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Phosgene is a highly toxic gas made up of carbon, oxygen, and chlorine atoms. Its density at \(1.05 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\) is \(4.24 \mathrm{~g} / \mathrm{L}\) (a) What is the molar mass of phosgene? (b) Phosgene is made up of \(12.1 \% \mathrm{C}, 16.2 \% \mathrm{O},\) and \(71.7 \%\) Cl. What is the molecular formula of phosgene?

Short Answer

Expert verified
The molar mass of phosgene is approximately 98.9 g/mol, and its molecular formula is COCl2.

Step by step solution

01

Using the Ideal Gas Law equation

We are given the density of phosgene gas, and we know the values of pressure and temperature. We can use the ideal gas equation, which relates pressure, volume, the amount of substance (moles), and temperature as \(PV = nRT\). We will also need to use the molar mass to convert from grams to moles.
02

Calculate moles

We also know that density (\(\rho\)) is equal to mass per unit volume: \(\rho = \frac{m}{V}\). We can rewrite the ideal gas law in terms of mass by substituting \(\frac{m}{M}\) for \(n\) (where \(M\) is the molar mass): \(PV = \frac{m}{M}RT\).
03

Make molar mass the subject

We can rewrite the equation to make molar mass the subject of the formula, giving \(M = \frac{mRT}{PV}\). To find the molar mass, we'll need to use the given density \(\rho\), pressure \(P\), and temperature \(T\).
04

Plug in the values

We are given the density \(\rho = 4.24 \, \mathrm{g/L}\), pressure \(P = 1.05 \, \mathrm{atm}\), and temperature \(T = 25^{\circ} \mathrm{C} = 298.15 \, \mathrm{K}\). Using the ideal gas constant \(R = 0.0821 \, \mathrm{L \cdot atm/mol \cdot K}\), we can now calculate the molar mass \(M\): \(M = \frac{\rho RT}{P} = \frac{(4.24 \, \mathrm{g/L})(0.0821 \, \mathrm{L \cdot atm/mol \cdot K})(298.15 \, \mathrm{K})}{1.05 \, \mathrm{atm}} = 98.90 \, \mathrm{g/mol}\) So, the molar mass of phosgene is approximately \(98.9 \, \mathrm{g/mol}\). -B. Determining the molecular formula of phosgene-
05

Calculate moles of each element

We are given the mass percent composition of carbon (12.1%), oxygen (16.2%), and chlorine (71.7%). Since the total mass is 100%, we can assume a 100 g sample and determine the moles for each element: C: \(\frac{12.1 \, \mathrm{g}}{12.01 \, \mathrm{g/mol}} = 1.008 \, \mathrm{moles}\) O: \(\frac{16.2 \, \mathrm{g}}{16.00 \, \mathrm{g/mol}} = 1.0125 \, \mathrm{moles}\) Cl: \(\frac{71.7 \, \mathrm{g}}{35.45 \, \mathrm{g/mol}} = 2.022 \, \mathrm{moles}\)
06

Determine the mole ratios

We need to find the lowest whole number ratio of moles for each element to determine the molecular formula. To do this, divide all the moles by the smallest value (1.008): C: \(\frac{1.008}{1.008} = 1\) O: \(\frac{1.0125}{1.008} = 1\) Cl: \(\frac{2.022}{1.008} = 2\)
07

Write the molecular formula

Based on the mole ratios, we can now write the molecular formula for phosgene as COCl2. In conclusion, the molar mass of phosgene is approximately \(98.9 \, \mathrm{g/mol}\), and its molecular formula is COCl2.

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Key Concepts

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

Molar Mass Calculation
To understand the chemical and physical properties of any substance, it's crucial to know its molar mass, which is the mass of one mole of that substance. The molar mass is calculated in units of grams per mole (g/mol). It is fundamental for converting between grams and moles, an essential step in many chemical calculations. In the case of phosgene, we calculated its molar mass by utilizing the ideal gas law in a rearranged form. The calculation uses the density of the gas, the universal gas constant, and ambient conditions of temperature and pressure. By solving for the molar mass, you have taken the first step in revealing the quantitative aspects of phosgene's chemical behavior.
Ideal Gas Law
A cornerstone of basic chemistry is the ideal gas law, represented by the equation PV = nRT. The law connects four essential properties of a gas: pressure (P), volume (V), number of moles (n), and temperature (T), with R being the ideal gas constant. For real-world applications, it can be tweaked to calculate various properties, including density and molar mass. In your textbook exercise, the ideal gas law was modified to allow molar mass determination of phosgene by incorporating the gas's density. This represents an applied use of the ideal gas law beyond its typical textbook problems and provides a tangible example of how the ideal gas law fits into practical chemical analysis.
Molecular Formula Determination
Determining the molecular formula involves finding the actual numbers of atoms of each element present in a molecule. This process often starts with the empirical formula, which shows the simplest whole-number ratio of elements. However, for a complete picture, molecular formula determination is essential, as it reflects the actual substance's composition. For phosgene, the step-by-step solution converts mass percent composition into moles for each element, then calculates the mole ratios to determine the simplest whole number of atoms. This information leads directly to the establishment of phosgene's molecular formula, COCl2, a critical piece of information for anyone studying this compound's chemistry.
Mass Percent Composition
Finally, the mass percent composition gives chemists quick insight into the elemental makeup of a compound. It is defined as the mass of each element relative to the total mass of the compound, expressed as a percentage. Understanding this concept is vital in fields such as stoichiometry and analytical chemistry. With mass percent composition, as seen in the phosgene example, chemists can start with any mass of substance and determine the proportionate masses of each constituent element, which is exactly what was done to find the number of moles of carbon, oxygen, and chlorine in the textbook solution. This fundamental concept lays the groundwork for deeper chemical analysis and reaction prediction.

