Question about calculating interatomic distances in metals

[ARS]

[Moderator's note:  This topic was originally split off from For Questions That Don't Merit Their Own Thread, since the ensuing discussion turned out to be a lengthy one.]

Is it possible to calculate the distance between molecules on a metal compound purely by mathematical calculation as long as we know the ratio of the metals used, density and molar mass of each metal as well as the resulting shape of the crystal matrix of the compound? Assuming that the metal constituent is evenly distributed in ideal condition

Edited December 18, 2020 by Snark
Added moderator's note

[kerbiloid]

Mass = total volume * density.

i-th Metal mass = Mass * i-th Metal ratio (or percentage / 100%).

i-th Metal  amount of substance = i-th Metal mass / i-th Metal molar mass

i-th Metal  amount of atoms = i-th Metal  amount of substance * Avogadro number (6.022*10²³ 1/mol)

Total amount of atoms = sum(i-th Metal  amount of atoms)

Average volume per atom = 
total volume /  total amount of atoms = 
total volume / sum(i-th Metal  amount of atoms) = 
total volume / sum(i-th Metal  amount of substance * Avogadro number) = 
total volume / sum(i-th Metal mass / i-th Metal molar mass * Avogadro number) = 
total volume / sum(Mass * i-th Metal ratio / i-th Metal molar mass * Avogadro number) = 
total volume / sum(total volume * density * i-th Metal ratio / i-th Metal molar mass * Avogadro number) = 
total volume / (total volume * density * Avogadro number * sum(i-th Metal ratio / i-th Metal molar mass)) = 
1 / (density * Avogadro number * sum(i-th Metal ratio / i-th Metal molar mass)) .

Average distance
= Average volume per atom ^1/3 
= 1 / (density * Avogadro number * sum(i-th Metal ratio / i-th Metal molar mass)) ^1/3.

Something like this (check the members before using).

Edited December 17, 2020 by kerbiloid

[ARS]

Thanks!

[K^2]

5 hours ago, ARS said:

Is it possible to calculate the distance between molecules on a metal compound purely by mathematical calculation as long as we know the ratio of the metals used, density and molar mass of each metal as well as the resulting shape of the crystal matrix of the compound? Assuming that the metal constituent is evenly distributed in ideal condition

In addition to composition and density, you need to know lattice structure. In a simple cubic lattice, volume of a single cell is distance³, and it contains one atom on average. In a face-centered cubic, there are 4 atoms per cell and the volume is 2 * sqrt(2) * distance³, so the density is significantly higher with the same interatomic distance.

Here's a good resource: Atomic Packing Factor. And the explicit formulas for interatomic distances for given lattice (distances in cm if density is in g/cm³):

• Simple cubic: distance = (atomic mass / (Avogadro's number * density))^1/3
• Face-centered cubic: distance = (4 * atomic mass / (Avogadro's number * density))^1/3 / sqrt(2)
• Body-centered cubic: distance = sqrt(3) * (2 * atomic mass / (Avogadro's number * density))^1/3 / 2
• Hexagonal: distance = (6 * atomic mass / (Avogadro's number * density))^1/3 /  (3 * sqrt(2))^1/3

This all assumes that you have just one type of atom in your lattice. If you have an alloy, things get complicated. There is no simple formula for the alloys, because there are a lot of ways atoms can be distributed within an alloy. And the variance in distances you get with different lattices should tell you that you can't simply use a one-size-fits-all formula unless you're going for a very rough number.

Edited December 17, 2020 by K^2

[kerbiloid]

If you need just an average distance, the positions and distribution of the atoms play no role, just their amount and occupied volume.

[K^2]

Average distance between two atoms in a 12 meter beam is 4 meters. There is no useful concept of just an average distance. And if you are looking instead at an average distance between nearest neighbors, which is a useful quantity, you absolutely need to know the crystalline structure. You can't just assume one atom per cube with the side equal to interatomic distance, because lattices aren't always simple cubic. And in fact, simple cubic gives you the absolute worst packing, and in practice, the distance between nearest neighbors will almost always be about 10% higher than the average you get assuming a cubic lattice.

