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# Chemistry Gas and STP problem? Help!!?

What is the molecular weight of a gas that occupies a volume of 278 mL, at a pressure of 680.0 mm Hg, a temperature of 21.7 degrees C, and a mass of 35.9 grams?

### 1 Answer

- Lexi RLv 79 years agoFavorite Answer
Using the STP method involves ...

1. Using the combined gas equation to calculate the volume of the gas at STP

2. using this volume to calculate moles of gas (since 1 mole any ideal gas at STP occupies 22.4 L

3. You now know the mass and moles of the sample (molar mass = mass / moles)

P1V1/T1 = P2V2/T2

P1 = 680.0 mmHg

V1 = 278 ml

T1 = 294.85 K

P2 = 760.0 mmhg

V2 = ? ml

T2 = 273.15 K

V2 = P1V1T2 / P2T1

= 680.0 mmHg x 278 ml x 273.15 K / (760.0 mmHg x 294.85 K)

= 230.4 ml

Now, at STP the volume of 1 mole of any ideal gas = 22.4 L

So moles gas = volume at STP / 22.4 L/mol

= 0.2304 L / 22.4 L/mol

= 0.01029 mol

moles = mass / molar mass

Therefore molar mass = mass / moles

= 35.9 g / 0.01029 mol

= 3488.8 g/mol

= 3490 g/mol (3 sig figs)

The alternative way of doing this problem is to use the equation

PV = mRT / Mw

where

Mw = molar mass

P = pressure

V = volume

m = mass

R = gas constant

T = temp in Kelvin

(obviously you have to be able to either remember or look up the gas constant (R) to do this one.)