Presentation on theme: "Unit 1: Gases and Stoichiometry Geoff Hackett and Adam Serck."— Presentation transcript:
Unit 1: Gases and Stoichiometry Geoff 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. Empirical Formula Molecular Formula Chemistry: The Central Science. Brown, Lemay, Bursten, Murphy. (105) C: (40.92g C)(1 mol C/ 12.01g C)= 3.407 mol C H: (4.58g H)(1 mol H/ 1.008g H)= 4.54 mol H O: (54.50g O)(1 mol O/ 16.00g O)= 3.406 mol O C: (3.407/ 3.406)= 1 H: (4.54/ 3.406)= 1.33 O: (3.406/ 3.406)= 1 C-H-O= 3(1-1.33-1)= 3-4-3C 3 H 4 O 3 C 3 H 4 O 3 : 88.062g
Limiting Reactant/ Excess Reactant There 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 Analysis In 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.
Hydrates A hydrate is a stoichiometrically consistent amount of water in a compound. XY3 nH2O XY 3 + nH 2 O (g)
Ideal Gas Laws PV = 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 CaCO 3 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 CO 2 gas were generated? V= 250mL = 0.250L P= 1.3atm T= 31°C= 304 K n= PV/RT n= (1.3atm)(o.250L)/(0.0821 L-atm/mol-K)(304 K) n= 0.013 mol CO 2
Gas Density Molar 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 mass R: 0.08206 = 8.314 (given) T: temperature (K)
Partial Pressure (Dalton) The partial pressure of a gas can be related to its mass P 1 = X 1 P total P 1 : partial pressure X 1 : mole fraction, n a /n total P total : total pressure P 1 = n 1 (RT/V), P 2 = n 2 (RT/V), …
Partial Pressure and Stoichiometry A gaseous mixture made from 6.00g O 2 and 9.00g CH 4 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? n O2 = (6.00g O 2 )(1 mol O 2 / 32.0g O 2 )= 0.188mol O 2 n CH4 = (9.00g CH 4 )(1 mol CH 4 /16.0g CH 4 )= 0.563mol CH 4 P O2 = (0.188 mol)(0.0821 L-atm/mol-K)(273 K)/(15.0 L)= 0.281 atm P CH4 = (0.563 mol)(0.0821 L-atm/mol-K)(273 K)/(15.0 L)= 0.841 atm P total = 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.)