5.04 Gas Calculations

STP: at 1 atm and 273.15 K

1 mol of gas / 22.4 L or 22.4 L / 1 mol of gas These ratios can be used within a stoichiometry calculation whenever the conditions are at STP. This relationship between volume and moles replaces the need to use the ideal gas law.

Standard Temperature and Pressure (STP)

1) 1 standard temperature = 0*C 2) 1 standard temperature = 273 K 3) 1 standard pressure = 1 atm 4) 1 standard pressure = 760 torr 5) 1 standard pressure = 14.7

Practice 1: Assuming all volume measurements are made at the same temperature and pressure, how many milliliters of carbon dioxide gas can be produced when 206.5 milliliters of oxygen gas react with excess carbon monoxide?

2 CO (g) + O2 (g) --> 2 CO2 (g) *413.0 mL carbon dioxide* 206.5 mL O2 x 2 mL CO2 / 1 mL O2 = 413.0 mL CO2 can be produced Notice that the coefficients for CO2 and O2 can be used in volume ratio, in the same way we have set up mole ratios in stoichiometry problems in the past. This works for any unit of volume, as long as the unit is the same in the numerator and denominator, as long as both substances are gases under the same conditions.

Ratios Explained

2 CO (g) + O2 (g) --> 2 CO2 (g) The coefficients in this balanced equation provide a volume ratio between gases when they are all at the same temperature and pressure. A given volume of oxygen gas will react with twice that volume of carbon dioxide, because of the ratio of O2 to CO is 1:2 (given by the coefficients). The coefficients provide volume ratios (only for gaseous reactants and products) that can be used in stoichiometry calculations involving volume.

Practice 2: If 38.5 grams of lithium metal react with excess water, how many liters of hydrogen gas will be produced at 27.5°C and 1.35 atmospheres?

2 Li (s) + 2 H2O (l) --> 2 LiOH (aq) + H2 (g) *50.7 L H2* 38.5 g Li x 1 mol Li / 6.94 g Li x 1 mol H2 / 2 mol Li = 2.77 mol H2 (n) Now, plug moles, pressure, and temperature into the ideal gas law to solve for volume of hydrogen gas. V (H2) = (2.77 mol H2)(0.0821)(300.5 K) / 1.35 atm V (H2) = 50.7 L H2

Practice 1: How many grams of sodium metal are needed to react completely with 25.8 liters of chlorine gas at 293 Kelvin and 1.30 atmospheres?

2 Na (s) + Cl2 (g) --> 2 NaCl (s) *63.9 g Na* n (Cl2) = (1.30 atm)(25.8 L) / (0.0821)(293 K) n (Cl2) = 1.39 mol Cl2 Now that you have the amount of chlorine gas in moles, you can complete the stoichiometry problem. 1.39 mol Cl2 x 2 mol Na / 1 mol Cl2 x 22.99 g Na / 1 mol Na = 63.9 g Na

Example One: (The molar mass of an unknown gas can be determined by comparing its rate of effusion to the rate of effusion of a known gas. This is a technique sometimes used by chemists to help identify unknown gases)

A sample of hydrogen gas (H2) effuses through a porous container 7.96 times faster than an unknown gas. What is the molar mass of the unknown gas? rate of enffusion of H2 / rate of enffusion of X = sqrt Molar Mass X / sqrt Molar Mass H2 The individual rates of effusion of hydrogen gas and the unknown gas are not provided, but the ratio of the rates is given (H2 is 7.96 times faster than the unknown gas). 7.96 rate of H2 / 1 rate of X = sqrt Molar Mass X / sqrt Molar Mass H2 Referring to the periodic table, we can determine the molar mass of hydrogen gas to be 2.02 grams per mole. This leaves one variable, the molar mass of X, left for us to solve for. 7.96 rate of H2 / 1 rate of X = sqrt Molar Mass X / sqrt 2.02 g/mol H2 multiply both sides by the molar mass of H2 sqrt 2.02 g/mol H2 x 7.96 rate of H2 / 1 rate of X = sqrt Molar Mass X / sqrt 2.02 g/mol H2 x sqrt 2.02 g/mol H2 Molar mass of X is isolated to one side of the equation. 7.96 x sqrt 2.02 g/mol H2 / 1 rate of X = sqrt Molar Mass X Solve for molar mass. 11.3 = sqrt Molar Mass X Square both sides to get rid of the square root. (11.3)^2 = (sqrt Molar Mass X)^2 128 g/mol = Molar Mass of X

Did you know?

