Calculate the surface area of 1 g of TiO2 powder

AI Thread Summary
The discussion focuses on calculating the surface area of 1 gram of TiO2 powder, assuming the particles are spherical with a diameter of 100 micrometers. The surface area of a single particle is calculated using the formula for the surface area of a sphere, and the total surface area for 1 gram is derived from the number of particles, which is determined by the mass of a single particle based on its volume and density. Participants clarify that the diameter of the particles is 100 micrometers, not nanometers, and emphasize the importance of calculating the number of particles correctly to find the total surface area. For part two, the discussion involves estimating the number of TiO2 molecules at the surface by considering a spherical shell of a specified thickness. The calculations are confirmed to be on the right track, with guidance on how to approach the surface layer's mass and molecular count.
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Homework Statement


Calculate the surface area of 1g of powder with a size of 100micrometer. Assume the particles are spherical. Then estimate the percentage of TiO2 molecules at the surface of the particle (relative to the total number of TiO2 molecules in the volume of the particle) for the same size. Assume the thickness of surface TiO2 is the diameter of O-2 (.3nm)

Known:
1g of TiO2 powder
Diameter = 100nm

FromWiki:
Density: 4.23 g/cc
Molar Mass: 77.9g


Homework Equations


Surface Area = Pi*D2
Volume = Pi*D3/6


The Attempt at a Solution



Surface Area of 1 particle = Pi*(100)^2 = 31415.9 nm^2
Surface Area of 1g = 31415.9 * 1/77.9 * 6.022*10^23 atoms = 1.89*10^28 nm^2

Part 2


Volume @ surface = Pi*D^2*thickness = 9424.78 nm^3
Mass = Density*Volume = 9424.78 * 4.23 *10^-21 g/nm^3 = 3.98*10^-17 g powder on surface thickness

3.98*10^-17/77.98*6.022*10^23 = 307871 atoms on surface?

What i think is wrong: Part 1, not sure how to take in account # of units of TiO2 are in 1g, Part 2 i just don't know.
 
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So let me see if I've got this straight. You've got a bunch of TiO2 powder that has a total mass of 1 g. This powder consists of individual spherical particles, each of which has a diameter of 100 μm. Is that right so far?

If so, then I would say that the total surface area would just equal the sum of the surface areas of all of the individual particles. So, the first thing you need to figure out the mass of a single particle. You can get this from its volume, and the density of TiO2. Once you have the mass of a single particle, you can figure out how many particles are needed to get a total mass of 1 g.

The problem with your solution to part 1 is that the surface area of a sphere is not given by πD². EDIT: No, sorry. That equation is fine. I am used to thinking of things in terms of radius rather than diameter.

The other problem seems to be that you tried to figure out the number of particles by figuring out what fraction of a mole corresponds to 1 g. That doesn't work, because that calculation gives you the total number of TiO2 molecules present. But that's not the number of spherical particles in the powder. Each particle is a sphere consisting of MANY TiO2 molecules.
 
Last edited:
So let me see if I've got this straight. You've got a bunch of TiO2 powder that has a total mass of 1 g. This powder consists of individual spherical particles, each of which has a diameter of 100 μm. Is that right so far?
Yes

If so, then I would say that the total surface area would just equal the sum of the surface areas of all of the individual particles. So, the first thing you need to figure out the mass of a single particle. You can get this from its volume, and the density of TiO2. Once you have the mass of a single particle, you can figure out how many particles are needed to get a total mass of 1 g.

Mass = Volume * Density = (Pi*(100*10^-7 cm)^3)/6 * 4.23 g/cm^3
mass 1 particle= 2.214 * 10^-15 g
mass 1 g = 2.214 * 10^-15 / 79
so 1.68 *10^7 particles to get 1g mass?
 
kpx001 said:
Mass = Volume * Density = (Pi*(100*10^-7 cm)^3)/6 * 4.23 g/cm^3

This looks good, except that in your problem statement, it says that the diameter of a particle is 100 μm, not 100 nm.

kpx001 said:
mass 1 g = 2.214 * 10^-15 / 79

This step doesn't make sense. If I have "x" grams per particle, then how many particles are required to make a total of 1 gram?

Example: If each particle has a mass of 0.1 g (1/10 of a gram per particle), then the number of particles in a 1 gram sample is 10 particles:

(1 gram) / (1/10 grams/particle) = (1 gram)*(10/1 particles/gram) = 10 particles.

The molar mass doesn't come into this at all.
 
Mass = Volume * Density = (Pi*(100*10^-4 cm)^3)/6 * 4.23 g/cm^3 = 2.214 * 10^-6 g per particle
1g / 2.214 *10^-6 g / particle = 451671 particles in 1 gram that are 100 micrometers.
SA = pi*D^2 = pi*(100*10^-6 m)^2 * 451671 = .01419 m^2 is the SA of 1gram ?

still a bit iffy about part 2 if all this is correct so far.
 
kpx001 said:
Mass = Volume * Density = (Pi*(100*10^-4 cm)^3)/6 * 4.23 g/cm^3 = 2.214 * 10^-6 g per particle
1g / 2.214 *10^-6 g / particle = 451671 particles in 1 gram that are 100 micrometers.
SA = pi*D^2 = pi*(100*10^-6 m)^2 * 451671 = .01419 m^2 is the SA of 1gram ?

still a bit iffy about part 2 if all this is correct so far.

This all looks good to me. As for part 2...

kpx001 said:
Assume the thickness of surface TiO2 is the diameter of O-2 (.3nm)

I have no idea what this O-2 business is, but it seems to me like you are supposed to consider the surface layer of molecules on a particle to be a sphereical shell of thickness 0.3 nm. Therefore, it seems like you have to figure out how many molecules are in that shell, and divide by the number of molecules in the entire sphere.

One way to do this would be to figure out the mass of the shell by figuring out its volume, and then using the density to find the mass. Once you know the mass, you can combine that with your knowledge of the molar mass to figure out how many molecules the shell has. Apply the same procedure to the particle as a whole (easier since you already know its mass). Et voila!

EDIT: It sort of seems like you were on track and doing that in the first place...so, follow it through
 
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