What is the molar mass of an ideal gas if a 0.622 g sample of this gas occupies a volume of 300 mL at 35 °C and 789 mm Hg?

1 Answer
Dec 7, 2016

Answer:

51 g/mol

Explanation:

Whenever you see molar masses in gas law questions, more often than not density will be involved. This question is no different. To solve this, however, we will first need to play with the combined ideal gas equation #PV = nRT# to make it work for density and molar mass. The derivation is simple but for the sake of time and space, I will skip it. Hence, just take my word for it that you will end up with the equation:

#M = (dRT)/P#

M = molar mass (g/mol)
d = density (g/L)
R = Ideal Gas Constant (#~~0.0821 (atm*L)/(mol*K)#)
T = Temperature (In Kelvin)
P = Pressure (atm)

As an aside, note that because calculations with this equation involve molar mass, this is the only variation of the ideal gas law in which the identity of the gas plays a role in your calculations. Just something to take note of.

Back to the problem: Now, looking back at what we're given, we will need to make some unit conversions to ensure everything matches the dimensions required by the equation:

#T = 35^oC + 273.15 =# 308.15 K
#V = 300 mL * (1000 mL)/(1 L) =# 0.300 L
#P = 789 mm Hg * (1 atm)/(760 mm Hg) =# 1.038 atm

So, we have almost everything we need to simply plug into the equation. The last thing we need is density. How do we find density? Notice we're given the mass of the sample (0.622 g). All we need to do is divide this by volume, and we have density:

#d = (0.622 g)/(0.300 L) =# 2.073 g/L

Now, we can plug in everything. When you punch the numbers into your calculator, however, make sure you use the stored values you got from the actual conversions, and not the rounded ones. This will help you ensure accuracy.

#M = (dRT)/P = ((2.073)(0.0821)(308.15))/(1.038) =# 51 g/mol

Rounded to 2 significant figures

Now if you were asked to identify which element this is based on your calculation, your best bet would probably be Vandium (molar mass 50.94 g/mol).

Hope that helped :)