16 (12 points) Given in the cube $A B C D$ $A_{1} B_{1} C_{1} D_{1}$, $O, E, F, G$ are the midpoints of $B D$, $B B_{1}$, $A_{1} D_{1}$, $D_{1} C_{1}$ respectively, and $A B=1$. Find the volume of the tetrahedron $O E F G$.
16 Connect $B_{1} D_{1}$ intersecting $F G$ at $H$, and connect $A_{1} C_{1}$, then $B_{1} D_{1} \perp A_{1} C_{1}$. Since $F, G$ are the midpoints of $A_{1} D_{1}$, $D_{1} C_{1}$ respectively, $F G / / A_{1} C_{1}$, thus $F G \perp B_{1} D_{1}$. Also, because $B B_{1} \perp$ plane $A_{1} B_{1} C_{1} D_{1}$, and $F G$ is in plane $A_{1} B_{1} C_{1} D_{1}$, so $B B_{1} \perp F G$.

From this, we get $F G \perp$ plane $B B_{1} D_{1} D$. Since $F H = G H$, we have
$$
V_{O-E F G}=V_{F-O E H}+V_{G-O E H}=2 V_{F-O E H}=\frac{2}{3} S_{O E H} \cdot F H.
$$

In trapezoid $O B B_{1} H$
$$
\begin{aligned}
S_{\triangle O E H} & =S_{\text{trapezoid } O B B_{1} H}-S_{\triangle E B_{1} H}-S_{\triangle O B E} \\
& =\frac{5 \sqrt{2}}{8}-\frac{\sqrt{2}}{8}-\frac{3 \sqrt{2}}{16}=\frac{5 \sqrt{2}}{16}.
\end{aligned}
$$

Therefore, the volume of tetrahedron $O E F G$ is
$$
V_{O-E F G}=2 \times \frac{1}{3} \times \frac{5 \sqrt{2}}{16} \times \frac{\sqrt{2}}{4}=\frac{5}{48}.
$$