Question

A sample of a liquid hydrocarbon known to consist of molecules with five carbon atoms is vaporized in a 0.204 -L flask by immersion in a water bath at \(101^{\circ} \mathrm{C}\). The barometric pressure is 767 torr, and the remaining gas weighs \(0.482 \mathrm{~g}\). What is the molecular formula of the hydrocarbon?

Short answer

The molecular formula of the hydrocarbon is \( \text{C}_5\text{H}_{12} \).

Step-by-step solution

Convert pressure to atm

First, convert the barometric pressure from torr to atm. Use the conversion factor 1 atm = 760 torr. \[ P = \frac{767 \text{ torr}}{760 \text{ torr/atm}} = 1.0092 \text{ atm} \]

Convert temperature to Kelvin

The temperature given is in Celsius (\(101^{\textdegree} \text{C}\)). Convert it to Kelvin using the conversion formula: \( T(K) = T(C) + 273.15 \). \[ T = 101 + 273.15 = 374.15 \text{ K} \]

Use the Ideal Gas Law to find moles of gas

Use the Ideal Gas Law \( PV = nRT \) to find the number of moles (n) of the gas. Given: \[ V = 0.204 \text{ L}, R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \] Solve for n: \[ n = \frac{PV}{RT} = \frac{(1.0092 \text{ atm})(0.204 \text{ L})}{(0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1})(374.15 \text{ K})} = 0.00669 \text{ mol} \]

Calculate the molar mass of the gas

Use the mass of the gas and the moles calculated to find the molar mass. \[ Molar \text{ mass} = \frac{\text{mass}}{\text{moles}} = \frac{0.482 \text{ g}}{0.00669 \text{ mol}} = 72.08 \text{ g/mol} \]

Determine the molecular formula

Since it is given that the hydrocarbon consists of molecules with five carbon atoms, denote the molecular formula as \( \text{C}_5\text{H}_x \). The molar mass of \( \text{C}_5 \) is: \[ 5 \times 12.01 \text{ g/mol} = 60.05 \text{ g/mol} \] Subtract the mass of carbons from the molar mass of the hydrocarbon: \[ 72.08 - 60.05 = 12.03 \text{ g/mol} \] Since the remaining mass is approximately the mass of hydrogen required: \[ \frac{12.03 \text{ g/mol}}{1.01 \text{ g/mol}} \text{(mass of H)} \rightarrow \text{ approximately } 12 \text{ hydrogen atoms} \]Therefore, the molecular formula is \( \text{C}_5\text{H}_{12} \).

Key concepts

Ideal Gas Law

The Ideal Gas Law is crucial for solving many problems in chemistry involving gases. It relates the pressure (P), volume (V), number of moles (n), and temperature (T) of a gas, with the gas constant (R) serving as a bridge. The law is expressed as \( PV = nRT \). Here’s how each variable impacts the calculation:
- **Pressure (P)**: Measured in atmospheres (atm) or other units like torr or pascals.
- **Volume (V)**: Usually measured in liters (L).
- **Number of moles (n)**: Represents the quantity of the gas in moles, indicating the amount of substance.
- **Temperature (T)**: Always converted to Kelvin (K) in these calculations.
- **Gas constant (R)**: Commonly used value in the equation is 0.0821 L atm K^-1 mol^-1.
By rearranging the Ideal Gas Law, you can solve for any of the unknowns if the other variables are known. For example, to find the number of moles, rearrange to \( n = \frac{PV}{RT} \). This fundamental equation is an essential tool for figuring out gas behaviors under different conditions.

Molar Mass Calculation

Molar mass is a measure of the mass of one mole of a substance, typically expressed in grams per mole (g/mol). In the given exercise, calculating the molar mass of a hydrocarbon involves determining the ratio of its mass to the amount in moles. The steps are:
1. **Determine the mass**: This is usually given directly in grams.
2. **Calculate the number of moles**: Using the Ideal Gas Law, \( n = \frac{PV}{RT} \), determine the moles of the gas.
3. **Compute the molar mass**: Divide the mass by the moles; \( Molar \text{ mass} = \frac{\text{mass}}{\text{moles}} \).
For instance, if 0.482 g of a gas is equal to 0.00669 moles, the molar mass is \( \frac{0.482 \text{ g}}{0.00669 \text{ mol}} = 72.08 \text{ g/mol} \). This step is key to identifying the substance, as each compound has a unique molar mass.

Conversion of Units

Understanding unit conversion is vital in chemistry. In this exercise, there are two main conversions: pressure (from torr to atm) and temperature (from Celsius to Kelvin).
- **Pressure**: Given in torr, converted to atm using \( 1 \text{ atm} = 760 \text{ torr} \). Hence, \( \frac{767 \text{ torr}}{760 \text{ torr/atm}} = 1.0092 \text{ atm} \).
- **Temperature**: Given in Celsius, converted to Kelvin using the formula \( T(K) = T(\degree C) + 273.15 \). For 101°C, we get \( 101 + 273.15 = 374.15 \text{ K} \).
These conversions ensure consistency in units across all calculations, as the Ideal Gas Law requires specific units for correct application. Always check and convert units to their standard forms before using them in equations.

Hydrocarbons

Hydrocarbons are organic compounds composed entirely of carbon and hydrogen atoms. They can be either aliphatic (straight or branched chains) or aromatic (ring structures). In this exercise, we deal with a hydrocarbon known to have five carbon atoms.
- **Types**: Hydrocarbons are broadly classified into alkanes, alkenes, and alkynes, depending on the type of bonds between carbon atoms.
- **Alkanes**: Saturated hydrocarbons with single bonds only, following the general formula \( C_nH_{2n+2} \).
- **Hydrocarbon with five carbons**: Given it has five carbon atoms, the molecular formula is likely to be \( C_5H_x \).
Determining the exact formula involves knowing the molar mass: for \( C_5 \), the molar mass from carbon alone is 60.05 g/mol. Considering the given total molar mass of 72.08 g/mol, the remaining weight attributed to hydrogen is 12.03 g/mol, suggesting 12 hydrogen atoms. Therefore, the molecular formula matches \( C_5H_{12} \), a clear example of an alkane. This understanding reinforces the importance of hydrocarbon classification in organic chemistry.