Stoichiometry IV : A levels H2 Chemistry 2008 Paper 1 Q2: Lord Rayleigh and his “Atmospheric Nitrogen”.
- April 01, 2018
- Stoichiometry and Mole Concept
Recall from the calculation of relative atomic mass of chlorine :
For a mixture of of 35Cl and 37Cl isotopes in an abundance ratio of 3:1,
RAM of Cl = (3(35)+1(35))/(3+1) = ¾(35)+1/4(37)= 35.5
This is generally an acceptable method of calculating the average property of a mixture, provided the components of the mixture do not react.
That is :
Average property = ∑(mole fraction of i) x (property of i)
- Average molecular weight of a polymer mixture of different polymeric chain length.
- Average density or viscosity of a mixture of liquid or gases.
- Average Mr of air
This mathematical method is called the weightage average calculation.
You might ask : but what is the application of this concept in H2 “A” Level Chemistry exams?
Let’s take a look at a 2008 A levels question (paper 1, Q2)
In 1892, Lord Rayleigh made “atmospheric nitrogen” by removing carbon dioxide, water vapour and oxygen from a sample of air. He found the density of this nitrogen to be 1.2572g/dm3 at s.t.p. Chemically pure nitrogen has a density of 1.2505g/dm3 at s.t.p.
Which gas present in “atmospheric nitrogen” caused this discrepancy?
A look through the internet will show you that this question has been raised before on popular forums, such as the SGforum.
However, the solution provided is far from satisfactory.
Let’s take a look at how the solution can be worked out using the weighted average concept:
ρ (mix of N2 & impurity) = XN2 (ρN2) + Ximpurity (ρimpurity)
or : 1.2572 = XN2 (1.2505) + Ximpurity (ρimpurity) ……..(1)
where X = mole fraction ; ρ= density
Since the impurity is present in very low concentration, Ximpurity → 0 and XN2 → 1
For the 2nd term of equation (1) to make any significant contribution to the average density of the mixture, ρimpurity > XN2
ρimpurity > ρN2
This is a very important conclusion for it leads us to the last step before the answer.
Remember we learnt for ideal gases,
PV = nRT
Density, ρ = mass/volume = PMr/RT ………..(2)
To compare densities of 2 gases, it is important to compare at similar T and P, because both T and P affects the density of a gas, as seen in the equation above.
Thus, for a similar T and P, if
ρimpurity > ρN2
Then it is not difficult to conclude, from (2),
Mrimpurity > MrN2
Thus a look at the 4 options given in the question, only Argon’s atomic mass (40) is more than the molecular mass of N2 (28), and thus is the answer.
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