For the compound iron (II) sulfide, answer the following questions: a. Write the correct chemical formula for iron (II) sulfide? b. How many atoms are in one formula unit of this substance? c. What is the formula weight in amu? d. What is the percentage of sulfur atoms in iron (II) sulfide? e. What is the molar mass in grams? f. How many moles of iron (II) sulfide are present in 156.7 grams of iron (II) sulfide ionic salt?
The Correct Answer and Explanation is:
. Write the correct chemical formula for iron (II) sulfide
The correct formula for iron (II) sulfide is FeS. In this compound, iron has a 2+ charge (Fe²⁺), and sulfur has a 2- charge (S²⁻). Since the charges are equal and opposite, one atom of iron combines with one atom of sulfur to form the neutral compound.
b. How many atoms are in one formula unit of this substance?
One formula unit of iron (II) sulfide consists of two atoms: one iron (Fe) atom and one sulfur (S) atom.
c. What is the formula weight in amu?
To calculate the formula weight of FeS in atomic mass units (amu), we sum the atomic weights of iron and sulfur:
- The atomic weight of iron (Fe) is approximately 55.85 amu.
- The atomic weight of sulfur (S) is approximately 32.07 amu.
Formula weight of FeS = 55.85 + 32.07 = 87.92 amu
d. What is the percentage of sulfur atoms in iron (II) sulfide?
The percentage of sulfur in iron (II) sulfide is calculated by dividing the atomic weight of sulfur by the formula weight of the compound and then multiplying by 100: Percentage of sulfur=(32.0787.92)×100≈36.5%\text{Percentage of sulfur} = \left( \frac{32.07}{87.92} \right) \times 100 \approx 36.5\%Percentage of sulfur=(87.9232.07)×100≈36.5%
So, sulfur makes up approximately 36.5% of the total mass of iron (II) sulfide.
e. What is the molar mass in grams?
The molar mass of iron (II) sulfide is numerically equivalent to its formula weight in grams per mole. Therefore, the molar mass of FeS is 87.92 g/mol.
f. How many moles of iron (II) sulfide are present in 156.7 grams of iron (II) sulfide ionic salt?
To find the number of moles in 156.7 grams of iron (II) sulfide, we use the formula: Moles=MassMolar Mass=156.787.92≈1.78 moles\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{156.7}{87.92} \approx 1.78 \text{ moles}Moles=Molar MassMass=87.92156.7≈1.78 moles
So, 1.78 moles of iron (II) sulfide are present in 156.7 grams of the substance.
In summary:
a. FeS
b. 2 atoms
c. 87.92 amu
d. 36.5% sulfur
e. 87.92 g/mol
f. 1.78 moles
