posted by John on .
In one experiment, the reaction of 1.00 g mercury and an excess of sulfur yielded 1.16 g of a sulfide of mercury as the sole product. In a second experiment, the same sulfide was produced in the reaction of 1.50 g mercury and 1.08 g sulfur . What mass of the sulfide of mercury was produced in the second experiment?
find the formula of the sulfide
You know 1.00 g of the compound is Hg.
Hg = 1/molmassHg = number moles H=
= 1/200 moles= 0.005 moles Hg
S= .16/32 moles= .005 moles S
To the mole ratio of Hg to S is 1:1, so the empirical formula is HgS
Now, in the second, you have 1.08g S, figure the number of moles of that. Then, you have the same number of moles of Hg. What is the mass of that number of moles of Hg? The mass of the sulfide then is the Hg mass, plus 1.08g