Unit 1: Gases and StoichiometryGeoff Hackett and Adam Serck

Stoichiometry

Chemistry: The Central Science. Brown, Lemay, Bursten, Murphy. (105)

Percent Composition and Empirical Formulae

40.92% Carbon, 4.58% Hydrogen, and 54.50% Oxygen by mass.

If the molar mass is 176.124g, you have to guess and check to find the molecular formula.

EmpiricalFormula

MolecularFormula

Chemistry: The Central Science. Brown, Lemay, Bursten, Murphy. (105)

C: (40.92g C)(1 mol C/ 12.01g C)= 3.407 mol CH: (4.58g H)(1 mol H/ 1.008g H)= 4.54 mol HO: (54.50g O)(1 mol O/ 16.00g O)= 3.406 mol OC: (3.407/ 3.406)= 1H: (4.54/ 3.406)= 1.33O: (3.406/ 3.406)= 1

C-H-O= 3(1-1.33-1)= 3-4-3C3H4O3

C3H4O3: 88.062g

C3H4O2: 176.124g

Limiting Reactant/ Excess ReactantThere are two methods to find the limiting reactant:

Use reactant A and convert it into reactant B to see how much of reactant B is needed using stoichiometry.

In this method, if the amount of reactant B needed is less than the amount of reactant B that you have, then reactant B is excess and reactant A is the limiting reactant.

Use both reactants A and B to find out how much product is produced.

If reactant A produces less product, it is the limiting reactant and reactant B is in excess.

Percent Yield

The percent yield is the relation between the two using the equation:

Chemistry: The Central Science. Brown, Lemay, Bursten, Murphy. (105)

Chemical AnalysisIn a chemical analysis, use the masses given to find the mass of all the elements in a compound. Then using the mass of the compound, find the percent composition.

HydratesA hydrate is a stoichiometrically consistent amount of water in a compound.

XY3 nH2O XY3 + nH2O (g)

Gases

Ideal Gas LawsPV = nRT

P = Pressure (1.000 atm = 101.325 kPa = 101 325 Pa = 14.69 psi = 760.0 Torr)

V = Volume of Container

n = Number of Moles

R = 0.08206 L∙atmmol∙K = 8.314 Jmol∙K

T = Absolute Temperature (K)

Gas Laws and Stoichiometry

A sample of CaCO3 is decomposed, and the carbon dioxide is collected in a 250mL flask. After the decomposition is complete, the gas has a pressure of 1.3 atm at a temperature of 31°C. How many moles of CO2 gas were generated?V= 250mL = 0.250LP= 1.3atmT= 31°C= 304 K

n= PV/RTn= (1.3atm)(o.250L)/(0.0821 L-atm/mol-K)(304 K)n= 0.013 mol CO2

Gas DensityMolar mass and pressure have a direct relationship with density and temperature is inversely related.

d = d: density of the gas (g/L)P: pressure (L)M: molar massR: 0.08206 = 8.314 (given)T: temperature (K)

Partial Pressure (Dalton)

The partial pressure of a gas can be related to its mass

P1 = X1Ptotal

P1: partial pressure

X1: mole fraction, na/ntotal

Ptotal: total pressure

P1= n1(RT/V), P2= n2(RT/V), …

Partial Pressure and Stoichiometry

A gaseous mixture made from 6.00g O2 and 9.00g CH4 is placed in a 15.0-L vessel at 0°C. What is the partial pressure of each gas, and what is the total pressure in the system?

nO2= (6.00g O2)(1 mol O2/ 32.0g O2)= 0.188mol O2

nCH4= (9.00g CH4)(1 mol CH4/16.0g CH4)= 0.563mol CH4

PO2= (0.188 mol)(0.0821 L-atm/mol-K)(273 K)/(15.0 L)= 0.281 atmPCH4= (0.563 mol)(0.0821 L-atm/mol-K)(273 K)/(15.0 L)= 0.841 atm

Ptotal= 0.281 atm + 0.841 atm= 1.122 atm

Kinetic Molecular Theory

Gases consist of large numbers of molecules (or atoms) in continuous, random motion.

Gas molecules have negligible volume compared to the total volume the gas occupies.

There are negligible attractive or repulsive forces between these molecules.

At constant temperature, the average kinetic energy of these molecules remains constant over time. (There is no net loss of kinetic energy due to their collisions, which are perfectly elastic.)

Calculation of r.m.s. speed

Gas molecules’ average kinetic energy is proportional to their absolute temperature. (Equipartition of energy.)

Use urms = to calculate the speed of a gas particle

A less massive molecule will have a higher rms speed

Rates of Diffusion and Effusion

(graham)Effusion: gas molecules escape into a vacuum

Diffusion: gas molecules mix into another gas

= r: rate of effusionM: molar mass

Non-Ideal Gases(Van Der Waals)[P + a()[V-bn] = nRT

P: real gas pressurea: Van der Waals constant (given)V: real container volumeb: Van der Waals constant (given)n: number of molesR: 0.08206 = 8.314 (given)T: temperature (K)