Homework Help: Problem with molecular separation calculations

1. May 20, 2014

emjay66

I am attempting to complete a basic question involving molecular separation from Alonso & Finn "Fundamental University Physics" Second Edition, Chapter 2, Question 9. The Problem states
"Using the data in table 2-1 and A-1, estimate the average separation in molecules of hydrogen at STP (gas), in water (liquid), and in iron (solid)."
Now, Table 2-1 contains density values (relative to water). The relevant values are:
Iron = 7.86
Water (4 Degrees C) = 1.000
Hydrogen = $8.988\times 10^{-5}$
Table A-1 is the periodic table of the elements and the relevant values for this are:
Hydrogen = 1.00797
Oxygen = 15.9994
Iron = 55.847

I will only show my attempt to compute the answer for Hydrogen since I feel I am (probably) making the same mistake throughout.
An inspection of the answers in the back showed that for Hydrogen they obtained:
3.39*10^-7 cm for spherical geometry and 1.14 * 10^-8 cm for cubic geometry

Firstly I assume a Hydrogen "molecule" is two atoms of Hydrogen
I also assume the calculation for a mass of 1kg of the relevant substance

the molecular mass of a Hydrogen molecule = 2 * 1.00797 = 2.01594 g/mol
Number of mole of Hydrogen in 1kg = 1000/2.01594 496.047 mol
Number of Hydrogen molecules = 496.047 * (6.02*10^23) = 2.986*10^26 molecules
Number of Hydrogen molecules at STP = 2.986*10^26 * (8.988*10^(-5)) = 2.68*10^22 molecules

At this point (for a cube) I know 1kg = 1000 cubic centimetres
which means the cube root of 2.68*10^22 (= 29940610.46) would give the number of molecules along one edge (10cm)
which (to me) would mean that 10/29940610 = 3.34*10^(-7) cm, which is clearly wrong.
For the Spherical geometry I reasoned $1000 = \frac{4}{3}\pi\,r^3$ which would mean $r = (\frac{3000}{4\pi})^{1/3}$ = 6.2 cm which means the diameter would be
12.4 cm. This would give a separation of 4.4*10^(-7) cm, which is also wrong.
I understand that my thinking is clearly faulty, so some pointers in the right direction would be nice. Thank you in advance.

2. May 20, 2014

Staff: Mentor

Why is that clearly wrong?

3. May 20, 2014

emjay66

I only feel my approach is wrong or incomplete because my answers do not match those contained at the back of the text (also stated in the original email above). Unfortunately I have no other assertion to base it on. (ps I was a little rushed with my OP so I apologise if that wasn't entirely clear)

Last edited: May 20, 2014
4. May 21, 2014

BvU

Did you notice that the book answers to b) and c) both have spherical < cubic ?
Perhaps you should try those and see what you find!

5. May 21, 2014

Staff: Mentor

What does the book give as an answer? I get a slightly different answer than yours when considering that each molecule occupies a sphere, but you have the correct answer when considering the volume to be cubic.

6. May 21, 2014

emjay66

Ok the answer for each (in the back of the book) are as follows (in spherical, cubic) order (all numbers are cm):
Hydrogen: 3.29*10^-7, 1.14*10^-8
Water: 3.85*10^-8, 6.21*10^-8
Iron: 2.82*10^-8, 4.46*10^-8

Some rethinking about the problem gave me the following:
mass of Hydrogen molecule = (2*1.00797)* 1.6606*10^-27 = 3.35*10-27kg = 3.35*10^-24g
if Density of Hydrogen at STP = 8.988*10^-5 g cm^-3 then Volume = 3.35*10^-24/8.988*10^-5 = 3.72*10^-20cm-3
Assuming the Hydrogen molecule is a sphere, then 3.72*10^-20 = 4/3 * pi * r^3 which means r (radius of Hydrogen molecule) = 2.07*10^-7 cm which means the diameter is approximately 4.14*10^-7 cm
Number of molecules of Hydrogen in 1kg = 2.684*10^19 (as before)
which is equivalent to (2.684*10^19)^(1/3) which is approximately 3*10^6 along one 10cm edge
This means that 3*10^6 molecules would occupy (with no gaps) 3*10^6*4.14*10-7 which is app 1.24 cm
leaving 8.758cm to be used as gaps for 3*10^6 molecules. This would mean that the inter molecule gap is 8.758/3*10^6 = 2.9*10^-6 cm for a cube
For a sphere I use the Wigner-Seitz approximation (with n = 3.72*10^-10cm^-3) which would give a inter molecule gap of 2.07*10^-7 cm.
I will post my other rework later.