A piece of gypsum, whose empirical formula is $\mathrm{CaO}+\mathrm{SO}_{3}+2 \mathrm{HO}$, weighs exactly 100 grams; how many grams of each element are in this salt, how many grams of $Ca$ and how many grams of $O$ in the base; how many grams of $S$ and $O_{3}$; how many grams of $H$ and $O$ in the $2 H O$, if the chemical equivalent weight of

$$
\begin{gathered}
Ca- \text{ is }=20 \\
O- \text{ is }=8 \\
S- \text{ is }=16 \\
H- \text{ is }=1
\end{gathered}
$$
From the combination of the chemical formula of gypsum and the measurement values of the chemical equivalent weights, it follows that

$$
20+8+16+3 \cdot 8+2 \times 1+2 \times 8=86
$$

In gypsum, 20 weight parts of $Ca$, 40 weight parts of $O$, 16 weight parts of $S$, and 2 weight parts of $H$ are present. Therefore, in 100 grams,

$$
\begin{aligned}
& \frac{2000}{86} \text { grams } Ca \\
& \frac{4000}{86} \text { grams } O \\
& \frac{1600}{86} \text { grams } S
\end{aligned}
$$

and

$$
\frac{200}{86} \text { grams } H \text { are present. }
$$

From this, the following amounts fall to the bottom:

$$
\begin{gathered}
\frac{2000}{86}=23.256 \text { grams } Ca \\
\frac{800}{86}=9.303 \text { grams } O
\end{gathered}
$$

for the acid anhydride

$$
\begin{aligned}
& \frac{1600}{86}=18.605 \text { grams } S \\
& \frac{2400}{86}=27.907 \text { grams } O
\end{aligned}
$$

and for the crystal water

$$
\begin{aligned}
& \frac{200}{86}=2.326 \text { grams } H \\
& \frac{1600}{86}=\underline{18.604} \text { grams } O
\end{aligned}
$$

In total

$$
100 \text{ grams. }
$$

(Béla Grünhut, real school, 5th grade, Pécs)

The problem was also solved by: Domonkos Berzenczy, real school, 7th grade, Déva; Sándor Suták, grammar school, 8th grade, Nyíregyháza; Aladár Visnya, real school, 7th grade, Pécs; Lipót Weisz, real school, 6th grade, Pécs.