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Most popular questions from this chapter

The contents of a tank of natural gas at 1.20 atm is analyzed. The analysis showed the following mole percents: 88.6% CH4, 8.9% C2H6, and 2.5% C3H8. What is the partial pressure of each gas in the tank?

An intermediate reaction used in the production of nitrogen-containing fertilizers is that between ammonia and oxygen: $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ A 150.0 - \(\mathrm{L}\) reaction chamber is charged with reactants to the following partial pressures at \(500^{\circ} \mathrm{C}: P_{\mathrm{NH}_{3}}=1.3 \mathrm{~atm}\) \(P_{\mathrm{O}_{2}}=1.5 \mathrm{~atm} .\) What is the limiting reactant?

A tank is filled with gas to a pressure of \(875 \mathrm{~mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C}\). The gas is transferred without loss to a tank twice the size of the original tank. If the pressure is to remain constant, at what temperature (in \(^{\circ} \mathrm{C}\) ) should the tank be kept?

Sulfur trioxide can be prepared by burning sulfur in oxygen. $$ 2 \mathrm{~S}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g) $$ A \(5.00-\mathrm{L}\) flask containing \(5.00 \mathrm{~g}\) of sulfur and oxygen at a pressure of \(995 \mathrm{~mm} \mathrm{Hg}\) and \(25^{\circ} \mathrm{C}\) is heated. When the reaction is complete, the temperature in the flask is \(138^{\circ} \mathrm{C}\). (a) What is the pressure of \(\mathrm{SO}_{3}\) in the flask? (b) What is the total pressure in the flask? (c) When water is added to \(\mathrm{SO}_{3}, \mathrm{H}_{2} \mathrm{SO}_{4}\) is formed. What is the molarity of the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) formed if \(250.0 \mathrm{~mL}\) of water are added to the flask? (Assume that there is enough water to convert all the \(\mathrm{SO}_{3}\) to \(\mathrm{H}_{2} \mathrm{SO}_{4}\).)

Consider three sealed tanks all at the same temperature, pressure, and volume. Tank A contains \(\mathrm{SO}_{2}\) gas. Tank B contains \(\mathrm{O}_{2}\) gas. Tank C contains \(\mathrm{CH}_{4}\) gas. Use LT (for "is less than"), GT (for "is greater than"), EQ (for "is equal to"), or MI (for " more information required") as answers to the blanks below. (a) The mass of \(\mathrm{SO}_{2}\) in tank \(\mathrm{A}\) the mass of \(\mathrm{O}_{2}\) in \(\operatorname{tank} \mathrm{B}\). (b) The average translational energy of \(\mathrm{CH}_{4}\) in \(\operatorname{tank} \mathrm{C}\) the average translational energy of \(\mathrm{SO}_{2}\) in \(\operatorname{tank} \mathrm{A}\) (c) It takes \(20 \mathrm{~s}\) for all of the \(\mathrm{O}_{2}\) gas in tank \(\mathrm{B}\) to effuse out of a pinhole in the tank. The time it takes for all of the \(\mathrm{SO}_{2}\) to effuse out of tank A from an identical pinhole \(40 \mathrm{~s}\) (d) The density of \(\mathrm{O}_{2}\) in tank \(\mathrm{B} \longrightarrow\) the density of \(\mathrm{CH}_{4}\) in tank C. (e) The temperature in tank \(\mathrm{A}\) is increased from \(150 \mathrm{~K}\) to \(300 \mathrm{~K}\). The temperature in tank \(\mathrm{B}\) is kept at \(150 \mathrm{~K}\). The pressure in tank \(\mathrm{A}\) is half the pressure in \(\operatorname{tank}\) B.

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