Molybdenum crystallizes with the body-centered unit cell. The radius of a molybdenum atom is 136 pm. Calculate the edge length of the unit cell and the density of molybdenum

Video transcript

hey everyone in this example we need to calculate the edge length and the density of our unit cell of tungsten. Given that it has a body centered unit cell and radius of 137 PK meters. So what we want to recognize is first to calculate for edge length, we would recall our formula where we take four times our radius divided by three to the point fifth power. And so what that would give us is four times our radius from the prompt of 1 kilometers Divided by 3 to the Point 5th power. And this is going to give us an edge length equal to 316. m. Our next step is to because this is our first answer here for edge length. Now we want to find our density. However, we don't have volume or mass of our cubic unit cell. So first let's find our volume. Recall that volume for our cubic unit cell is the edge length to the third power. And so this means that we would take 316. m and we want to go ahead and raise this to the third power. However, we're also going to multiply by our conversion factor to go from PICO meters two centimeters. And so we would recall that our prefix PICO tells us that we have or rather our prefix anti tells us that we have 10 to the 10th power PICO meters for one centimeter. And so now we're able to cancel out our units of PICO meters. We're left with centimeters. And when we cube this, we're going to have cubic centimeters which is going to give us our volume as a value of 3.16 times 10 to the negative 23rd power cubic centimeters. When we cube everything. And this volume is interpreted as cubic centimeters per unit cell. Now that we have volume, we just need mass. So we are going to refer to the molar mass for tungsten. So recall that tungsten on our periodic table as a symbol. W and it has a molar mass according to our periodic table of 183.84 g per mole. So we're going to utilize this as a conversion factor to solve for the mass of our cubic unit cell. So we should recall that for one unit cell, We can show two atoms or hold two atoms. Now we're going to multiply by our molar mass of tungsten as a conversion factor To get grams for our mass. So we should recall that according to the periodic table for one mole of Tungsten, we have a molar mass of 183.84 g. Now we're going to go ahead and multiply by our last conversion factor to cancel out the term atoms, saying that for one mole we have six point oh 22 times 10 to the 23rd power, which we recall is avocados number atoms. So this allows us to cancel our units of atoms. Were also able to cancel out moles. And we're left with grams per unit cell as our unit for mass, which is what we want. So this gives us our mass of our cubic unit cell equal to a value of 6.11 times 10 to the negative 22nd power. This is in units of grams per unit cell. And now that we have our mass, we can now find density and so we should recall that density is mass divided by volume. So we're going to take our mass 6.11 times 10 to the negative 22nd power grams per unit cell. And we're going to divide by our volume which we found above as 3. times 10 to the negative 23rd power cubic centimeters per unit cell. So now we're able to go ahead and cancel our units of unit cell leaving us with grams per cubic centimeter as our final units for density. And we should get a value for this quotient equal to 19.3 grams per cubic centimeter as our final value for density. So, this would be our final our second final answer here highlighted in yellow. So everything highlighted in yellow shows our edge length as well as our density for our cubic unit cell of tungsten. So I hope that everything I explained was clear. If you have any questions, please leave them down below. And I will see everyone in the next practice video.