Without doing detailed calculations, explain which of the following has the greatest mass percent of sulfur. $\mathrm{SO}_{2}, \mathrm{S}_{2} \mathrm{Cl}_{2}, \mathrm{Na}_{2} \mathrm{S}, \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}, \mathrm{or} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{SH}$

To decide on the greatest percent of sulfur, let's begin by roughly estimating the molar masses of the key elements: Sulfur (S) about 32, Oxygen (O) about 16, Chlorine (Cl) about 35.5, Sodium (Na) about 23, Carbon (C) about 12, and Hydrogen (H) about 1.

We can now evaluate each compound for its mass percent of Sulfur. In \(\mathrm{SO}_{2}\), S is about 32 and the rest of \(2 \times \mathrm{O}\) is around 32, which makes S about 50%. In \(\mathrm{S}_{2} \mathrm{Cl}_{2}\), \(2 \times \mathrm{S}\) = 64 and the rest about \(2 \times 35.5\), which keeps S less than 50%. For \(\mathrm{Na}_{2} \mathrm{S}\), S is about 32 and the rest about \(2 \times 23\), keeping S over 50%. For \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\), \(2 \times \mathrm{S}\) = 64 and the rest far more, hence, we quickly see that S is less than 50%. Finally, in \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{SH}\), S is about 32, but the rest is far less. So S is definitely less than 50% here too.

From the analysis of each compound, we can tell that the compound with the highest mass percent of Sulfur is Na2S (sodium sulfide).