Info: PV = nRT R = 0.0821 L atm, mole k or 62.4L torr mole^K^STP = 1 atm = 101 k
ID: 531479 • Letter: I
Question
Info: PV = nRT R = 0.0821 L atm, mole k or 62.4L torr mole^K^STP = 1 atm = 101 kPa = 760 mm Hg 273K Q = m c delta T 1 cal g^-1 k^or 4.184 J g^K^Aveg = 6.022 times 10^molar volume of gas, STP = 22.4 L Electronegativities: B:2.0, H:21, C:2.5, N: 3.0, O: 3.5 F = 4.0. CL= 3.0, S:2.5, Br:2.8. 1: 2.5 Show a neat complete setup showing units for each problem involving calculations; show answers with all Units, to the appropriate a of significant figures. No answer without a setup will be credited a) A hydrocarbon compound contains 92.3% C and 7.70 % H. What is its empirical formula? b) This same substance is a gas with a density of 1.25 g/L at 127 degree c and a pressure of 400 torr. What is its molar mass? c) What is the molecular formula of the compound How many liters of CO_2 can be produced by combustion of 5.80 g butane with 15.0 liters of oxygen. All gases measured at STP.Explanation / Answer
Q1)
a) Let the mass of the compound be 100g
Mass of carbon = 92.3gms
Number of moles of Carbon in compound = 92.03/12.01 = 7.685
Mass of Hydrogen = 7.7 gms
Number of moles of carbon in compound = 7.7/1.008 = 7.685
Hence mole ratio is 1:1, therefore the empirical formula of the compound will be CH
b)
Using the ideal gas equation
PV = nRT
PV = m/M * RT
m/V = PM/RT = density
M = density * RT/P = 1.25g * 0.0821 * (273+127)/(400/760) = 77.995 gm/mol
c) Formula of the compound will be 77.995/13 = 6
Molecular formula will be C6H6 or Benzene
2)
C4H10 + 6.5O2 ------> 4CO2 + 5H2O
molar mass of butane = 4 * 12 + 10 = 58 gm/mol
number of moles of butane = 5.80/580 = 0.10 moles
1 mole at STP contains 22.4L
Hence 0.10 moles of C4H10 will require 0.65 moles of O2 which is 0.65 * 22.4 = 14.56L
Hence C4H10 is the limiting reagent
number of moles of CO2 = 4 * moles of C4H10 = 0.4 moles
Volume of CO2 produced = 0.4 * 22.4 = 8.96L