And anyone who'd bother to even try plugging this into real world numbers would know this. Atomic weight of copper is 63.546amu. Density of copper is 8.96g/cm³.

distance(?) = (63.546amu / (8.96g/cm³ * 6.022 * 10²³))^1/3 = 2.28A

What's the actual distance? 2.55A (see atomic diameter, which assumes atoms are touching nearest neighbors). Oh, look at that, wrong by more than 10%. How did that happen? Well, copper has a face-centered cubic lattice. So lets use correct formula.

distance = (4 * 63.546amu / (8.96g/cm³ * 6.022 * 10²³))^1/3 / sqrt(2) = 2.55A

And that's right on the money.

This is a simple case, because copper has a nice, clean structure. Alloys often do not, so figuring this out for alloys is going to be complicated.

[kerbiloid]

When the task asks for an average distance between atoms, it does ask about the average distance between atoms [snip].

Edited December 18, 2020 by Snark
Redacted by moderator

[K^2]

8 hours ago, kerbiloid said:

When the task asks for an average distance between atoms, it does ask about the average distance between atoms [snip].

Cool. Go ahead and tell me what YOU are getting for average distance between copper atoms, then explain why your AVERAGE is smaller than 2.55A atomic diameter of copper.

[kerbiloid]

8 hours ago, K^2 said:

Cool. Go ahead and tell me what YOU are getting for average distance between copper atoms, then explain why your AVERAGE is smaller than 2.55A atomic diameter of copper.

I have no idea why the task author needs an average distance between the atoms, like I have no idea why is a car moving from A to B at 50 km/h speed, and how old is its carburetor.

Also I'm not going to go ahead and invent the task prequel and sequel [snip].

Edited December 18, 2020 by Snark
Redacted by moderator

[K^2]

14 minutes ago, kerbiloid said:

Also I'm not going to go ahead and invent the task prequel and sequel

What's the average distance between copper atoms in copper metal? That's a simple question. It's the exact question ARS asked, just for a specific metal.

[kerbiloid]

8 hours ago, K^2 said:

What's the average distance between copper atoms in copper metal? That's a simple question. It's the exact question ARS asked, just for a specific metal.

Apply the formulas above and serve yourself.

P.S.
Do you at all understand the term "average"?

[snip]

Edited December 18, 2020 by Snark
Redacted by moderator

[K^2]

Just now, kerbiloid said:

Apply the formulas above and serve yourself.

I did. I'm getting 2.28A. Are you getting something different?

1 minute ago, kerbiloid said:

Do you at all understand the term "average"?

Do you? Because 2.28A < 2.55A, which is the MINIMUM distance between copper atoms. Do you have an explanation for how an average can be less than the smallest value?

[kerbiloid]

[snip]

8 hours ago, K^2 said:

Do you? Because 2.28A < 2.55A, which is the MINIMUM distance between copper atoms. Do you have an explanation for how an average can be less than the smallest value?

"Average" means "how much pere every item if the total divide by total".
Because neither total volume, nor total count depend on the atomic structure.

[snip]

Edited December 18, 2020 by Snark
Redacted by moderator

[K^2]

@ARS sorry about the diversion. The moral is, if somebody gives you a formula on the internet, always plug in some real world values to see if they actually work. There is usually a simple case for which you can find all the data.

Edited December 18, 2020 by K^2

[kerbiloid]

Exactly. And when you need to divide a constant number (volume) by a constant value (number), there is no need to invent complicated structures.

Because when the task needs exact value, it doesn't ask about the average one.

[K^2]

Average value is always bigger or equal to the smallest value. If formula gives you average that's smaller than the smallest value, it's not an approximation error. It's just wrong.

And I don't think it even matters that the underlying reason it's wrong is because average number density and average distance aren't the same thing at all, but really, it should be obvious to anyone that a formula estimating an average should not ever give a value that's smaller than the minimum.

[kerbiloid]

Volume = mass / density

Atom count = (mass / molar mass) * Avogadro number

Average volume per atom =  Volume / Atom count = mass / density /  ((mass / molar mass) * Avogadro number) = 1 / density /  ((1 / molar mass) * Avogadro number) = 1 / (density / molar mass) * Avogadro number).

