HELP - Molar mass changes at stp?

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The discussion centers on discrepancies in calculating the molar mass of butane at standard temperature and pressure (STP) based on experimental data. The initial calculations yielded values of 72.71 g/mol and 70.42 g/mol, which raised concerns about consistency. A correction was made to the mass of butane from 0.15g to 0.13g, significantly impacting the results. Participants emphasize that the molar mass should remain constant regardless of temperature and pressure changes, suggesting a possible error in the equations used. The conversation highlights the importance of accurate data and proper application of gas laws in calculations.
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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
 
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OMG! Sorry, i made the dumbest error. The mass of butane is 0.13g not 0.15...i subtracted the data wrong by 0.02 and it changed the answer by around 9g/mol!

but still, should i be getting different answers for each case? Why is this or is it not?

-jon
 
Last edited:
No, you should not be getting different answers. One mol of a quantity contains a specific number of molecules (Avogadro's number) and so has a specific mass. That doesn't change just because you change temperature and pressure (although the volume might.)
 
rocketboy said:
(snip)I won't bother you with error propagation as it is easy to caculate, (snip)
M = 72.71g/molButane (snip)

M = 70.42g/molButane

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

Thanks you SO much,
-Jon

You might try the "easy" part.
 
Oh I did the "easy" part in my lab notebook of course. Since i made the correction (post 2 of mine) i have values approx 61 and 64. But I think I'm using the wrong equation or something..because ivy is right, molar mass should always stay the same.
 
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