An element has ccp packing with a face-centered cubic unit cell. its density is 8920 kg/m3 and the unit cell volume is 4.72 x 10-26 l. calculate the molar mass (g/mol) of the element to three significant figures.

63.4 g/mol
   First, let's determine how many atoms per unit cell in face-centered cubic. There is 8 corners, each of which has 1 atom, and each of those atoms is shared between 8 other unit cells. So 8*1/8 = 1 atom per unit cell. Additionally, there are 6 faces, each of which has 1 atom that's shared between 2 unit cells. So 6*1/2 = 3 atoms per unit cell. So each unit cell has the mass of 1+3 = 4 atoms. Since there is 1000 liters per cubic meter, the mass per liter is 8920 kg/1000 = 8.920 kg/L. Now the mass per unit cell is 8920 g * 4.72x10^-26 = 4.21024x10^-22 g per unit cell. The mass per atom is 4.21024x10^-22 g / 4 = 1.05256x10^-22 g/atom, Finally, multiply by Avogadro's number, getting 1.05256x10^-22 g/atom * 6.0221409x10^23 atom/mol = 63.38664625704 g/mol.
   Rounding to 3 significant digits gives 63.4 g/mol.

the catch is that the top numbers are atomic masses the bottom numbers are the numbers of protons and electrons in an uncharged atom. they are the number of that atoms as you would find them on the periodic table. further, the numbers on the left are used to determine the missing numbers on the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer