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Accelerated ChemistryChapter 12

Molecular Composition of Gases

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First, Let’s Review Chapter 11…

The coefficients in the balanced equation give the ratio of moles of reactants and products

Stoichiometric Calculations

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Stoichiometric Calculations

• From the mass of Substance A you can use the ratio of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant)

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Stoichiometric Calculations

Starting with 1.00 g of C6H12O6…

we calculate the moles of C6H12O6…

use the coefficients to find the moles of H2O…

and then turn the moles of water to grams

C6H12O6 + 6 O2 6 CO2 + 6 H2O

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Let’s Review…

• How Many Cookies Can I Make?– You can make cookies

until you run out of one of the ingredients

– Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat)

Limiting Reactants

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How Many Cookies Can I Make?

• In this example the sugar would be the ________________, because it will limit the amount of cookies you can make

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Limiting Reactants

• The limiting reactant is the reactant present in the smallest stoichiometric amount– In other words, it’s the reactant you’ll run out of

first (in this case, the H2)

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Limiting Reactants

• In the example below, the O2 would be the ______________________

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12.1 Volume - Mass Relationships of Gases

• Last chapter, we related the volume and the mass of a gas:– _____________________________

• Note: a “stoich” problem can be recognized by the fact that information from one substances is given and the problem is asking about a different substance.

– _____________________________

• Let’s compare the volumes of gases in two example problems…

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…Reaction Stoichiometry at standard conditions

• Example 1 - How many L of Oxygen are needed to react with 50.0L of Hydrogen at STP?

2H2(g) + O2(g) ---> 2 H2O(g)

– But...we can skip steps 1 and 3.– Why? All gases take up the same amount of space at

STP.

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• Example 2 - Calculate the volume of one mole of H2 at 20ºC and 1000 torr.

…The Combined Gas Law at nonstandard conditions

• In both examples, the relationship between the volume and mass of a gas can be compared because…

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Avogadro’s Principle

• 1 mole of gas at STP = ____________• “Equal volumes of gases at equal temperatures and

equal temperatures and equal pressure contain an equal number of molecules.”

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In general…

• The combined gas law, PoTnVo = PnToVn, is used for changing conditions.

• But, a new gas law, the Ideal Gas Law, can be introduced when you have problems containing:– one set of _________________– solving for _________________– solving for _________________– calculating ____________________________– calculating _________________– involving _______________________________

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12.2 The Ideal Gas Law

• A relationship between pressure, volume, temperature and the # of moles of a gas

• A new formula that can help us solve all kinds of gas law problems more easily.

PV = nRT

.

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The Ideal Gas Law

• Again, the Ideal Gas Law, can be introduced when you have problems containing:– one set of conditions– solving for grams– solving for moles– calculating molecular weight (molar mass)– calculating density– involving stoichiometry and non-STP conditions

• Let’s try sample problems for each of the above conditions.

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One Set of Conditions

Example: Calculate the volume of 1.00 mole of Hydrogen at 20.0 ˚C and 1000 torr

Step 1: Convert your pressure to atm.

Step 2: Write the ideal gas law and derive Volume.

Step3: Using V = nRT/P, we have

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Solving for Grams

Example: Calculate the number of grams of helium in a 6.0 liter cylinder at 27˚C and 800 torr.

Step 1: Convert your pressure to atm.

Step 2: Write the ideal gas law and derive for moles.

PV = nRT

n = PV/RT

800 torr x 1 atm / 760 torr = 1.05 atm

Step 3: Calculate the moles.

n = (1.05 atm)(6.0 L)/(.0821 L atm/mol K)(300K)

n = 0.26 molesStep 4: Convert to grams.

.26 moles x 4.0 g/mole = 1.04 g He

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Solving for Moles

Example: A sample of CO2 in a 10.0 L container at 293K exerts a pressure of 50,000 torr. How many moles of CO2 are in your sample?

Step 1: Convert your pressure to atm.

Step 2: Write the ideal gas law and derive for moles.

Step 3: Calculate the moles.

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Calculating M.W.

Example: If 18.0 grams of a gas at 380 torr and 546.0 K occupies 44.8 L, what is the molecular weight of the gas?

Step 2: Identify the m.w. formula. So, plug in 18.0 grams into the formula.

Step 3: Now, use PV = nRT to solve for n (the number of moles).

Step 1: Convert your pressure to atm.

Step 4: Plug this value into the molecular weight (m.w.) equation.

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Step 2: If it is assumed that one mole of CO2 is present, then the mass can be calculated from the periodic table and the volume at STP is 22.4 L.

