The formula of a compound of iron and sulfur was determined using the method of

ID: 535707 • Letter: T

Question

The formula of a compound of iron and sulfur was determined using the method of mass measurements. A sample of iron was weighed into a crucible and covered with finely powdered sulfur. The crucible was covered and heated to allow the iron and sulfur to react. Assume all the iron reacted with the sulfur. Additional heating burned off all unreacted sulfur. The crucible was then cooled and weighed. The following data were collected. Mass of crucible and cover, 27.889 g. Mass of crucible, cover and iron, 32.025 g. Mass of crucible, cover and compound formed, 35.587 g. a. Calculate the mass of iron reacted. b. Calculate the mass of compound that formed. c. Calculate the mass of sulfur reacted. d. Calculate the moles of iron reacted. e. Calculate the number of moles of sulfur reacted. f. Find the simplest whole-number ratio of the number of moles of iron to the number of moles of sulfur in the compound formed. g. Write the empirical formula for the compound formed. Write a balanced chemical reaction showing the reaction of carbon with oxygen gas to give carbon dioxide gas. How many grams of oxygen are needed to react completely with 0.234 g of carbon? A sample of 0.234 g of carbon is placed in a closed stainless steel bomb with 0.502 g of oxygen gas, O_2. The mixture is ignited and the carbon reacts to give carbon dioxide. How many grams of carbon dioxide should be obtained upon combustion?

Explanation / Answer

Ans. 5a.

Mass of Fe reacted = (Mass of crucible+ cover+ iron) - (Mass of crucible+ cover)

= 32.025 g – 27.889 g

= 4.136 g

#5b. Mass of compound formed =

(Mass of crucible+ cover+ compound) - (Mass of crucible+ cover)

= 35.587 g – 27.889 g

= 7.698 g

#5c. Mass of S reacted =

(Mass of crucible+ cover+ compound) - (Mass of crucible + Cover+ Iron)

= 35.587 g – 32.025 g

= 3.562 g

#5d. Mole of Fe reacted = Mass of Fe/ molar mass of Fe

= 4.136 g / (55.847 g/mol)

= 0.074059 mol

#5e. Mole of S reacted = Mass of S/ molar mass of S

= 3.562 g / (55.847 g/mol)

= 0.111083 mol

#5f. Simplest whole number ratio = Moles of Fe/ Moles of S

= 0.074059 mol / 0.111083 mol

= 0.666702 : 1

Multiplying a ratio with any factor does not affect the ratio. Multiply the above ratio by 3.

= (0.666702 : 1) x 3 = 2.000 : 3

= 2 : 3

#5g. we have, Fe: S = 2: 3 - from #5f.

Empirical formula = Fe2S3

#6. Balanced reaction: C(s) + O2(g) -------> CO2(g)

Stoichiometry : 1 mol carbon reacts with 1 mol O2.

Moles of C consumed = Mass of C / Molar mass

= 0.234 g/ (12.011 g/mol)

= 0.0195 mol

So,

Using stoichiometry of balanced reaction-

Moles of O2 required = Moles of C consumed = 0.0195 mol

Mass of O2 required = Moles of O2 required x Molar mass

= 0.0195 mol x (31.9988 g/mol)

= 0.624 g

#7. Moles of C = 0.234 g/ (12.011 g/mol) = 0.0195 mol

Moles of O2 = 0.502 g/ (31.9988 g/mol) = 0.0157 mol

Note that moles of O2 is less than that of C when compared to the 1:1 theoretical ration of reactants in balanced stoichiometry (C: O2 = 1: 1).

So, O2 is the limiting reactant. The product formation follows limiting reactant.

So, moles of CO2 formed = Moles of limiting reactant (O2) consumed = 0.0157 mol

Mass of Co2 obtained = Moles of CO2 x Molar mass

= 0.0157 mol x (44.00 g/mol)

= 0.6908 g

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