Question

A certain anesthetic contains 64.9 percent \(\mathrm{C}, 13.5\) percent \(\mathrm{H}\), and 21.6 percent \(\mathrm{O}\) by mass. At \(120^{\circ} \mathrm{C}\) and \(750 \mathrm{mmHg}, 1.00 \mathrm{~L}\) of the gaseous compound weighs \(2.30 \mathrm{~g}\). What is the molecular formula of the compound?

Short answer

The molecular formula is \( \mathrm{C}_4\mathrm{H}_{10}\mathrm{O} \).

Step-by-step solution

Convert Percentages to Grams

Assume you have 100 grams of the compound. Then, you will have 64.9 grams of \( \mathrm{C} \), 13.5 grams of \( \mathrm{H} \), and 21.6 grams of \( \mathrm{O} \).

Convert grams to moles

Calculate the number of moles for each element using their molar masses: \( n_{\mathrm{C}} = \frac{64.9}{12.01} \), \( n_{\mathrm{H}} = \frac{13.5}{1.008} \), \( n_{\mathrm{O}} = \frac{21.6}{16.00} \).

Calculate moles

The moles are calculated as follows: \( n_{\mathrm{C}} = 5.405 \), \( n_{\mathrm{H}} = 13.393 \), \( n_{\mathrm{O}} = 1.35 \).

Determine the simplest mole ratio

Divide each of the moles by the smallest number obtained: \( \frac{5.405}{1.35} = 4.0 \) for C, \( \frac{13.393}{1.35} = 9.9 \approx 10 \) for H, \( \frac{1.35}{1.35} = 1 \) for O. The empirical formula is \( \mathrm{C}_4\mathrm{H}_{10}\mathrm{O} \).

Use the Ideal Gas Law to find molar mass

First, find the conditions in appropriate units: \( P = 750 \mathrm{mmHg} = 0.987 atm \), \( V = 1 \mathrm{~L} \), \( T = 120^{\circ}C = 393 \mathrm{~K} \). Use the ideal gas law: \( PV = nRT \) to find moles \( n = \frac{m}{M} = \frac{PV}{RT} \).

Solve for Molar Mass

Using \( PV = nRT \), \( M = \frac{(0.987) \times (1)}{(0.0821) \times (393)} \times 2.30 \approx 74 \mathrm{~g/mol} \).

Determine the Molecular Formula

The empirical formula \( \mathrm{C}_4\mathrm{H}_{10}\mathrm{O} \) has a molar mass of \( 74 \mathrm{~g/mol} \) which matches the calculated molar mass. Therefore, the molecular formula is also \( \mathrm{C}_4\mathrm{H}_{10}\mathrm{O} \).

Key concepts

Empirical Formula Calculation

To determine the molecular formula of a compound, it is first necessary to find the empirical formula. The empirical formula is the simplest whole-number ratio of atoms in a compound. Here’s how it's done:

• Start with the percent composition of each element in the compound. Assume you have a 100 g sample. This simplifies calculations since the percentages can be directly interpreted as grams.
• Convert these masses to moles using each element's molar mass. Moles can be calculated with the formula: \( n = \frac{mass}{Molar\, Mass} \).
• The next step involves determining the simplest mole ratio by dividing all moles by the smallest calculated mole value. This provides the subscripts for the empirical formula.

For instance, using the data provided, carbon, hydrogen, and oxygen percentages convert to moles as follows. Hydrogen and oxygen, with their calculated mole values, will help keep the ratio and bring us to the empirical formula. The empirical formula derived in our problem is \( \mathrm{C}_4\mathrm{H}_{10}\mathrm{O} \). This formula gives a basic structure, but the molecular formula is sought.

Molar Mass Calculation

Molar mass is a crucial concept in transitioning from an empirical to a molecular formula. The molar mass of a compound is the sum of the masses of all atoms present in one mole of that compound. Here's how you calculate it:

• First, you identify the mass of the empirical formula unit by adding together the atomic masses of its constituent elements in grams per mole.
• For this problem, we already have the molar mass calculation from the gas laws that suggest approximately 74 g/mol. This sticks as our reference to verify the empirical formula.

The empirical formula mass is determined by multiplying the number of each type of atom by its atomic mass and then summing these amounts. If this value perfectly matches the molar mass we calculated from physical conditions (gas measurements), the empirical formula is also the molecular formula, as seen in this example.

Ideal Gas Law

The ideal gas law is an important tool for calculating the molar mass of a gaseous compound under specific conditions of temperature and pressure. The equation can be expressed as:\[PV=nRT\]Here is how you use it:

• \(P\) is the pressure in atmospheres, \(V\) is the volume in liters, \(n\) is the number of moles of gas, \(R\) is the ideal gas constant \(0.0821 \, \text{L atm mol}^{-1}\text{K}^{-1}\), and \(T\) is the temperature in Kelvin.
• To find the molar mass, first use the ideal gas law to determine the number of moles \(n\), then relate these moles to the sample mass.
• Given \(P\), \(V\), \(T\), and the mass of the gas, use \(PV=nRT\) rearranged as \(n = \frac{PV}{RT}\) to find the moles. The molar mass \(M\) is calculated by \(M=\frac{mass}{n}\).

This technique allows the determination of the molar mass for gases under known conditions, providing a foundation for molecular formula calculations.

Percent Composition

Percent composition gives insight into the ratio of each element present in a compound, displayed as a percentage by mass. You can derive it as follows:

• The mass percentage of each element can be calculated using the formula: \(\text{Percent Composition} = \frac{\text{Mass of the element}}{\text{Total mass of the compound}} \times 100\).
• In our problem, the direct conversion of percent composition to the mass of a 100 g sample simplifies flash calculations, translating directly to grams.
• This method helped determine empirical findings for carbon, hydrogen, and oxygen, working backward to establish their ratios and moles in the initial compound sample.

Mastering percent composition provides a solid foundation for identifying the elemental framework within a compound before any other calculations proceed. This leads straight into empirical and molecular formula derivation.