Copper, wiki

Spoiler

Copper
Appearance	red-orange metallic luster
Standard atomic weight Ar, std(Cu)	63.546(3)[1]
Copper in the periodic table
Hydrogen   Helium Lithium Beryllium   Boron Carbon Nitrogen Oxygen Fluorine Neon Sodium Magnesium   Aluminium Silicon Phosphorus Sulfur Chlorine Argon Potassium Calcium Scandium   Titanium Vanadium Chromium Manganese Iron Cobalt Nickel Copper Zinc Gallium Germanium Arsenic Selenium Bromine Krypton Rubidium Strontium Yttrium     Zirconium Niobium Molybdenum Technetium Ruthenium Rhodium Palladium Silver Cadmium Indium Tin Antimony Tellurium Iodine Xenon Caesium Barium Lanthanum Cerium Praseodymium Neodymium Promethium Samarium Europium Gadolinium Terbium Dysprosium Holmium Erbium Thulium Ytterbium Lutetium Hafnium Tantalum Tungsten Rhenium Osmium Iridium Platinum Gold Mercury (element) Thallium Lead Bismuth Polonium Astatine Radon Francium Radium Actinium Thorium Protactinium Uranium Neptunium Plutonium Americium Curium Berkelium Californium Einsteinium Fermium Mendelevium Nobelium Lawrencium Rutherfordium Dubnium Seaborgium Bohrium Hassium Meitnerium Darmstadtium Roentgenium Copernicium Nihonium Flerovium Moscovium Livermorium Tennessine Oganesson – ↑ Cu ↓ Ag nickel ← copper → zinc
Atomic number (Z)	29
Group	group 11
Period	period 4
Block	d-block
Element category	Transition metal
Electron configuration	[Ar] 3d10 4s1
Electrons per shell	2, 8, 18, 1
Physical properties
Phase at STP	solid
Melting point	1357.77 K (1084.62 °C, 1984.32 °F)
Boiling point	2835 K (2562 °C, 4643 °F)
Density (near r.t.)	8.96 g/cm3
when liquid (at m.p.)	8.02 g/cm3

Density = 8960 kg/m³, molar mass ~= 0.0635 kg/mol.

Average volume per atom = 1 / (8960 * 6.022*10²³ / 0.0635) ~= 1.18*10^-29 m³.

As it's always written in these books, "d ~V^1/3", because when we need an average distance, we suppose the geometry insufficient and just use a cubic approximation.

Average distance between atoms ~ _{(sorry, can't find a bigger font for the tilda to emphasize that this is not "=")} (1.18*10^-29)^1/3 ~= 2.27*10^-10 m = 2.28 A.

If remember that we use a cubic approximation, then the cube diagonal is 2.28 * 2^1/3 ~= 2.87 A, which is greater than the mentioned 2.55 A.

So, to calculate exact average value we should perform a 2-factor integration of a cube.

That's because in such tasks they use "tilda" instead of "equal" and talk about approximate value.

For the task which doesn't bring additional constraints (like the state, temperature, and so on)  it's important to bring a proper methodology, rather than theoretical purity.

Also before talking about types of lattice, one should first mention the temperature, because thermal expansion would bring its own error.

So, more proper answer is "between 2.3  and 2.9 A", but actually just the estimated cubic root is required, otherwise the task should bring an additional constraint.

When they need an actual value, they should bring more additional conditions as constrains. The task obviously asks for an estimation.

Especiallym since the task asks about an arbitrary mix of unknown elements, so without additional info it's just impossible to say anything about its geometrical structure.

P.S.
Now feel free to integrate the distance between two points on arbitrary cube edges, because (2.28+2.87) / 2 .58, so unlikely the 2.55 can be a proper answer, it's too close to the average of the edge length and the cube diagonal length.

***

It's very important to avoid excessive accuracy, a false friend of physical calculations.

The experimental model itself should suppress the accuracy error.

Compensation and attenuation is our motto!

Edited December 18, 2020 by kerbiloid

[K^2]

6 hours ago, kerbiloid said:

Average distance between atoms ~ _{(sorry, can't find a bigger font for the tilda to emphasize that this is not "=")} (1.18*10^-29)^1/3 ~= 2.27*10^-10 m = 2.28 A.

Funny. The AVERAGE is still smaller than the diameter of the atoms. Hm.

6 hours ago, kerbiloid said:

If remember that we use a cubic approximation, then the cube diagonal is 2.28 * 2^1/3 ~= 2.87 A

So the diagonal of a unit cube is 2^1/3? Are you sure about that?

6 hours ago, kerbiloid said:

So, to calculate exact average value we should perform a 2-factor integration of a cube.

Why just the cube? Why aren't you including next atom in the (2, 1, 1) direction? Or a (3, 2, 1) direction? Or one on the other side of the metal plate? Average should be over all distances, no?

