Laws
1. Gas Chemical analysis of a gaseous compound showed that it contained 33.0% Si and 67.0% F by mass. At 35.00C, 0.210 L of the compound exerted a pressure of 1.70 atm. If the mass of 0.210 L of the compound was 2.38 g, calculate the molecular formula of the compound. Si2F6 BLB text: p. 96 for empirical formula then calculate empirical weight PV = nRT solve for moles MW = mass/moles look at multiple 2. A chemist synthesized a greenish-yellow gaseous compound of chlorine and oxygen and finds that its density is 7.71 g/L at 36.0 0C and 2.88 atm. Calculate the molar mass of the compounds and determine its molecular formula. ClO2 PV = nRT = mol RT/MW solve for MW and substitute in 3. A mixture of gases contains 4.46 moles of neon, 0.74 argon and 2.15 moles of xenon. Calculate the partial pressure of the gases if the total pressure is 2.00 at a constant temperature. P(Ar) = 0.20 atm P(Xe) = 0.586 atm (each component) = mol(component) / mol(total) P(each component) = (each component) (Total pressure)
4. The decomposition of potassium chlorate generates 128 mL of oxygen at gas 24.00C and 762 mm Hg. Calculate the grams of oxygen obtained. The vapor pressure of water vapor at 24.00C is 22.4 mm Hg. 0.164 g oxygen 3 KClO3(s) 2 KCl(s) + 3 O2(g) P(O2) = P(T) P(H2O) g = PV(MW)/RT 5. Calculate the root mean square speed (m/s) of helium atoms at 25.0 0C. 1.36 x 103 m/s 1/2 [ 3RT/MW] put MW in kg; R =8.314 J/mol K = 8.314 kg m2/ s2mol K 6. A flammable gas made up of only carbon and hydrogen is found to effuse through a porous barrier at 1.50 min. Under the same conditions, it takes an equal volume of bromine vapor 4.73 min to effuse through the same barrier. Calculate the molar mass of the unknown gas. Note: the flammable gas must be heavier as it diffuses slower than bromine r(CH) = [ MW(Br2) ] r(Br2) MW(CH) 1.50 min [ 159.8 g/mol ]1/2 = 1.59 x 103 g/mol 4.73 min = MW(CH)
7. Given that 3.50 moles of ammonia occupy 5.20 L at 47.00C, calculate the pressure of a gas (atm) using (a) the ideal gas equation and (b) the van der Waals equation. a = 4.17 atm L2/mol2 b = 0.0371 L/mol a) PV = nRT P = 17.7 atm b) use van der Waals Eqn P = 16.2 atm