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Chapter 5: Problem 144

A large flask with a volume of \(936 \mathrm{~mL}\) is evacuated and found to have a mass of \(134.66 \mathrm{~g} .\) It is then filled to a pressure of \(0.967\) atm at \(31^{\circ} \mathrm{C}\) with a gas of unknown molar mass and then reweighed to give a new mass of \(135.87 \mathrm{~g}\). What is the molar mass of this gas?

Short Answer

Expert verified
The molar mass of the unknown gas is approximately \(32.88 \frac{g}{mol}\).

Step by step solution

01

Convert given information to proper units

First, we need to convert the given information into proper units. The temperature should be in Kelvin, and the volume should be in Liters. Given: - Volume: \(936~mL\) Convert the volume to liters: \(936~mL = 0.936~L\) - Temperature: \(31^\circ C\) Convert the temperature to Kelvin: \(31^\circ C = 304~K\)
02

Calculate the mass of the gas

We are provided with the mass of the evacuated flask and the mass of the flask with the gas. To find the mass of the gas, we subtract the mass of the empty flask from the mass of the flask with the gas: Mass of gas = (Mass of flask with gas) - (Mass of empty flask) Mass of gas = \(135.87~g - 134.66~g\) Mass of gas = \(1.21~g\)
03

Apply Ideal Gas Law to find number of moles

We will now use the ideal gas law to find the number of moles (\(n\)) of the gas: \(PV = nRT\) Rearrange the equation to solve for n: \(n = \frac{PV}{RT}\) Given: - Pressure: \(0.967~atm\) - Volume: \(0.936~L\) - Temperature: \(304~K\) - Gas constant (R): \(0.0821~ \frac{L \cdot atm}{K \cdot mol}\) Plug in the values and solve for n: \(n = \frac{0.967~atm \cdot 0.936~L}{0.0821~ \frac{L \cdot atm}{K \cdot mol} \cdot 304~K}\) \(n \approx 0.0368~mol\)
04

Calculate molar mass

To find the molar mass of the gas, we will use the relationship between mass, moles, and molar mass: Molar mass = \( \frac{mass}{moles}\) Given: - Mass of gas: \(1.21~g\) - Moles of gas: \(0.0368~mol\) Plug in the values and solve for molar mass: Molar mass = \( \frac{1.21~g}{0.0368~mol}\) Molar mass = \(32.88 \frac{g}{mol}\) So, the molar mass of the unknown gas is approximately \(32.88 \frac{g}{mol}\).

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental equation in chemistry and physics that relates the pressure, volume, temperature, and number of moles of an ideal gas. This law is represented by the equation
\[ PV = nRT \].
Here,
  • P represents the pressure of the gas
  • V is the volume
  • n is the number of moles of the gas
  • R is the universal gas constant, which is \(0.0821 \frac{L \cdot atm}{K \cdot mol}\)
  • T is the temperature in Kelvin
When using the Ideal Gas Law to find the molar mass of a gas, like in our textbook problem, we first determine the number of moles of the gas using the given conditions of pressure, volume, and temperature, and then we can relate these moles to the mass to find the molar mass.
Conversion of Units
Solving chemistry problems often requires converting between different units of measurement to ensure consistency. This is crucial because the Ideal Gas Law requires that we use specific units, like liters for volume and Kelvin for temperature.
For the temperature, we convert Celsius to Kelvin by adding 273.15, reflecting the absolute scale used by Kelvin. For volume, we often convert milliliters to liters as the Ideal Gas Law uses liters, where \(1 L = 1000 mL\). These conversions are essential for accurate calculations in chemistry and are one of the reasons why the original problem requires these first steps.
Molar Mass
Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (\(g/mol\)). It is a bridge between the mass of a substance and the amount of substance (moles).

