What is the density

(0.540 g)(6.02 1023 atoms) 27.0 g 1.20 1022 atoms

Because density equals mass per unit volume, the mass of the cube is m V (2.70 g/cm3)(0.200 cm3) 0.540 g write this relationship twice, once for the actual sample of aluminum in the problem and once for a 27.0-g sample, and then we divide the first equation by the second: m sample m 27.0 g kN sample kN27.0 g

m sample m 27.0 g

Nsample N27.0 g

To solve this problem, we will set up a ratio based on the fact that the mass of a sample of material is proportional to the number of atoms contained in the sample. This technique of solving by ratios is very powerful and should be studied and understood so that it can be applied in future problem solving. Let us express our proportionality as m kN, where m is the mass of the sample, N is the number of atoms in the sample, and k is an unknown proportionality constant. We

Notice that the unknown proportionality constant k cancels, so we do not need to know its value. We now substitute the values: 0.540 g 27.0 g Nsample 6.02 N sample 1023 atoms