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Chapter 5: Problem 70

A \(2.30-\mathrm{g}\) sample of white solid was vaporized in a \(345-\mathrm{mL}\) vessel. If the vapor has a pressure of \(985 \mathrm{mmHg}\) at \(148^{\circ} \mathrm{C},\) what is the molecular weight of the solid?

Short Answer

Expert verified
The molecular weight of the solid is approximately 190.08 g/mol.

Step by step solution

01

Convert Units

First, convert the pressure from mmHg to atm, and the temperature from Celsius to Kelvin. For pressure: \[ 985 \text{ mmHg} \times \frac{1 \text{ atm}}{760 \text{ mmHg}} = 1.296 \text{ atm} \] For temperature: \[ 148 + 273.15 = 421.15 \text{ K} \]
02

Use the Ideal Gas Law

Use the ideal gas law, \(PV = nRT\), to find the number of moles of the vapor. We know: - \(P = 1.296 \text{ atm}\) - \(V = 0.345 \text{ L}\) - \(R = 0.0821 \text{ L atm} / \text{mol K}\) - \(T = 421.15 \text{ K}\)Plug these into the equation: \[1.296 \times 0.345 = n \times 0.0821 \times 421.15\] Solve for \(n\):\[ n = \frac{1.296 \times 0.345}{0.0821 \times 421.15} \approx 0.0121 \text{ moles}\]
03

Calculate Molecular Weight

The molecular weight \(M\) (in g/mol) is calculated by dividing the mass of the sample by the number of moles. Given mass \(m = 2.30 \text{ g}\), and moles \(n = 0.0121\):\[ M = \frac{m}{n} = \frac{2.30}{0.0121} \approx 190.08 \text{ g/mol}\]

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Key Concepts

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

Molecular Weight Calculation
Understanding the molecular weight of a substance is crucial in chemistry. It helps in identifying the chemical formula and properties of a compound. Molecular weight, also known as molecular mass, is the mass of a given molecule. It is usually expressed in atomic mass units (amu) or grams per mole (g/mol).

To calculate molecular weight, divide the mass of the compound by the number of moles of the substance. In our exercise, the solid had a mass of \(2.30\) grams and there were \(0.0121\) moles of it in the vaporized state. As calculated:
  • Molecular Weight \( M = \frac{2.30}{0.0121} \approx 190.08 \text{ g/mol}\).
  • This tells us that each mole of this substance has a mass of approximately \(190.08\) grams.
Understanding this concept aids in determining the actual elements or compounds present in a substance, based on their molecular weights.
Pressure Conversion
Pressure is a measurement of force exerted per unit area. Sometimes, it's necessary to convert pressure into different units when applying gas laws.
In this exercise, the pressure measured in millimeters of mercury (mmHg) was converted to atmospheres (atm), a common unit in chemical calculations.
The conversion factor is:
  • 1 atm = 760 mmHg
Thus, to convert \(985 \text{ mmHg}\) to atm:
  • \(985 \times \frac{1\ \text{atm}}{760\ \text{mmHg}} = 1.296\ \text{atm}\)
Performing correct pressure conversion is crucial for accurate application of the Ideal Gas Law, ensuring units are consistent with the gas constant \(R\). This ensures reliable calculation results.
Temperature Conversion
Temperature conversion is vital for chemical reactions and processes. Often, temperatures are given in Celsius (掳C) and need to be converted to Kelvin (K) when dealing with gas laws. The Kelvin scale is an absolute temperature scale starting at absolute zero.
A simple conversion formula between Celsius and Kelvin is:
  • \( K = ^\circ C + 273.15 \)
For this exercise, the vapor's temperature \(148^\circ C\) was converted to Kelvin:
  • \(148 + 273.15 = 421.15\ \text{K}\)
Using Kelvin in calculations involving gases is crucial because it reflects absolute temperature, impacting the volume and pressure relationships within the Ideal Gas Law.
Mole Calculation
Calculating the number of moles in a gas sample helps us understand the quantity of substance present. The Ideal Gas Law, \( PV = nRT \), relates pressure, volume, and temperature to the number of moles \( n \), with \( R \) being the gas constant.
The procedure for solving for \( n \) includes:
  • Identify the known variables: pressure \( P = 1.296 \text{ atm} \), volume \( V = 0.345 \text{ L} \), temperature \( T = 421.15 \text{ K} \), and \( R = 0.0821 \text{ L atm/mol K}\).
  • Rearrange the equation to solve for \( n \): \( n = \frac{PV}{RT} \)
  • Substitute the values: \( n = \frac{1.296 \times 0.345}{0.0821 \times 421.15} \approx 0.0121 \text{ moles} \)
Understanding mole calculations helps in determining the molecular weight and in stoichiometric computations for reactions.