Finding Molar Mass

To calculate the molar mass from experimental data, as in our textbook problem, you need to know the mass of the sample and the number of moles that the sample contains. The formula for molar mass (M) is given by \[ M = \frac{mass}{moles} \].In our exercise, the molar mass calculation lets us identify the gas in question by comparing the calculated value to known molar masses of elements or compounds.
Stoichiometry
Stoichiometry is the area of chemistry that pertains to the quantitative relationships between the reactants and products in a chemical reaction. It involves calculations based on the molar ratios as determined from a balanced chemical equation.
While our textbook problem does not involve a chemical reaction, the principles of stoichiometry apply to any context where we deal with the quantitative aspects of substances. For example, stoichiometry allows us to use the molar mass as a conversion factor between the mass of a gas and the number of moles, which is a crucial step in solving our textbook problem.

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Most popular questions from this chapter

In the "Méthode Champenoise," grape juice is fermented in a wine bottle to produce sparkling wine. The reaction is $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g) $$ Fermentation of \(750 . \mathrm{mL}\) grape juice (density \(=1.0 \mathrm{~g} / \mathrm{cm}^{3}\) ) is allowed to take place in a bottle with a total volume of \(825 \mathrm{~mL}\) until \(12 \%\) by volume is ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\). Assuming that the \(\mathrm{CO}_{2}\) is insoluble in \(\mathrm{H}_{2} \mathrm{O}\) (actually, a wrong assumption), what would be the pressure of \(\mathrm{CO}_{2}\) inside the wine bottle at \(25^{\circ} \mathrm{C}\) ? (The density of ethanol is \(0.79 \mathrm{~g} / \mathrm{cm}^{3} .\).)

One way of separating oxygen isotopes is by gaseous diffusion of carbon monoxide. The gaseous diffusion process behaves like an effusion process. Calculate the relative rates of effusion of \({ }^{12} \mathrm{C}^{16} \mathrm{O},{ }^{12} \mathrm{C}^{17} \mathrm{O}\), and \({ }^{12} \mathrm{C}^{18} \mathrm{O}\). Name some advan- tages and disadvantages of separating oxygen isotopes by gaseous diffusion of carbon dioxide instead of carbon monoxide.

An organic compound containing only C, H, and N yields the following data. i. Complete combustion of \(35.0 \mathrm{mg}\) of the compound produced \(33.5 \mathrm{mg} \mathrm{CO}_{2}\) and \(41.1 \mathrm{mg} \mathrm{H}_{2} \mathrm{O}\). ii. A \(65.2\) -mg sample of the compound was analyzed for nitrogen by the Dumas method (see Exercise 129), giving \(35.6 \mathrm{~mL}\) of dry \(\mathrm{N}_{2}\) at 740 . torr and \(25^{\circ} \mathrm{C}\). iii. The effusion rate of the compound as a gas was measured and found to be \(24.6 \mathrm{~mL} / \mathrm{min}\). The effusion rate of argon gas, under identical conditions, is \(26.4 \mathrm{~mL} / \mathrm{min}\). What is the molecular formula of the compound?

Concentrated hydrogen peroxide solutions are explosively decomposed by traces of transition metal ions (such as Mn or Fe): $$ 2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g) $$ What volume of pure \(\mathrm{O}_{2}(\mathrm{~g})\), collected at \(27^{\circ} \mathrm{C}\) and 746 torr, would be generated by decomposition of \(125 \mathrm{~g}\) of a \(50.0 \%\) by mass hydrogen peroxide solution? Ignore any water vapor that may be present.

Natural gas is a mixture of hydrocarbons, primarily methane \(\left(\mathrm{CH}_{4}\right)\) and ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\). A typical mixture might have \(\chi_{\text {methane }}=\) \(0.915\) and \(\chi_{\text {ethane }}=0.085\). What are the partial pressures of the two gases in a \(15.00\) - \(\mathrm{L}\) container of natural gas at \(20 .{ }^{\circ} \mathrm{C}\) and \(1.44\) atm? Assuming complete combustion of both gases in the natural gas sample, what is the total mass of water formed?
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