Question

Calculate the density for each of the following gases at STP: (a) xenon, \(\mathrm{Xe}\) (b) fluorine, \(\mathrm{F}_{2}\) (c) propane, \(\mathrm{C}_{3} \mathrm{H}_{8}\) (d) butane, \(C_{4} H_{10}\)

Short answer

Xenon: 5.86 g/L, Fluorine: 1.70 g/L, Propane: 1.97 g/L, Butane: 2.59 g/L.

Step-by-step solution

Identify the formula for density

Density is defined as mass per unit volume. In this exercise, we will use the formula for calculating the density of gases: \[ \text{Density} = \frac{\text{Molar Mass}}{\text{Volume at STP}} \] where the volume of 1 mole of gas at Standard Temperature and Pressure (STP) is \(22.4\) liters.

Calculate molar mass of each gas

- For xenon (Xe), the molar mass is \(131.29 \, \text{g/mol}\).- For fluorine (\(\text{F}_2\)), the molar mass is \(2 \times 18.998 \, \text{g/mol} = 37.996 \, \text{g/mol}\).- For propane (\(\text{C}_3\text{H}_8\)), the molar mass is \((3 \times 12.01) + (8 \times 1.008) \, \text{g/mol} = 44.094 \, \text{g/mol}\).- For butane (\(\text{C}_4\text{H}_{10}\)), the molar mass is \((4 \times 12.01) + (10 \times 1.008) \, \text{g/mol} = 58.124 \, \text{g/mol}\).

Calculate the density of the gases at STP

- For xenon, density \[ = \frac{131.29 \, \text{g/mol}}{22.4 \, \text{L}} = 5.86 \, \text{g/L} \]- For fluorine, density \[ = \frac{37.996 \, \text{g/mol}}{22.4 \, \text{L}} = 1.697 \, \text{g/L} \]- For propane, density \[ = \frac{44.094 \, \text{g/mol}}{22.4 \, \text{L}} = 1.968 \, \text{g/L} \]- For butane, density \[ = \frac{58.124 \, \text{g/mol}}{22.4 \, \text{L}} = 2.594 \, \text{g/L} \].

Key concepts

Molar Mass

The concept of molar mass is fundamental in chemistry, especially when dealing with gases. Molar mass refers to the mass of a given substance (chemical element or chemical compound) divided by the amount of substance.
This means it is essentially a measure of how much one mole of a chemical weighs, usually expressed in grams per mole (g/mol). Understanding molar mass helps you determine how much one mole of any substance weighs, which is crucial for calculating things like density or when measuring specific quantities in a reaction. For example, to determine the molar mass of xenon (Xe), you only need to look at the periodic table, which shows xenon has a molar mass of 131.29 g/mol. However, when you're dealing with compounds, like fluorine gas (F₂), you must consider each atom in the molecule: - Fluorine consists of two atoms of fluorine, and each has a molar mass based on atomic weight which is approximately 19 g/mol. - Hence, the molar mass for F₂ would be twice that of a single fluorine atom. Knowing how to calculate the molar mass is a building block for further calculations, helping you transition into applying the density formula.

Standard Temperature and Pressure (STP)

Standard Temperature and Pressure (STP) is a set point that is widely used to refer to a specific set of conditions. This generally makes scientific calculations easier because it creates consistency across experiments and equations. At STP, the temperature is 0°C (or 273.15 K), and the pressure is 1 atm. Under these conditions, one mole of any ideal gas occupies a volume of 22.4 liters. The usage of STP allows scientists to compare and predict how gases behave under these standard conditions, ensuring accuracy. When you are performing calculations involving gas density or molar volumes, being able to rely on STP values simplifies the processes and gives predictable results for comparisons.

Density Formula

The density formula for gases links the concepts of molar mass and the ideal gas volume at STP to provide a value for gas density. Density, in general, is defined as mass per unit volume. For gases at STP, specifically, the density can be calculated by: \[ \text{Density} = \frac{\text{Molar Mass}}{\text{Volume at STP}} \]Where:- Molar Mass is the mass of one mole of gas, expressed in g/mol.- Volume at STP is the volume of one mole of gas, commonly 22.4 liters per mole.Calculating the density of a gas involves finding the molar mass using previous steps and dividing it by 22.4 L (volume of gas at STP).
For example, with xenon gas, the formula calculation: - Molar Mass of Xe is 131.29 g/mol divided by the STP volume 22.4 L.- Calculating this gives a density of 5.86 g/L for xenon at STP.This process highlights the utility of the density formula in converting theoretical concepts into tangible values.

Chemical Calculation

Chemical calculations are integral to understanding how different elements and compounds interact. They allow you to predict the outcomes of reactions, determine quantities required, and calculate properties like density or concentration. When it comes to gases, these calculations often revolve around: - Determining molar masses. - Relating moles to grams through molar conversion. - Using the ideal gas law to find volumes or pressures. In the context of our exercise: 1. First, you must calculate the molar mass of the gas in question. 2. Using the volume at STP, you find its density. 3. This is done by dividing the molar mass by the standard molar volume (22.4 L). These steps showcase how chemical calculations are executed from start to finish, underpinning many chemical principles taught in higher education.

Gases in Chemistry

Gases in chemistry present unique attributes and behaviors that can be challenging but fascinating to study. They are all governed by the gas laws, which describe how gases respond to changes in pressure, temperature, and volume. Some quick facts about gases: - They have neither a fixed shape nor a fixed volume. - They will expand to fill the size and shape of their container. - Under identical conditions (STP), gases behave in a predictable and uniform manner, enabling the use of concepts like a universal molar volume for calculations. Understanding gases is critical because it lays the groundwork for further exploration into states of matter and kinetic molecular theory. By exploring these characteristics, students gain insights that are crucial when progressing to more advanced topics like reaction kinetics or thermodynamics. Ensuring a grasp of these concepts will aid both basic and advanced chemical investigations.