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Chapter 10: Problem 28

A \(0.0130-\mathrm{g}\) sample of a gas with an empirical formula of \(\mathrm{C}_{4} \mathrm{H}_{5}\) is placed in a 165 -mL flask. It has a pressure of \(13.7 \mathrm{mm}\) Hg at \(22.5^{\circ} \mathrm{C}\). What is the molecular formula of the compound?

Short Answer

Expert verified
The molecular formula of the compound is \( \text{C}_{8}\text{H}_{10} \).

Step by step solution

01

Convert Pressure to Atmospheres

First, convert the pressure from mm Hg to atmospheres because the ideal gas constant often uses this unit. Use the conversion 1 atm = 760 mm Hg.\[ P = \frac{13.7 \, \text{mm Hg}}{760 \, \text{mm Hg/atm}} = 0.0180 \, \text{atm} \]
02

Convert Volume to Liters

Convert the volume from milliliters to liters. Knowing there are 1000 mL in 1 L:\[ V = \frac{165 \, \text{mL}}{1000 \, \text{mL/L}} = 0.165 \, \text{L} \]
03

Convert Temperature to Kelvin

Convert the temperature from degrees Celsius to Kelvin by adding 273.15:\[ T = 22.5 + 273.15 = 295.65 \, \text{K} \]
04

Calculate Moles using Ideal Gas Law

Use the ideal gas law \( PV = nRT \) to determine the number of moles \( n \). Using \( R = 0.0821 \, \text{L atm/mol K} \):\[ n = \frac{PV}{RT} = \frac{0.0180 \, \text{atm} \times 0.165 \, \text{L}}{0.0821 \, \text{L atm/mol K} \times 295.65 \, \text{K}} \approx 1.21 \times 10^{-4} \, \text{mol} \]
05

Calculate Molar Mass of the Gas

Use the mass of the sample and the moles just calculated to find the molar mass.\[ \text{Molar Mass} = \frac{0.0130 \, \text{g}}{1.21 \times 10^{-4} \, \text{mol}} \approx 107.44 \, \text{g/mol} \]
06

Find Empirical Formula Mass

Determine the empirical formula mass of \( \text{C}_4\text{H}_5 \): \[ 4 \times 12.01 \, \text{g/mol (C)} + 5 \times 1.008 \, \text{g/mol (H)} = 53.06 \, \text{g/mol} \]
07

Determine the Molecular Formula

Divide the molar mass by the empirical formula mass to find the ratio:\[ \text{Ratio} = \frac{107.44 \, \text{g/mol}}{53.06 \, \text{g/mol}} \approx 2.02 \]This suggests that the molecular formula is twice the empirical formula:\( \text{Molecular Formula} = \text{C}_{8}\text{H}_{10} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Empirical Formula
The empirical formula of a compound provides the simplest whole number ratio of the elements present. It doesn't reveal the actual number of atoms, but gives a foundational guideline of proportions.
For example, in the empirical formula \(\text{C}_4\text{H}_5\), this means for every four carbon atoms, there are five hydrogen atoms. To compute an empirical formula, you need to:
  • Determine the mass percentage of each element in the compound.
  • Convert these percentages to moles by dividing by the atomic mass of each element.
  • Reduce the mole ratios to the smallest whole numbers.
This provides the empirical framework to further build on, especially when combined with other data like molecular weight.
Ideal Gas Law
The Ideal Gas Law is a critical tool in chemistry and physics, relating pressure, volume, temperature, and number of moles of a gas. It's expressed in the formula:\[ PV = nRT \]where:
  • \( P \) is the pressure of the gas.
  • \( V \) is the volume.
  • \( n \) is the number of moles.
  • \( R \) is the ideal gas constant \(0.0821 \, \text{L atm/mol K}\).
  • \( T \) is the absolute temperature in Kelvin.
In the given problem, the Ideal Gas Law allows us to calculate the number of moles by rearranging the formula:\[ n = \frac{PV}{RT} \]This step is crucial because it translates the physical conditions of the gas into quantifiable data, necessary for inferring properties such as molar mass.
Molar Mass Calculation
Molar mass is the mass attributed to one mole of a given substance and is an essential component in determining molecular formulas.To calculate it, you divide the mass of the sample in grams by the moles calculated:\[ \text{Molar Mass} = \frac{\text{mass} (\text{g})}{\text{moles} (\text{mol})} \]From the exercise, the molar mass is estimated as \(107.44 \, \text{g/mol}\). With the empirical formula mass known (\(53.06 \, \text{g/mol}\)), you can determine how many empirical units fit into the molar mass:
Divide the molar mass by the empirical formula mass:\[ \text{Ratio} = \frac{107.44}{53.06} \approx 2.02 \]This tells you that the molecular formula is essentially twice the empirical formula, leading us to derive the molecular formula \(\text{C}_8\text{H}_{10}\). This concept links the abstract calculation to a real-world compound's structure.