As you have learned from Graham's law, the rate of effusion of a gas increases as the mass of the gas particles decrease (mass and rate are inversely related). The English physicist Francis William Aston (1877-1945) was the first to use this relationship to separate isotopes, atoms of the same element with different masses. Although the technology has improved over the years, Aston's application of Graham's law is still used to separate isotopes today.

Practice 2: How many liters of propane gas (C3H8) need to be combusted to produce 3.8 liters of carbon dioxide, if all measurements are taken at the same temperature and pressure?

C3H8 (g) + 5 O2 (g) --> 3 CO2 (g) + 4 H2O (g) *1.3 L* 3.8 L CO2 x 1 L C3H8 / 3 L CO2 = 1.3 L C3H8 would need to combust In this problem, liters were used as the unit of volume. Any unit of volume can be used with the coefficients in the volume ratio, as long as the unit is the same for both gases in the ratio.

Example Problem: STP

How many grams of calcium oxide (lime) would be produced with 50.0 liters of carbon dioxide is produced at STP? CaCO3 (s) --> CaO (s) + CO2 (g) Solution: 50.0 L CO2 x 1 mol of CO2 / 22.4 L CO2 x 1 mol of CaO / 1 mol CO2 x 56.08 g CaO / 1 mol CaO = 125 g CaO Notice that when a reaction is conducted at standard temperature and pressure (STP), the mole-to-volume ratio given above can be used within the stoichiometry calculation in place of using the ideal gas law. This is a useful shortcut, but it is not necessary because you can still use the ideal gas law to solve a problem at STP. The choice is yours.

Example 1: Ideal gas law before stoichiometry If you are given the volume, pressure, and temperature of a gaseous reactant, you can use the ideal gas law to solve for moles before solving a stoichiometry problem. n = PV / RT

How many grams of sodium chloride can react with 18.3 liters of fluorine gas at 1.2 atmospheres and 299 Kelvin? F2 (g) + 2 NaCl (s) --> Cl2 (g) + 2 NaF (s) The given information (volume, pressure and temperature) about the fluorine gas can be used to solve for moles of fluorine, which can then be used to start the stoichiometry problem. n (F2) = (1.2 atm)(18.3 L) / (0.0821)(299 K) n (F2) = 0.895 mol F2 Now that you have the amount of fluorine gas in moles, you can start the stoichiometry problem. 0.895 mol F2 x 2 mol NaCl / 1 mol F2 x 58.44 g NaCl / 1 mol NaCl = 105 g NaCl

Example 2: Ideal gas law after stoichiometry If you are asked to solve for the volume or pressure of a gaseous product, you can first solve for moles of that product using stoichiometry and then plug that value into the ideal gas law. V = nRT / P or P = nRT / V

If 52.0 grams of magnesium metal react with excess hydrochloric acid, how many liters of hydrogen gas can be produced at 27°C and 0.97 atmospheres? Mg (s) + 2 HCl (aq) --> H2 (g) + MgCl2 (aq) In this problem, you can start with a stoichiometry problem and solve for moles of hydrogen gas formed. Then plug that value (n) into the ideal gas law with the temperature and pressure to solve for the volume of hydrogen gas. 52.0 g Mg x 1 mol Mg / 24.3 g Mg x 1 mol H2 / 1 mol Mg = 2.14 mol H2 (plugged into ideal gas law equation as "n") Now, plug moles, pressure, and temperature into the ideal gas law to solve for volume of hydrogen gas. Remember to convert the temperature to Kelvin before plugging it into the equation. V = (2.14 mol H2)(0.0821)(300 K) / 0.97 atm V (H2) = 54.3 L H2

Demonstration of Diffusion

In this demonstration, we will compare the rate of diffusion of ammonia gas (NH3) and hydrogen chloride gas (HCl). When these 2 colorless gases come in contact with each other, they will react to form a white solid, ammonium chloride, according to the reaction: NH3 (g) + HCl (g) --> NH4Cl (s) Cotton balls on each end of this glass tubing will be soaked with 2 different compounds, one with an ammonia (NH3) solution and the other with a hydrochloric acid (HCl) solution. As both of these liquids evaporate, they produce gas particles inside the tube. When the 2 colorless gases come in contact with each other, they will react to produce the white solid ammonium chloride. As the liquids evaporate, the gas particles will move through the glass tube. Based on the molar masses of the 2 gases, NH3 has a mass of around 17 g/mol and HCl has a mass of about 36.5 g/mol, where do you expect the gas particles to come in contact with each other and react? A white solid will form where the 2 gases collide and react.

