Tungsten coated molybdenum powder

[johnsonb.engr]

Hello Everybody,

I would greatly appreciated your help. I have been working this problems for the past couple hours and can't seem to find the answer. I tried Google, and opening up my old textbooks but to no avail. I could really use some help.

I have sample of tungsten coated molybdenum powder. The material is 20% Tungsten by volume and 80 % Molybdenum by weight.

the density of Tungsten = ρw=19.25 g/cm^-3
the density of Molybdenum = ρmo=10.28 g/cm^-3

Obviously ρ=M/V(eqn 1) and Vtotal= Vw + Vmo (eqn 2)

If the material is 20% W by volume, then Vw=0.2*(Vtotal)

Therefore; Vtotal=5Vw

and using eqn 2: 5Vw=Vmo + Vw

4Vw=Vmo

using eqn 1: ρ=M/V; V=M/ρ

And by plunging ρ I obtain,

Vw=Mw/19.25
Vmo=Mmo/10.28

Therefore; 4(Mw/19.25)=Mmo/10.28

Which equals Mw=0.468Mo-------> which makes since because the density of W is ~2x of that of Mo.

However, this is where I am stuck, how do I find the % mass from here using?


FYI...I am trying to solve this theoretically...if this helps the powder I have is 20 % by mass w, has and has an apparent density of 3.0 g.cm^3. Also, when I subjected this powder to a hand held XRF gun, I obtain 94 % Mo and 6 % W.

 

[Navin A S]

I understand that the % mass of W should be ~20 %, but I am trying to solve this theoretically and not just use the XRF results.Thanks in advance for your help!</code>Since you already have the apparent density of the powder (3 g/cm^3), you can calculate the total mass of the powder with the equation:M = ρ*VWhere M is the total mass of the powder, ρ is the density, and V is the volume.Once you have the total mass, you can then calculate the individual masses of tungsten and molybdenum with the equation you derived earlier:Mw = 0.468*MoNow you can use these values to calculate the % mass of tungsten and molybdenum in the powder:%mass(W) = (Mw/M)*100%mass(Mo) = (Mo/M)*100