An element has ccp packing with face-centered cubic unit cell. Its density is 3.78 g/cm^3 and the unit cell volume is 2.31 x 10^-25 L. Calculate the molar mass (g/mol) of the element to 3 sig figs.
the answer is 131 g/mol
i first changed 2.31 x 10^-25 L to 2.31 x 10^-22 mL and multiplied that by 3.78 g/mL
after that i got m = 8.73 x 10^-22 g
& multiplied that by 6.022 x 10^23 = 525.7
and the only way i get 131 g/mol is when i divide 525.7/4....my question is WHY DO YOU DIVIDE BY 4????
pleasee help thank youu :)
You did it correctly. The four (4) comes from the fact that there are four atoms to the unit cell; therefore, 4*mass of 1 atom = mass of the unit cell.
To determine the molar mass of the element, you need to know the number of atoms in the unit cell of the face-centered cubic (FCC) lattice. For an FCC unit cell, there are 4 atoms located at the corners of the cube, and each atom is shared by 8 adjacent unit cells.
When you multiplied the unit cell volume, 2.31 x 10^-22 L, by the density, 3.78 g/mL, you obtained the mass of the unit cell, which is correct.
So, m = (2.31 x 10^-22 L) x (3.78 g/mL) = 8.73 x 10^-22 g
Next, you multiplied this mass by Avogadro's number, 6.022 x 10^23 molecules/mol, which is also correct.
m x (6.022 x 10^23 molecules/mol) = 525.7
Since the unit cell of the FCC lattice contains 4 atoms, you need to divide this number by 4 to get the molar mass of a single atom in grams/mole.
525.7 / 4 = 131.425
When rounded to three significant figures, the molar mass is 131 g/mol.
Therefore, dividing by 4 is necessary because there are 4 atoms in the unit cell, and you want to find the molar mass per atom.