Prelaboratory Assignment: Molar Mass of a Volatile Liquid l. A sample of air wit
ID: 1042752 • Letter: P
Question
Prelaboratory Assignment: Molar Mass of a Volatile Liquid l. A sample of air with a volume of 125 mL at a pressure of 745 mm Hg a tem found to weigh 0.437 As perature of 22°C is grams. Assuming air is only composed of N, and O, determine the mole suming air is only fraction of each in the sample? 2. What is the density of carbon dioxide gas in grams/Liter at 127°C and 763 mm Hg? 3 Cyclopropane gas is 85.7%C and 14.3% H by mass and has a density of 1.71 g L at 755 mm Hg and 25°C: a) What is the molar mass of cyclopropane? b) What is the chemical formula for cyclopropane? 4. A 3.35 g sample of a gas SX, has a volume of 329.5 cm' at 1.00 atm and 20°C. What element is X?Explanation / Answer
2) T=127C= 127+273=400K
P = 763mm= 763/760 =1.004 atm
molar mass of CO2 = 44.0gram/mole
R= 0.0821 L-atm/mole-K
PM=dRT
d= PM/RT = 1.004x44.0/0.0821x400 =1.345 grasm/L
density of CO2 = 1.345 gras/L
3) d= 1.71geam/L
P=755mm=755/760=0.993 atm
T=25C= 25+273=298K
PM=dRT
M= dRT/P = 1.71x0.0821x298/0.993=42.13 gram/mole
molar mass of Cyclopropane = 42.13gram/mole
Element % by mass atomic weight relative number simple ratio
C 85.7 12.0 85.7/12.0=7.142 7.142/7.142=1.0
H 14.3 1.0 14.3/1.0- 14.3 14.3/7.142= 2.0
Emperical formula= CH2
emperical formula mass= 14
Molar mass= 42.13 gram
n= molar mass/emperical formula mass = 42.13/14= 3.00
Molecular formula = nxemperical formula
Molecular formula = 3xCH2 = C3H6
Molecular formula of cyclopropane= C3H6
4)
W= 3.35 grams
V= 329.5cm3= 0.3295L
P=1atm
T=20C=20+273=293K
PV=nRT
n=PV/RT= 1x0.3295/0.0821x293=0.0137 moles
number of moles = mass/molar mass
molar mass= mass/nuber of moles = 3.35/0.0137=244.52 grams/mole
The compound is SCl6
The element X is Cl.