Question

Cinnabar (HgS) is a prominent ore of mercury. What is the percentage of mercury present in 225 grams of pure HgS? (molar mass of Hg and S is 200.6 gm mol$^{-1}$ and 32 gm mol$^{-1}$ respectively).
a.) 194.04 gms
b.) 200.6 gms
c.) 320.4 gms
d.) 260.8 gms

Answer 1

Hint: The percentage of mercury can be found by mass=moles$\times$molecular mass. We are given with the molar mass of Hg, and S. The moles of HgS present in 225 gms by dividing the given mass to the molecular mass of HgS.

Complete step by step solution:
Firstly, let us calculate the molar mass of HgS i.e. the sum of molar mass of Hg (200.6 gm mol$^{-1}$) and the molar mass of S (32 gm mol$^{-1}$). It can be written as: Molar mass of HgS = 200.6 + 32 = 232.6 gm mol$^{-1}$
Now, we will calculate the moles present in given mass i.e. 225 gm = $\dfrac{225}{molar mass of HgS}$ = $\dfrac{225}{232.6}$= 0.94 moles
We know, in the 1 mole of HgS, there is one mole of Hg; so, 0.94 moles of HgS will contain 0.94 moles of Hg.
As mentioned, mass = moles $\times$ molecular mass = 0.94$\times$200.6 = 194.04 gms.
Therefore, the percentage of mercury is 19.04 gms. The correct option is A.

Note: Don’t get confused between the formula, the percentage of mercury in 225 gms of HgS can also be calculated by the simple mathematics i.e. mass of Hg in 225 gms = $\dfrac{225\times200.6}{232.6}$ gms = 194.04 gms. This method is also valid.