Answer
The mineral is cinnabar with density 8.10 $\frac{g}{cm^{3}}$ .
Work Step by Step
1- Find the density of the mineral with mass 5.94 g and volume 0.73 $cm^{3}$.
d = $\frac{m}{v}$
d = $\frac{5.94 g}{0.73cm^{3}}$
d= 8.14 $\frac{g}{cm^{3}}$
2- Compare the density that we found d= 8.14 $\frac{g}{cm^{3}}$ with the given densities of minerals of :
sphalerite 4.0 $\frac{g}{cm^{3}}$
cassiterite 6.99 $\frac{g}{cm^{3}}$
cinnabar 8.10 $\frac{g}{cm^{3}}$
The density is closer to the density of cinnabar as 8.14 $\frac{g}{cm^{3}}$ $\approx$ 8.10 $\frac{g}{cm^{3}}$.
Taking into consideration the experimental error , we arrive at conclusion that the mineral is cinnabar.
