Question

Which of the following gases has a density of $5.40\mathrm{g}/\mathrm{L}$ at $\mathrm{STP}:{\mathrm{H}}_2\mathrm{S},{\mathrm{SO}}_3,{\mathrm{CF}}_2{\mathrm{Cl}}_2?$

Short answer

The gas with a density of 5.40 g/L at STP is $C{F}_{2}C{l}_2$.

Step-by-step solution

Understand the Problem

Express the problem of finding the density of each gas under STP conditions, which are: the temperature is 0°C or 273 Kelvin and the pressure is 1 atm. You'll need to find which gas has a density equal to 5.40 g/L already given in the problem statement.

Use the Ideal Gas Law

The Ideal Gas Law is given as: $PV=nRT$, where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the ideal gas constant $0.0821L\cdot atm/(mol\cdot K)$, and $T$ is temperature in Kelvin. At STP, $P=1atm$ and $T=273K$.

Calculate Molar Volume at STP

Under STP conditions, one mole of any gas occupies 22.4 L. Knowing the molar volume is crucial because it helps connect the mass of the gas to its volume and thus its density.

Determine the Molar Mass of Each Gas

Calculate the molar mass of each gas: - $H_{2}S$: $2(1)+32=34g/mol$- $S{O}_3$: $32+3(16)=80g/mol$- $C{F}_{2}C{l}_2$: $12+2(19)+2(35.5)=121g/mol$

Calculate Densities at STP

Find the density of each gas using the relation: Density ($D$) = Molar Mass / Molar Volume:- $H_{2}S$: $34g/mol÷22.4L/mol=1.52g/L$- $S{O}_3$: $80g/mol÷22.4L/mol=3.57g/L$- $C{F}_{2}C{l}_2$: $121g/mol÷22.4L/mol=5.40g/L$

Identify the Gas with Given Density

Compare the calculated densities to the given 5.40 g/L. From our calculations, $C{F}_{2}C{l}_2$ has a density of 5.40 g/L, which matches the problem's given density.

Key concepts

Density Calculation

Density is a measure of how much mass is contained in a given volume. For gases, density can be determined using the relationship between molar mass and molar volume. This concept is vital when we need to compare densities under Standard Temperature and Pressure (STP) conditions.To calculate the density of a gas, use the formula:$$D=\frac{M}{V}$$where:

• $D$ is the density.
• $M$ is the molar mass of the gas, typically in grams per mole $g/mol$.
• $V$ is the molar volume at a given condition, usually in liters per mole $L/mol$.

At STP, the molar volume of a gas is $22.4L/mol$. By knowing the molar mass, one can easily compute the density. This method is straightforward and offers a clear understanding of how gases behave under standard conditions. For example, in the problem, calculating the density of $C{F}_{2}C{l}_2$ provided the exact value of $5.40g/L$, confirming its match.

Molar Mass Determination

Molar mass is the mass of one mole of a substance, generally expressed in $g/mol$. It is essential in bridging the mass and the amount of a substance in chemical reactions, especially for gases.To determine the molar mass of a compound, add the atomic masses of each element, multiplied by their respective number of atoms in the molecule:- For $H_{2}S$: The atomic masses are 1 for Hydrogen and 32 for Sulfur, resulting in a molar mass of $2(1)+32=34g/mol$.- For $S{O}_3$: Sum up 32 for Sulfur and 16 for each of three Oxygen atoms, making it $32+3(16)=80g/mol$.- For $C{F}_{2}C{l}_2$: Add 12 for Carbon, 19 for each of two Fluorines, and 35.5 for each of two Chlorines, leading to $12+2(19)+2(35.5)=121g/mol$.Knowing the molar mass allows us to calculate density and resolve various gas-related problems, crucial under conditions like STP.

STP Conditions

Standard Temperature and Pressure (STP) is a set of conditions used to standardize measurements in chemistry, ensuring experiments are replicable and comparable. STP is defined as a temperature of $0^{\circ }C$ or $273K$ and a pressure of $1atm$.Under STP, gases exhibit predictable behavior, where:

• The volume occupied by one mole of any ideal gas is $22.4L$.
• The Ideal Gas Law $PV=nRT$ simplifies calculations, with $P=1atm,$ $V$ easy to determine, and $R$ being the gas constant $0.0821L\cdot atm/(mol\cdot K)$.

These conditions are crucial for comparing gases like in our problem, which involves finding a gas with a certain density at STP. Thus, understanding STP helps in identifying appropriate gases based on known or calculated properties like density, as shown through the calculation of $C{F}_{2}C{l}_2$ in this exercise.