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Phosgene is a highly toxic gas made up of carbon, oxygen, and chlorine atoms. Its density at \(1.05\) 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 \(71.7 \% \mathrm{Cl}\). What is the molecular formula of phosgene?

Short Answer

Expert verified
Question: Determine the molar mass and molecular formula of phosgene, given its density is 4.24 g/L at 25°C and 1.05 atm, and its percentage composition by mass is 12.1% C, 16.2% O, and 71.7% Cl. Answer: The molar mass of phosgene is approximately 98.91 g/mol, and its molecular formula is COCl₂.

Step by step solution

01

Use the Ideal Gas Law to find the molar mass of phosgene.

The Ideal Gas Law is given by \(PV=nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the gas constant and \(T\) is the temperature. We can rearrange the formula to find the molar mass of phosgene: \(PV=nRT \Rightarrow P=\frac{nRT}{V}\) Since the molar mass \(M\) is the ratio between mass \(m\) and number of moles \(n\), \(M=\frac{m}{n}\), we can rewrite the equation in terms of mass and molar mass as: \(P=\frac{m}{M}\frac{RT}{V} \Rightarrow M=\frac{mRT}{PV}\) Now, we plug in the given values: \(M=\frac{(4.24\times 0.0821\times (273+25))}{(1.05\times 1)}\)
02

Calculate the molar mass of phosgene

From the previous step, we can now calculate the molar mass: \(M=\frac{(4.24\times 0.0821\times 298)}{1.05} \approx 98.91 \, \mathrm{g/mol}\) So, the molar mass of phosgene is approximately \(98.91\, \mathrm{g/mol}\).
03

Use the percentage composition by mass to find the molecular formula of phosgene

We are given the percentages of C, O, and Cl in phosgene. First, we convert these percentages into grams assuming \(100 \, \mathrm{g}\) sample of phosgene: - Mass of C = \(12.1\% \times 100 = 12.1 \, \mathrm{g}\) - Mass of O = \(16.2\% \times 100 = 16.2 \, \mathrm{g}\) - Mass of Cl = \(71.7\% \times 100 = 71.7 \, \mathrm{g}\) Then, we divide each mass by the corresponding molar mass of each element to find the mole ratio: - Moles of C = \(\frac{12.1}{12.01} \approx 1.01\) - Moles of O = \(\frac{16.2}{16.00} \approx 1.01\) - Moles of Cl = \(\frac{71.7}{35.45} \approx 2.02\) Finally, we divide each mole ratio by the smallest mole ratio to get the molecular formula: - Ratio of C:O:Cl = \(\frac{1.01}{1.01} : \frac{1.01}{1.01} : \frac{2.02}{1.01} = 1:1:2\) So, the molecular formula of phosgene is \(\mathrm{COCl_2}\).

