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Calculate the density in grams per liter of the following gases at STP. (a) NO (b) \(\mathrm{NO}_{2}\) (c) \(\mathrm{O}_{2}\)

Short Answer

Expert verified
The density of NO at STP is 1.34 g/L, the density of \(\mathrm{NO}_{2}\) at STP is 2.05 g/L, and the density of \( \mathrm{O}_{2} \) at STP is 1.43 g/L.

Step by step solution

01

Calculate the Molar Mass of the Gases

Calculate the molar mass of each gas using the periodic table. The molar mass of NO (Nitrogen monoxide) is approximately \(14.01 g/mol (N) + 15.999 g/mol (O) = 30.01 g/mol\). The molar mass of \(\mathrm{NO}_{2}\) (Nitrogen dioxide) is \(14.01 g/mol (N) + 2*15.999 g/mol (O) = 46.01 g/mol \). And the molar mass of \( \mathrm{O}_{2} \) (Oxygen) is \(2*15.999 g/mol (O) = 31.998 g/mol\).
02

Calculate the Density of the Gases

Use the formula for density which is mass per unit volume. The volume at STP is 22.4 liters. So, the density of NO is \(30.01 g/mol / 22.4 L/mol = 1.34 g/L\). The density of \(\mathrm{NO}_{2}\) is \(46.01 g/mol / 22.4 L/mol = 2.05 g/L\). And the density of \( \mathrm{O}_{2} \) is \(31.998 g/mol / 22.4 L/mol = 1.43 g/L\).

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Key Concepts

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

Molar Mass Calculation
Calculating the molar mass is a crucial step in determining the density of a gas. The molar mass tells us how much one mole of a substance weighs and is derived from the periodic table of elements. For instance,
  • The molar mass of Nitrogen monoxide (NO) is determined by adding the atomic masses of Nitrogen (N) and Oxygen (O). Using the periodic table, Nitrogen has an atomic mass of approximately 14.01 g/mol and Oxygen 15.999 g/mol. Thus, the molar mass of NO is 14.01 + 15.999 = 30.01 g/mol.
  • Similarly, for Nitrogen dioxide ( O_2), you have one Nitrogen and two Oxygens. This gives us a molar mass calculation as follows: 14.01 + 2 imes 15.999 = 46.01 g/mol.
  • For diatomic Oxygen ( O_2), it simply means we are doubling the weight of one Oxygen atom to get 2 imes 15.999 = 31.998 g/mol.
Understanding how to calculate molar mass is foundational in chemistry because it enables the calculation of other important properties, such as density, which will be discussed next.
Standard Temperature and Pressure (STP)
Standard Temperature and Pressure (STP) is a set of conditions for analyzing and comparing gases. STP is defined as a temperature of 0 degrees Celsius (273.15 K) and a pressure of 1 atmosphere (atm). At STP, gases have certain predictable properties, which simplify calculations of their density or volume.
  • At these conditions, one mole of any ideal gas occupies a volume of exactly 22.4 liters.
  • This predictability at STP enables chemists to easily convert between the volume and the amount of gas.
Using STP conditions as a baseline, the density of gases can be calculated using the molar mass and volume occupied at STP, leading to consistent and comparable results.
Density Formula for Gases
The density of a gas is defined as its mass divided by its volume. At standard temperature and pressure (STP), we have the advantage of knowing that the volume is consistently 22.4 liters per mole for any ideal gas. This allows us to calculate gas density using the simplified formula:\[\text{Density (g/L)} = \frac{\text{Molar Mass (g/mol)}}{22.4 \text{ L/mol}}\]For clarity, the density calculations for our example gases are:
  • For Nitrogen monoxide (NO), with a molar mass of 30.01 g/mol, the density is \(\frac{30.01}{22.4} = 1.34 \text{ g/L}\).
  • For Nitrogen dioxide (NO2), at 46.01 g/mol, the density is \(\frac{46.01}{22.4} = 2.05 \text{ g/L}\).
  • For Oxygen (O_2), at 31.998 g/mol, it amounts to \(\frac{31.998}{22.4} = 1.43 \text{ g/L}\).
Knowing how to use the density formula at STP will greatly aid in completing calculations for various scenarios in chemistry.

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Most popular questions from this chapter

What does Boyle's law tell us about the effect of pressure on the volume of a gas?

Assume that the volume of a fixed amount of gas in a rigid container does not change. Calculate the temperature in degrees Celsius to which the gas would have to be changed to achieve the change in pressure shown in the following table. $$ \begin{array}{|c|c|c|c|} \hline \begin{array}{c} \text { Initial } \\ \text { Temperature } \end{array} & \begin{array}{c} \text { Initial } \\ \text { Pressure } \end{array} & \begin{array}{c} \text { Final } \\ \text { Pressure } \end{array} & \begin{array}{c} \text { Final } \\ \text { Temperature } \end{array} \\ \hline 30.0^{\circ} \mathrm{C} & 1525 \text { torr } & 915 \text { torr } & ? \\\ \hline 250.0^{\circ} \mathrm{C} & 0.70 \mathrm{~atm} & 1042 \text { torr } & ? \\\ \hline 355 \mathrm{~K} & 500.0 \text { torr } & 1000.0 \text { torr } & ? \\ \hline \end{array} $$

For a fixed amount of gas held at constant pressure, calculate the temperature in degrees Celsius to which the gas would have to be changed to achieve the change in volume shown in the following table. $$ \begin{array}{|c|c|c|c|} \hline \begin{array}{c} \text { Initial } \\ \text { Temperature } \end{array} & \begin{array}{c} \text { Initial } \\ \text { Volume } \end{array} & \begin{array}{c} \text { Final } \\ \text { Volume } \end{array} & \begin{array}{c} \text { Final } \\ \text { Temperature } \end{array} \\ \hline 0.0^{\circ} \mathrm{C} & 70.0 \mathrm{~mL} & 140.0 \mathrm{~mL} & ? \\ \hline-37^{\circ} \mathrm{C} & 2.55 \mathrm{~L} & 85 \mathrm{~mL} & ? \\ \hline 165 \mathrm{~K} & 87.5 \mathrm{~L} & 135 \mathrm{~L} & ? \\ \hline \end{array} $$

Given a fixed quantity of a gas at constant temperature, calculate the new pressure the gas would exert if the volume were changed as shown in the following table. $$ \begin{array}{|c|c|c|c|} \hline \begin{array}{c} \text { Initial } \\ \text { Pressure } \end{array} & \begin{array}{c} \text { Initial } \\ \text { Volume } \end{array} & \begin{array}{c} \text { Final } \\ \text { Volume } \end{array} & \begin{array}{c} \text { Final } \\ \text { Pressure } \end{array} \\ \hline 845 \text { torr } & 155 \mathrm{~mL} & 1.55 \mathrm{~L} & ? \\ \hline 5.30 \mathrm{~atm} & 2.85 \mathrm{~L} & 4.50 \mathrm{~L} & ? \\ \hline 755 \text { torr } & 2.00 \mathrm{~L} & 5500 \mathrm{~mL} & ? \\ \hline \end{array} $$

A steel tank contains acetylene gas at a pressure of \(7.25 \mathrm{~atm}\) at \(18.5^{\circ} \mathrm{C}\). What is the pressure at \(37.2^{\circ} \mathrm{C}\) ?

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