Example Problem: Ideal Gas Law

Let's compare the problem you just saw with using the ideal gas law at STP. PV = nRT n = PV / RT How many grams of calcium oxide (lime) would be produced when 50.0 liters of carbon dioxide is produced at STP? CaCO3 (s) --> CaO (s) + CO2 (g) n (CO2) = (1.0 atm)(50.0 L) / (0.0821)(273.15 K) n (CO2) = 2.23 mol CO2 2.23 mol CO2 x 1 mol of CaO / 1 mol CO2 x 56.08 g CaO / 1 mol CaO = 125 g CaO

Solving for Pressure (in atmospheres)

P = nRT / v

Solving for Temperature (in Kelvin)

T = PV / nR

Mole Ratios and Coefficients

The coefficients in a balanced equation represent a mole ratio between the reactants and products. If the entire reaction occurs at a constant temperature and pressure, then Avogadro's law tells us that volume and moles are directly related, no matter the identity of the gas. Because of the relationships between volume and moles, the coefficients in a balanced equation can represent a volume ratio whenever pressure and temperature remain constant throughout the reaction. This means that you can convert from volume of one gaseous reactant or product to the volume of another by using the coefficients as a volume ratio.

Diffusion and Effusion

The constant motion of gas particles causes them to spread out and fill any container in which they are placed. When two gases gradually mix together because of this random motion, it is called diffusion. A similar term, effusion, describes when a gas sealed in a container with a small hole gradually leaks out of the hole as the motion of the gas particles results in the particles randomly encountering, and passing through, the hole. Because the rates of both diffusion and effusion are dependent on the velocity of gas particles, another gas law can be used to compare the rates of diffusion or effusion of two different gases. The velocity of gas particles is determined by two things: temperature and the molar mass of the gas. Temperature is a measure of the average kinetic energy of a sample of particles. The higher the temperature, the higher the velocity of the particles. However, mass also affects the velocity of the particles. Particles with a lower mass move faster than heavier particles at the same temperature. This relationship between mass and velocity is expressed by the gas law known as Graham's law.

Standard Temperature and Pressure

To aid in the comparisons of volumes or moles of gases, scientists have agreed upon conditions known as standard temperature and pressure, commonly abbreviated STP. The conditions of standard temperature and pressure are exactly one atmosphere pressure and 0°C (273.15 Kelvin). Avogadro's principle states that equal volumes of gases at the same temperature and pressure contain equal numbers of particles. At standard temperature and pressure (STP), one mole of any gas occupies a volume of 22.4 liters. This mole-to-volume relationship can be used as conversion factor in calculations pertaining to measurements and reactions conducted at STP (1 atmosphere and 273.15 Kelvin).

Ideal Gas Law and Stoichiometry

Using the relationships described in the gas laws above, scientists developed an equation that can be used to determine one property of a gas sample when the other conditions are known. Ideal gas law: PV = nRT The new variable in this equation, R, represents a constant known as the ideal gas constant. Its value depends on the units being used for pressure, volume, moles, and temperature. R = 0.0821 L atm / mol K We will use the above value of the ideal gas constant as long as volume (V) is in the unit liters, pressure (P) is in atmospheres, temperature (T) is in kelvin, and (n) is in moles. The ideal gas law can be used to solve for any of the conditions of a gas when the other three conditions are known. Although the general setup for the ideal gas law is given as PV = nRT, this equation can be rearranged to solve for any of the four properties (P, V, n, or T) of a gas. The ideal gas law can be combined with the stoichiometry you have already learned when a reactant and/or product is in the gas phase.

Solving for Volume (in liters)

V = nRT / P

Example Two

We can also make comparisons of velocity, rate of effusion, or rate of diffusion based on the general rule presented by Graham's law: Gas particles with less mass have higher velocities (and rates of effusion and diffusion) than gas particles with greater mass. Which sample of gas would effuse the fastest under the same temperature and pressure conditions? Use the periodic table to help answer this question. CH4 Cl2 O2 Ne Answer: CH4 would effuse the fastest because it has the lowest molar mass out of the gases above.

Solving for Number of Particles (in moles)

n = PV / RT

Graham's Law

rate of effusion of A / rate of effusion of B = sqrt Molar Mass B / sqrt Molar Mass A This equation shows that the rate at which two different gases (A and B) effuse from the same container is dependent on their molar masses. Notice that in the equation the molar mass of A is diagonal from the rate of A. Also notice that to actually calculate and compare the rates of effusion, the square root of each molar mass must be taken. Graham's law can be used to calculate the molar mass of an unknown gas or to predict the rate of effusion or diffusion of a gas. The concept behind Graham's law—lighter gas particles move faster at a given temperature—can also be used to make general predictions or comparisons of the velocities of two different gases without calculations.