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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 valuable principle in chemistry that allows us to relate the pressure, volume, temperature, and number of moles of a gas. This law is represented by the formula: \(PV = nRT\), where:
  • \(P\) is the pressure of the gas in atmospheres (atm).
  • \(V\) is the volume of the gas in liters (L).
  • \(n\) is the number of moles of the gas.
  • \(R\) is the ideal gas constant \(0.0821 \text{ L atm/mol K}\).
  • \(T\) is the temperature in Kelvin (K).
To solve for the molar mass, which is the mass of one mole of a substance, we can rearrange the Ideal Gas Law to \(M = \frac{mRT}{PV}\). Here, \(m\) is the mass of the gas. Using the values provided in the exercise, this formula helps us determine the molar mass of phosgene by considering how much of the gas weighs in a specific volume under known conditions of temperature and pressure.
Molar Mass Calculation
The calculation of molar mass involves understanding the concept of the amount of a substance in terms of moles. Given the weight of the gas per liter and the conditions under which it is measured, we can use the rearranged Ideal Gas Law formula:\[M = \frac{mRT}{PV}\]By plugging in given values like the density \(4.24 \text{ g/L}\), the pressure \(1.05 \text{ atm}\), and the temperature \(25^{\circ}\text{C}\), converted to Kelvin \((298 \text{ K})\), we find the molar mass. It simplifies the process to solve for \(M\), resulting in approximately \(98.91 \text{ g/mol}\). This result indicates the weight of one mole of phosgene gas.
Molecular Formula Determination
The molecular formula gives the exact number of different atoms in a molecule. For phosgene, its molecular formula can be derived from its percentage composition by mass. Given:
  • \(12.1\%\) carbon (C),
  • \(16.2\%\) oxygen (O), and
  • \(71.7\%\) chlorine (Cl),
we assume a \(100 \text{ g}\) sample for simplicity. Thus:
  • \(12.1 \text{ g}\) of carbon,
  • \(16.2 \text{ g}\) of oxygen, and
  • \(71.7 \text{ g}\) of chlorine.
To find the moles of each element, divide their respective masses by their atomic masses: \(12.01 \text{ g/mol}\) for carbon, \(16.00 \text{ g/mol}\) for oxygen, and \(35.45 \text{ g/mol}\) for chlorine. Calculating these gives you:
  • \(\frac{12.1}{12.01} \approx 1.01\) moles of carbon,
  • \(\frac{16.2}{16.00} \approx 1.01\) moles of oxygen,
  • \(\frac{71.7}{35.45} \approx 2.02\) moles of chlorine.
The ratios \(1:1:2\) correspond to the atoms in the molecular formula \(\text{COCl}_2\), confirming the formula tells us exactly how these elements are combined in phosgene.

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Most popular questions from this chapter

A sample of \(\mathrm{CO}_{2}\) gas at \(22^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) has a volume of \(2.00 \mathrm{~L}\). Determine the ratio of the original volume to the final volume when (a) the pressure and amount of gas remain unchanged and the Celsius temperature is doubled. (b) the pressure and amount of gas remain unchanged and the Kelvin temperature is doubled.

Hydrogen is collected over water at \(25^{\circ} \mathrm{C}\) and \(748 \mathrm{~mm} \mathrm{Hg}\) in a 250-mL (3 significant figures) flask. The vapor pressure of water at \(25^{\circ} \mathrm{C}\) is \(23.8 \mathrm{~mm} \mathrm{Hg}\) (a) What is the partial pressure of hydrogen? (b) How many moles of water are in the flask? (c) How many moles of dry gas are collected? (d) If \(0.0186 \mathrm{~g}\) of \(\mathrm{He}\) are added to the flask at the same temperature, what is the partial pressure of helium in the flask? (e) What is the total pressure in the flask after helium is added?

A cylinder with a movable piston records a volume of \(12.6 \mathrm{~L}\) when \(3.0 \mathrm{~mol}\) of oxygen is added. The gas in the cylinder has a pressure of \(5.83\) atm. The cylinder develops a leak and the volume of the gas is now recorded to be \(12.1 \mathrm{~L}\) at the same pressure. How many moles of oxygen are lost?

Consider a vessel with a movable piston. A reaction takes place in the vessel at constant pressure and a temperature of \(200 \mathrm{~K}\). When reaction is complete, the pressure remains the same and the volume and temperature double. Which of the following balanced equations best describes the reaction? (a) \(\mathrm{A}+\mathrm{B}_{2} \longrightarrow \mathrm{AB}_{2}\) (b) \(\mathrm{A}_{2}+\mathrm{B}_{2} \longrightarrow 2 \mathrm{AB}\) (c) \(2 \mathrm{AB}+\mathrm{B}_{2} \longrightarrow 2 \mathrm{AB}_{2}\) (d) \(2 \mathrm{AB}_{2} \longrightarrow \mathrm{A}_{2}+2 \mathrm{~B}_{2}\)

The volume fraction of a gas \(\mathrm{A}\) in a mixture is defined by the equation $$ \text { volume fraction } \mathrm{A}=\frac{V_{\mathrm{A}}}{V} $$ where \(V\) is the total volume and \(V_{A}\) is the volume that gas A would occupy alone at the same temperature and pressure. Assuming ideal gas behavior, show that the volume fraction is the same as the mole fraction. Explain why the volume fraction differs from the mass fraction.

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