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Most popular questions from this chapter

You want to store \(165 \mathrm{g}\) of \(\mathrm{CO}_{2}\) gas in a \(12.5-\mathrm{L}\) tank at room temperature \(\left(25^{\circ} \mathrm{C}\right) .\) Calculate the pressure the gas would have using (a) the ideal gas law and (b) the van der Waals equation. (For \(\mathrm{CO}_{2}\) \(\left.a=3.59 \text { atm } \cdot \mathrm{L}^{2} / \mathrm{mol}^{2} \text { and } b=0.0427 \mathrm{L} / \mathrm{mol} .\right)\)

One of the cylinders of an automobile engine has a volume of \(400 . \mathrm{cm}^{3} .\) The engine takes in air at a pressure of 1.00 atm and a temperature of \(15^{\circ} \mathrm{C}\) and compresses the air to a volume of \(50.0 \mathrm{cm}^{3}\) at \(77^{\circ} \mathrm{C} .\) What is the final pressure of the gas in the cylinder? (The ratio of before and after volumesin this case, 400: 50 or \(8: 1-\) is called the compression ratio.

A 1.0 -L flask contains 10.0 g each of \(\mathrm{O}_{2}\) and \(\mathrm{CO}_{2}\) at \(25^{\circ} \mathrm{C}\) (a) Which gas has the greater partial pressure, \(\mathrm{O}_{2}\) or \(\mathrm{CO}_{2}\), or are they the same? (b) Which molecules have the greater rms speed, or are they the same? (c) Which molecules have the greater average kinetic energy, or are they the same?

Acetaldehyde is a common liquid compound that vaporizes readily. Determine the molar mass of acetaldehyde from the following data: Sample mass \(=0.107\) g Volume of gas \(=125 \mathrm{mL}\) Temperature \(=0.0^{\circ} \mathrm{C} \quad\) Pressure \(=331 \mathrm{mm} \mathrm{Hg}\)

A Chlorine gas \(\left(\mathrm{Cl}_{2}\right)\) is used as a disinfectant in municipal water supplies, although chlorine dioxide \(\left(\mathrm{ClO}_{2}\right)\) and ozone are becoming more widely used. \(\mathrm{ClO}_{2}\) is a better choice than \(\mathrm{Cl}_{2}\) in this application because it leads to fewer chlorinated by-products, which are themselves pollutants. (a) How many valence electrons are in \(\mathrm{ClO}_{2} ?\) (b) The chlorite ion, \(\mathrm{ClO}_{2}^{-},\) is obtained by reducing \(\mathrm{ClO}_{2}\). Draw a possible electron dot structure for \(\mathrm{ClO}_{2}^{-} .\) (Cl is the central atom.) (c) What is the hybridization of the central Cl atom in \(\mathrm{ClO}_{2}^{-}\) ? What is the shape of the ion? (d) Which species has the larger bond angle, \(\mathrm{O}_{3}\) or \(\mathrm{ClO}_{2}^{-} ?\) Explain briefly. (e) Chlorine dioxide, \(\mathrm{ClO}_{2},\) a yellow-green gas, can be made by the reaction of chlorine with sodium chlorite: $$2 \mathrm{NaClO}_{2}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NaCl}(\mathrm{s})+2 \mathrm{ClO}_{2}(\mathrm{g})$$ Assume you react \(15.6 \mathrm{g}\) of \(\mathrm{NaClO}_{2}\) with chlorine gas, which has a pressure of \(1050 \mathrm{mm} \mathrm{Hg}\) in a 1.45-L flask at \(22^{\circ} \mathrm{C}\). What mass of \(\mathrm{ClO}_{2}\) can be produced?
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