Molar Mass of Butane

Topics: Pressure, Gas, Ideal gas law, Gas laws / Pages: 4 (789 words) / Published: Aug 18th, 2013
Jennifer Jiang
Dr. Iobst
Chemistry I Honors
11 March 2013
Molar Mass of Butane: Applying the Gas Laws 1. Water bath temperature: 17.7° C or 291 K
Celsius to Kelvin temperature conversion: 17.7° C + 273 = 290.7 Kelvin (rounded to SF= 291 K) 2. 1 atm
1 atm
10 mm
10 mm
2.54 cm
2.54 cm
According to the digital barometer our teacher provided, the barometric pressure in the lab is 29.77 in Hg, which will need to be converted to atmospheric pressure.
760 mm Hg
760 mm Hg
1 cm
1 cm
1 in
1 in
29.77 in Hg x x x = 0.9949 atm
Barometric pressure: 756.2 mm Hg or 0.9949 atm 3. Volume of gas collected: 100.0 mL 4. Initial mass of butane lighter: 21.88 g 5. Final mass of butane lighter: 21.38 6. Mass of butane collected: 0.30 grams of butane collected 21.88
Find the difference of the initial mass and the final mass of the butane lighter ~ - 21.58 7. The two gases present in the graduated cylinder are water vapor and butane. 0.30 8. The vapor pressure (PH2O) depends on the temperature of the water because there is is a direct correlation between vapor pressure and temperature (E.g. ~ when temperature increases, vapor pressure increases).
1 atm
1 atm
According to Dr. John I. Gelder of Oklahoma State University, the vapor pressure of water at the water bath temperature of 17.7° C (rounded off to 18.0° C) is 15.5 mm Hg or 0.0204 atm.
760 mm Hg
760 mm Hg
15.5 mm Hg x = 0.020394737 atm (rounded to 3 SF = 0.0204 atm)

9. Partial pressure of butane: Pbut = Patm – PH2O 0.9949
0.9949 atm – 0.0204 atm = 0.9745 atm ~ partial pressure of butane - 0.0204 10. Used the combined gas law to determine the volume (in L) of butane at STP 0.9745
=
=
=
=
P1V1 P2V2 (0.9745 atm)(0.1000 L) (1 atm)(V L) 0.000335 = 0.00366V ~ V= 0.0915 L T1 T2 291 K 273 K 11. Use Avogadro’s law to determine the number of moles of butane gas. Assume that

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