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Most popular questions from this chapter

A glass tumbler containing \(243 \mathrm{~cm}^{3}\) of air at \(1.00 \times\) \(10^{2} \mathrm{kPa}\) (the barometric pressure) and \(20^{\circ} \mathrm{C}\) is turned upside down and immersed in a body of water to a depth of \(25.5 \mathrm{~m}\). The air in the glass is compressed by the weight of water above it. Calculate the volume of air in the glass, assuming the temperature and barometric pressure have not changed.

Methanol, \(\mathrm{CH}_{3} \mathrm{OH},\) can be produced in industrial plants by reacting carbon dioxide with hydrogen in the presence of a catalyst. Water is the other product. How many volumes of hydrogen are required for each volume of carbon dioxide when each gas is at the same temperature and pressure?

Gas Laws and Kinetic Theory of Gases I Shown here are two identical containers labeled \(\mathrm{A}\) and \(\mathrm{B}\). Container A contains a molecule of an ideal gas, and container B contains two molecules of an ideal gas. Both containers are at the same temperature. (Note that small numbers of molecules and atoms are being represented in these examples in order that you can easily compare the amounts. Real containers with so few molecules and atoms would be unlikely.) How do the pressures in the two containers compare? Be sure to explain your answer. Shown below are four different containers \((\mathrm{C}, \mathrm{D}, \mathrm{E}\) and \(\mathrm{F}\) ), each with the same volume and at the same temperature. How do the pressures of the gases in the containers compare? Container \(\mathrm{H}\) below has twice the volume of container G. How will the pressure in the containers compare? Explain your reasoning. How will the pressure of containers \(\mathrm{G}\) and \(\mathrm{H}\) compare if you add two more gas molecules to container \(\mathrm{H}\) ? Consider containers I and J below. Container J has twice the volume of container \(\mathrm{I}\). Container \(\mathrm{I}\) is at a temperature of \(100 \mathrm{~K},\) and container \(\mathrm{J}\) is at \(200 \mathrm{~K}\). How does the pressure in container I compare with that in container \(\mathrm{J} ?\) Include an explanation as part of your answer.

5.88 An aqueous solution of ammonium nitrite, \(\mathrm{NH}_{4} \mathrm{NO}_{2}\), decomposes when heated to give off nitrogen, \(\mathrm{N}_{2}\) $$ \mathrm{NH}_{4} \mathrm{NO}_{2}(s) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g) $$ This reaction may be used to prepare pure nitrogen. How many grams of ammonium nitrite must have reacted if \(4.00 \mathrm{dm}^{3}\) of nitrogen gas was collected over water at \(26^{\circ} \mathrm{C}\) and \(97.8 \mathrm{kPa}\) ?

Sodium hydrogen carbonate is also known as baking soda. When this compound is heated, it decomposes to sodium carbonate, carbon dioxide, and water vapor. Write the balanced equation for this reaction. What volume (in liters) of carbon dioxide gas at \(77^{\circ} \mathrm{C}\) and \(756 \mathrm{mmHg}\) will be produced from \(26.8 \mathrm{~g}\) of sodium hydrogen carbonate?
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