- #1

rocketboy

- 243

- 1

**PLZ HELP - Molar mass changes at stp??**

Hey, I have this lab report due tomorrow, and I seem to be getting some discrepencies in my calculations. I am calculating the molar mass of butane, without using the values from the periodic table, using data obtained by the experiment below:

Pressure: 101.09 kPa

Temperature: 23ºC --> 296 K

Initial Mass of Lighter: 18.19g

Secondary Mass of Lighter: 18.06g

Mass of Butane used: 0.15g

Volume of Butane: 50.2

I won't bother you with error propagation as it is easy to caculate, so disregard that. Below are my calculations for the molar mass of butane, at the temp and pressure given above and at STP (standard temp and pressure).

Am I doing this properly? Do my results make sense? (Ignore the "..." they are for placement)

**Molar Mass of Butane at Given Temperature and Pressure:**

50.2 +/- 1.0mL X

__1.0L__= 5.02 x 10^-2 L

......1000mL

P =

__nRT__--> n =

__PV__

...V.....RT

n =

__101.09kPa (5.02 x 10-2L)__

...8.31kPa.L/mol.K (296K)

n = 2.06 x 10^-3 molButane

M =

__m__

...n

M =

__0.15g__.

...2.06 x 10^-3 molButane

M = 72.71g/molButane

**Molar Mass of Butane at STP:**

T1 = 296K P1 = PTotal – PH20 P2 = 101.3kPa

T2 = 273K = 101.09kPa – 2.81kPa V1 = 50.2mL

= 98.28kPa

50.2 mL X

__273K__X

__101.3kPa__= 47.7mL = 4.77 x 10^-2 L

...296K...98.28kPa

P =

__nRT__--> n =

__PV__

...V...RT

n =

__101.3kPa(4.77 x 10^-2 L)__

...8.31kPa.L/mol.K (273K)

n = 2.13 x 10^-3 molButane

M =

__m__

...n

M =

__0.15g__.

...2.13 x 10^-3 molButane

M = 70.42g/molButane

I know I've done something wrong but don't know what.

Thanks you SO much,

-Jon