Question

To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 2.2-L bulb, then filled it with the gas at 1.70 atm and 26.0 ∘C and weighed it again. The difference in mass was 4.3 g . Identify the gas.

Answer 1

The gas is N₂

Step-by-step explanation:

To solve this problem we need to use PV=nRT. The problem gives us V, P and T, so first let's convert 26.0 °C into K:

26.0 + 273.16 = 299.16 K

There's no need to convert V and P as they already are in proper units.

• Calculating n:

PV = nRT

1.70 atm * 2.2 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 299.16 K

n = 0.1524 mol

From the difference in mass of the experiment, we know that the mass of the gass is 4.3 g. We also now know that 0.1524 moles of X₂ weigh 4.3 g. Now we calculate the molecular weight:

4.3 g / 0.1524 mol = 28.22 g/mol

• Two atoms of X weigh 28.22, so one atom weighs 14.11

Looking at the periodic table, the element with a similar atomic weight is N. So the gas is N₂.