Question

A sample of pure iron was covered with an excess of powdered elemental sulfur. The mixture was heated to a temperature where the excess sulfur was volatilized. The mass of the sample of Iron is 0.676g. The mass of the sample of Sulfur is 0.773g.

a. What is the mass ratio of Fe to S in a compound?

b. What is the mole ratio (empirical formula) of Fe to S?

c. What is the percent by mass of Fe and S?

Answer 1

a. The mass ratio of Fe to S in a compound is 0.876. b. The mole ratio (empirical formula) of Fe to S is 0.5. c. The percent by mass of Fe is 46.72% and the percent by mass of S is 53.28%.

Step-by-step explanation:

a. The mass ratio of Fe to S in a compound can be found by dividing the mass of Fe by the mass of S. In this case, the mass of Fe is 0.676g and the mass of S is 0.773g, so the mass ratio is 0.676g / 0.773g = 0.876.

b. The mole ratio (empirical formula) of Fe to S can be found by dividing the moles of Fe by the moles of S. To find the moles, we need to convert the masses of Fe and S to moles using their molar masses. The molar mass of Fe is 55.845 g/mol and the molar mass of S is 32.06 g/mol. With the given masses, the moles of Fe are 0.676g / 55.845 g/mol = 0.0121 mol and the moles of S are 0.773g / 32.06 g/mol = 0.0241 mol. Therefore, the mole ratio is 0.0121 mol / 0.0241 mol, which simplifies to 0.5.

c. To find the percent by mass of Fe and S, we need to calculate the mass of Fe and S as a percentage of the total mass of the compound. The total mass of the compound is the sum of the masses of Fe and S, which is 0.676g + 0.773g = 1.449g. The percent by mass of Fe is (0.676g / 1.449g) * 100% = 46.72%, and the percent by mass of S is (0.773g / 1.449g) * 100% = 53.28%.