Also, funny how I got 2.55A exactly with my formula without having to integrate anything. I just used geometry of FCC lattice and correctly computed the relevant diagonal.

[snip]

Edited December 18, 2020 by Snark
Redacted by moderator

[kerbiloid]

6 hours ago, K^2 said:

So the diagonal of a unit cube is 2^1/3? Are you sure about that?

Ok, my fault, 3^1/2. "Between 2.3 and 3.9 A".

6 hours ago, K^2 said:

Why just the cube? Why aren't you including next atom in the (2, 1, 1) direction? Or a (3, 2, 1) direction? Or one on the other side of the metal plate? Average should be over all distances, no?

Also, funny how I got 2.55A exactly with my formula without having to integrate anything. I just used geometry of FCC lattice and correctly computed the relevant diagonal.

Because if you read the original question, it is not about copper, iron, aluminium, or gold.
It's about a casual mixture of unknown metals. 

The cube is usually used in such kind of tasks just as the simplest approximation for qualitative assessment 
Everywhere you can read "d ~ V^1/3", because it's a qualitative assessment.
It's enough good if use same formula to compare various materials, or various states of the same material.
It's important to follow same methodics, to keep same ratio. The coefficients of proportionality are not essential in this task until they define the opposite.

And yes, the sample can't be composed of spheres. The cube is much better.

If the task was from the "solid body physics", it would contain additional constraints. It obviously isn't.

But even in this case you should use a cube, unless you live in a non-3d universe where the volume is not a cube, but a 2.453 degree of size.

[snip]

Edited December 18, 2020 by Snark
Redacted by moderator

[K^2]

16 minutes ago, kerbiloid said:

The cube is usually used in such kind of tasks just as the simplest approximation for qualitative assessment

No. No, it isn't. Because almost no metals or alloys have a simple cubic lattice. It is so exceptionally rare, that even given an unknown metal, nobody would ever use a simple cubic lattice for an estimate.

And you still aren't saying why you aren't considering any other neighbors in your average. Here's a Poisson grid of some points in 2D plane. It is amorphous, but since you insist that the structure doesn't matter, that shouldn't be a problem. I've circled one of the dots. Which of its neighbors are you going to consider when deciding what the average distance is?

There's a reason why there is no such thing as an average distance between atoms. There's an average nearest neighbor distance which is uniquely defined, and which does, very much, depend on structure.

[kerbiloid]

Lattice doesn't mean anything here.

It's an AVERAGE distance, total/total.

When in a 1000 m3 house live 10 persons, it's still 100 m3 per person, no matter in which rooms they live.

Edited December 18, 2020 by kerbiloid

[K^2]

1 minute ago, kerbiloid said:

When in a 1000 m3 house live 10 persons, it's still 100 m3 per person, no matter in which rooms they live.

Sure. But we're asking for average distance between persons, not average volume per person. So why aren't you counting the distance between two people who are on opposite sides of the house?

As I've said from the very beginning, if you take a narrow beam 12m long and consider all pair-wise distances, average distance between any two atoms in that beam is 4m. Exactly a third of total length, which doesn't tell you anything about the material. That's why people talk about average nearest-neighbor distances, which can be defined for atoms in a metal, molecules in a gas, or people in a house.

[kerbiloid]

Realize the difference between the cube as a volume finite element and a cube as a form of crystal lattice.

[K^2]

I mean, I can do that too. Realize the difference between number density and average distance! But I don't think that does anything. I prefer practical things. Numbers, charts. You know, stuff where you can actually point to a mistake. Anything else is demagoguery.

So lets work through your example. What's the average distance between these 10 people in the 1000m³ home? Can you provide me with a diagram of an example arrangement where that's the average distance? I'm sure you can manage placing 10 dots on a graph, right?

[kerbiloid]

The simple method brought by me has given a ~20%-accurate solution for a qualitative assessment task from the 1st year of the universe after a minute of googling for data and calculations.

(2.55-2.28)/2.55 *100% = 10.5%.

For the given task its almost perfect.

A true perfectionist would just say "Better consider d ~1.25 V^1/3 because the cubic finite element diameter is greater than it width,  so the 1.25 factor gives more accurate estimation.", and I would be applauding.

(1.25 here is taken randomly, just as an example)

[snip]

Edited December 18, 2020 by Snark
Redacted by moderator