Question

A sample of an unknown compound is vaporized at 190.°C. The gas produced has a volume of 2420.mL at a pressure of 1.00 atm, and it weighs 2.04 g. Assuming the gas behaves as an ideal gas under these conditions, calculate the molar mass of the compound. Be sure your answer has the correct number of significant digits.

Answer 1

Answer : The molar mass of unknown compound is, 32.0 g/mole

Explanation :

Using ideal gas equation:

$PV=nRT\\\\PV=\frac{w}{M}RT$

where,

P = pressure of gas = 1.00 atm

V = volume of gas = 2420 ml = 2.420 L      (1 L = 1000 ml)

T = temperature of gas = $190^oC=273+190=463K$

n = number of moles of gas

R = gas constant = 0.0821 L.atm/mole.K

w = mass of unknown compound = 2.04 g

M = molar mass of unknown compound = ?

Now put all the given values in the ideal gas equation, we get:

$(1atm)\times (2.420L)=\frac{2.04g}{M}\times (0.0821L.atm/mole.K)\times (463K)$

$M=32.0g/mole$

Therefore, the molar mass of the compound is, 32.0 g/mole

Answer 2

Taking into account the ideal gas law, the molar mass of the compound is 31.87 $\frac{g}{mole}$.

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P×V = n× R×T

In this case, you know:

• P= 1 atm
• V= 2420 mL= 2.420 L
• n=?
• R= 0.082 $\frac{atmL}{molK}$
• T= 190 °C= 463 K

Replacing in the ideal gas law:

1 atm×2.42 L= n×0.082 $\frac{atmL}{molK}$×463 K

Solving:

$n=\frac{1 atmx2.42 L}{0.082\frac{atmL}{molK}x463 K}$

n=0.064 moles

Molar mass is the amount of mass that a substance contains in one mole. So the molar mass can be calculated as:

$molar mass=\frac{mass}{number of moles}$

In this case:

• mass= 2.04 g
• number of moles=0.064 moles

Replacing in the definition of molar mass:

$molar mass=\frac{2.04 g}{0.064 moles}$

Solving:

molar mass= 31.87 $\frac{g}{mole}$

Finally, the molar mass of the compound is 31.87 $\frac{g}{mole}$.

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