Calculating Density at STP

Example: Find the density of carbon dioxide at STP.

Step 1: Identify the density formula.

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Step 2: Assume one mole of CO2. Thus, CO2 weighs 44.0 grams. Plug this into the Density formula.

Calculating Density (not at STP)

Example: Find the density of carbon dioxide at 546.0 K and 4.00 atm.

Step 1: Identify the density formula.

Step 3: Because the conditions are non-STP values, 22.4 L can’t be used. So, use PV=nRT and solve for V (assume one mole).

Step 4: Plug this value into the density equation.

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Deviations from Ideal Behavior

• Real gases do not behave according to the KMT - Why?– Real gases have molecules that occupy space– Real gases have attractive and repulsive forces

• Ideal gases conform exactly to the KMT– no such gas exists– gases only behave close to ideally at low P and high T.– At low T and high P, gases deviate greatly from ideal behavior.

• There is an equation that enables us to account for deviations in behavior - it contains correction factors that are specific for a gas - Van der Walls equation – you’ll see it in College Chemistry

• Some gases are close to ideal (if they are small and nonpolar):– H2, He, Ne– O2 and N2 are not too bad– NH3, H2O are not even close to ideal

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Deriving the Ideal Gas Law

• Let’s derive the ideal gas law and gas constant...– Volume is proportional to 1/P (as P is reduced, the V increases)

– V is proportional to T (as T increases, the V increases)

– V is proportional to n (as more moles are added, the V increases)

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12.3 Stoichiometry of Gases

• At STP, there is nothing new here. • The only thing new is that 1 mole = 22.4

L must be adjusted if not at STP.• Problems for Stoichiometry of Gases

include converting: – _________________________– _________________________– _________________________

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Grams to Liters

Example: How many liters of O2 are generated when 50.0 grams of sodium chlorate

decomposes at 0.950 atm and 20.0˚C?

Step 1: The conditions are non-STP values, 22.4 L can’t be used so, use PV=nRT and solve for V (assume one mole).

Step 2: Plug this value into the appropriate step of the stoichiometry problem.

2NaClO3(s) + heat ---> 2NaCl(s) + 3O2(g)

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Liter to Grams

Example: If a lawn mower engine generates 555.0 L CO2 on a lovely Sunday afternoon (.996 atm and 37.00 C0 - well it is kind of hot today...), how many grams of

octane were consumed?

Step 1: The conditions are non-STP values, 22.4 L can’t be used so, use PV=nRT and solve for V (assume one mole).

Step 2: Plug this value into the appropriate step of the stoichiometry problem.

2C8H18(l) + 25O2(g) ---> 16CO2(g) + 18H2O(g)

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Liter to Liter (at STP)

Example: How many L of CO2 can be made from the combustion of 2.00 L of

propane (C3H8) at STP?

– Again...we can skip steps 1 and 3.– Since, all gases take up the same amount of space at STP.

4C8H3(l) + 18O2(g) ---> 16CO2(g) + 6H2O(g)

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Liter to Liter (not at STP)

Example: How many L of carbon dioxide can be made from the combustion of 2.00 L

of propane (C3H8) at 500. K and 3.00 atm? 4C8H3(l) + 18O2(g) ---> 16CO2(g) + 6H2O(g)

Step 1: The conditions are non-STP values, 22.4 L can’t be used so, use PV=nRT and solve for V (assume one mole).

Step 2 : Plug this value into the appropriate step of the stoichiometry problem.

– Again...we can skip steps 1 and 3.

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12.4 Effusion and Diffusion

• Graham’s Law of Diffusion - (English - 1824) - Graham noticed that gases with low densities diffuse ____________ than gases with higher densities

• “Under the same conditions of temperature and pressure, gases diffuse at a rate ____________ proportional to the square roots of their ____________ ____________ “

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Graham’s Law Formulas

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• Gas “A” has a kinetic energy of ½ mv2 and Gas “B” has a kinetic energy of ½ mv2.

• At the same temperature, the two gases will have the same kinetic energy.

Derive Graham’s Law from Gas A ½ mv2 = Gas B ½ mv2

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In other words….

• If CO2 molecules travel at 200.0 mph, how fast do H2 molecules go?

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• If He atoms travel at 800.0 mph, how fast do nitrogen molecules go?

A second example….

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Citations

• Textbook– Nicholas D. Tzimopoulos, H. Clark

Metcalfe, Williams, Castka, Holt Modern Chemistry. Austin, Texas: Holt Rinehart & Winston, 1993.