition of molecules and is fundamental in calculations like those involving reactions and stoichiometry. The molecular weight can be found using a few known parameters through the ideal gas law, especially when the substance is in the gaseous state.
To find the molecular weight, use the equation:

• Calculate the number of moles ( n ) of the gas using the ideal gas law formula: \( PV = nRT \)
• Then, use the relationship: \( M = \frac{\text{mass}}{\text{moles}} \)

In this way, the molecular weight (\( M \)) represents the mass per mole of the substance. It is typically given in grams per mole (g/mol). In practical applications, accurate molecular weight helps in predicting physical properties and reaction outcomes.

Unit Conversion in Chemistry

Unit conversion is a fundamental skill in chemistry that ensures all units are compatible when plugged into equations. It often involves converting temperatures, pressures, volumes, and other measurements into accepted standard units.

• Temperature: Original exercise uses Celsius, which must be converted to Kelvin by adding 273.15, making calculations compatible with thermodynamic and gas laws.
• Pressure: Often reported in mmHg, must be converted to atm using the relation: \( 1 \ \text{atm} = 760 \ \text{mmHg} \), as this fits well into the ideal gas law.

Standardizing units facilitates smooth calculations and allows various physical constants like the ideal gas constant (\( R \)) to be consistently applied without errors.

Vaporization and Gases

The process of vaporization, especially in closed systems like flasks, is a key part of understanding gas behavior using the ideal gas law. When a liquid is vaporized, molecules overcome intermolecular forces and transition into a gaseous state. This can be explained using the following concepts:

• As the vaporized molecules occupy the flask’s volume, their pressure, volume, and temperature measurements can be inserted into the ideal gas law equation \( PV = nRT \).
• This allows determination of moles, a crucial step in finding molecular weight.

Comprehending the behavior of gases during vaporization helps predict how substances will respond under varying conditions, such as temperature and pressure, and reinforces the use of equations in practical applications.