What is the empirical formula of a compound that contains 28% iron, 24% sulfur, and 48% oxygen by mass?

Emprical formula = FeS206

Explanation:

Empirical formula is a formula that shows the ratio of elements present in a compound. Empirical formula does not give you the amount of the actual number of atom present in the compound but the proportion of the elements in the compound.

Calculating the empirical formula  for a compound with 28% of iron, 24% of sulfur and 48% of oxygen by mass simply indicate the compound is 100% by mass. Or one could denote the percentage as mass(gram)

Usually

Molarity(mol) = mass/ molar mass

we  look for the Molarity value to get the ratio of the elements present in the compound.

Fe(iron) = mass/molar mass = 28/55.845 = 0.50138776972 mol

S(sulfur) = mass /molar mass = 24 / 32.065 = 0.7484796507  mol

O2(oxygen) = mass/molar mass = 48 / 15.999 = 3.00018751172  mol

Divide through the mole value by the smallest mol number

0.50138776972/0.50138776972 = 1 mol iron

0.7484796507/ 0.50138776972 = 1.4928159319 ≈ 2 sulfur

3.00018751172/0.50138776972 = 5.98376684257 ≈ 6 oxygen

The value gotten is used as a subscript in the empirical formula.

Empirical formula = FeS206

c. 75.7

i think that is the right

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