6. A sample of pure iron was covered with an excess of powdered elemental sulfur

ID: 1036927 • Letter: 6

Question

6. A sample of pure iron was covered with an excess of powdered elemental sulfur. The ature where a reaction occurred and the excess sulfur was volatilized. The following mixture was heated to a temp data were colected. Compethe rt Sheet Record the calculated value with the correct number of significant figures. 1. Mass of crucible and lid (g) 2. Mass of crucible, lid, and iron sample (g) 3. Mass of Fe sample (g) 6. Final mass of crucible, lid, and Fe/S compound (g) 21.195 7. Mass of Fe/S compound (g) 8. Combination Reaction of Iron and Sulfur Calculation Zone 19.746Part 8b 20.422 Part 8c a. Mass of S in compound (g) b. Mass ratio of Fe to S in compound Show calculation Part 8d c. Moles of Fe in compound d. Moles of S in compound e. Mole ratio (empirical formula) of Fe to S f. Percent by mass (%) Show calculation Show calculation. Part 8e Show calculation. Le.t --%Fe %s

Explanation / Answer

Mass of crucible and lid (g)

19.746

Mass of crucible, lid, and iron sample (g)

20.422

Mass of Fe sample (g)

20.422-19.746=0.676

Final mass of crucible, lid, and Fe/S compound (g)

21.195

Mass of Fe/S compound (g)

21.195-19.746=1.449

8. Combination reaction of Iron and Sulfur

a. Mass of S in compound (g) = 1.449 – 0.676 = 0.773

b. Mass ratio of Fe to S in compound= 0.676g of Fe to 0.773g of S

=1: 1.43

c. Moles of Fe in compound = mass / molar mass

= 0.676 / 55.845 (molar mass of Fe =55.845)

=0.0121

d. Moles of S in compound = mass / molar mass

= 0.773 / 32.065 (molar mass of S=32.065)

= 0.0241

e. Mole ratio (empirical formula) of Fe to S= Divide each mole value by the smallest number of moles calculated

=(0.0121 /0.121) :: 0.0421/0.0121)

= 1: 2

Thus for every iron atom there are 2 sulphur atoms.

FeS2is the empirical formula

To find the percent by mass just divide the mass of the element by the mass of the compound and then multiply the answer by 100%

Percent by mass (%) = (0.676 / 1.449) X 100%

= 46.65% Fe

Percent by mass (%) = (0.773 / 1.449) X 100%

= 53.35% S