1. A gaseous hydrocarbon collected over water at a temperature of 21 degree C an
ID: 916334 • Letter: 1
Question
1. A gaseous hydrocarbon collected over water at a temperature of 21 degree C and barometric pressure of 753 torr occupied a volume of 48.1 mL. the hydrocarbon in this volume weighs 0.1133 g. Calculate the molar mass of the hydrocarbon. the vapor pressure of the water at 21 degree C is 18.7 torr.
2. Using the data from question 1, how would your result be affected if you do not account for the vapor pressure of water (assume vapor pressure of water is negligible)?
3. Using data from question 1 calculate the volume of the hydrocarbon at standard temprature and pressure (STP). Account for the vapor pressure of water.
Explanation / Answer
(1) The pressure of hydrocarbon = P = 753 - 18.7 = 734.3 torr
= 734.3/760 atm
= 0.966 atm
Volume = V = 48.1 mL = 0.0481 L
Temperature = T = 21 oC = 21+273 = 294 K
Mass of Hydrocarbon = m = 0.1133 g
Molar mass of Hydrocarbon = M = ?
Gas constant = 0.0821 Latm/(mol-K)
We know that PV = nRT
PV = (m/M) RT
M = (mRT)/(PV)
= (0.1133x0.0821x294)/(0.966x0.0481)
= 58.8 g/mol
Therefore the molar mass of hydrocarbon is 58.8 g/mol
(2) The pressure of hydrocarbon = P = 753 torr considering vapour pressure of water
= 753/760 atm
= 0.991 atm
Volume = V = 48.1 mL = 0.0481 L
Temperature = T = 21 oC = 21+273 = 294 K
Mass of Hydrocarbon = m = 0.1133 g
Molar mass of Hydrocarbon = M = ?
Gas constant = 0.0821 Latm/(mol-K)
We know that PV = nRT
PV = (m/M) RT
M = (mRT)/(PV)
= (0.1133x0.0821x294)/(0.991x0.0481)
= 57.4 g/mol
Therefore the molar mass of hydrocarbon will be decreased to 57.4 g/mol
(3) We know that PV = nRT
PV = (w/M) RT
The pressure of hydrocarbon = P =1atm
Volume = V = ?
Temperature = T = 0 oC = 0+273 = 273 K
Mass of Hydrocarbon = m = 0.1133 g
Molar mass of Hydrocarbon = M = 58.8 g/mol
Gas constant = 0.0821 Latm/(mol-K)
Plug the values we get V = (wRT) / (MP)
= (0.1133x0.0821x273)/(58.8x1)
= 0.0432 L
= 43.2 mL
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