Avogadro's Number — Question 2

Question 2

Equal masses do not imply equal molecular populations

Use $N_A=6.02214076\times10^{23}\,\mathrm{mol^{-1}}$, exact. Two samples each have mass 8.00 g. Sample A is helium with assigned molar mass 4.00 g/mol; sample B is oxygen molecules with assigned molar mass 32.0 g/mol. Helium is monatomic and oxygen is diatomic.

1. Find the ratio of entity counts and the ratio of total atom counts.

2. Find the oxygen mass needed to contain as many atoms as the 8.00 g helium sample. Compare the entity counts in that new pair.

Question 2 - Solution

Equal masses do not imply equal molecular populations

(a) Method and calculation

$$n_{\mathrm{He}}=8/4=2.00\,\mathrm{mol},\quad n_{\mathrm{O_2}}=8/32=.250\,\mathrm{mol}.$$$$\frac{N_{\mathrm{He}}}{N_{\mathrm{O_2}}}=8,\qquad\frac{N_{\mathrm{atoms,He}}}{N_{\mathrm{atoms,O_2}}}=\frac2{2(.250)}=4.$$

(b) Interpretation and further calculation

$$2n_{\mathrm{O_2,new}}=2.00\ \Rightarrow\ n_{\mathrm{O_2,new}}=1.00\,\mathrm{mol}.$$$$m_{\mathrm{O_2,new}}=32.0\,\mathrm g,\quad N_{\mathrm{He}}/N_{\mathrm{O_2,new}}=2.$$Matching atoms still leaves twice as many separate helium entities as oxygen molecules.

Physical check

At fixed mass, entity count is inversely proportional to molar mass; counting atoms introduces molecular composition.