Question

What is the empirical formula? A compound is used to treat iron deficiency in people. It contains 36.76% iron, 21.11% sulfur, and 42.13% oxygen. The empirical formula is Fe-S-O-.

Answer 1

Answer: The empirical formula of the compound is
`Fe_1S_1O_4`

Step-by-step explanation:

Empirical formula is defined formula which is simplest integer ratio of number of atoms of different elements present in the compound.

Percentage of iron in a compound = 36.76 %

Percentage of sulfur in a compound = 21.11 %

Percentage of oxygen in a compound = 42.13 %

Consider in 100 g of the compound:

Mass of iron in 100 g of compound = 36.76 g

Mass of iron in 100 g of compound = 21.11 g

Mass of iron in 100 g of compound = 42.13 g

Now calculate the number of moles each element:

Moles of iron=
`(36.76 g)/(55.84 g/mol)=0.658 mol`

Moles of sulfur=
`(21.11 g)/(32.06 g/mol)=0.658 mol`

Moles of oxygen=
`(42.13 g)/(16 g/mol)=2.633 mol`

Divide the moles of each element by the smallest number of moles to calculated the ratio of the elements to each other

For Iron element =
`(0.658 mol)/(0.658 mol)=1`

For sulfur element =
`(0.658 mol)/(0.658 mol)=1`

For oxygen element =
`(2.633 mol)/(0.658 mol)=4.001\approx 4`

So, the empirical formula of the compound is
`Fe_1S_1O_4`

Answer 2

Hello!

What is the empirical formula? A compound is used to treat iron deficiency in people. It contains 36.76% iron, 21.11% sulfur, and 42.13% oxygen. The empirical formula is Fe-S-O-.

data:

Iron (Fe) ≈ 55.84 a.m.u (g/mol)

Sulfur (S) ≈ 32.06 a.m.u (g/mol)

Oxygen (O) ≈ 16 a.m.u (g/mol)

We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100g solution)

Fe: 36.76 % = 36.76 g

S: 21.11 % = 21.11 g

O: 42.13 % = 42.13 g

The values ​​(in g) will be converted into quantity of substance (number of mols), dividing by molecular weight (g / mol) each of the values, we will see:

`Fe: (36.76\:\diagup\!\!\!\!\!g)/(55.84\:\diagup\!\!\!\!\!g/mol) \approx 0.658\:mol`

`S: (21.11\:\diagup\!\!\!\!\!g)/(32.06\:\diagup\!\!\!\!\!g/mol) \approx 0.658\:mol`

`O: (42.13\:\diagup\!\!\!\!\!g)/(16\:\diagup\!\!\!\!\!g/mol) \approx 2.633\:mol`

We realize that the values ​​found above are not integers, so we divide these values ​​by the smallest of them, so that the proportion does not change, let us see:

`Fe: (0.658)/(0.658)\to\:\:\boxed{Fe = 1}`

`S: (0.658)/(0.658)\to\:\:\boxed{S = 1}`

`O: (2.633)/(0.658)\to\:\:\boxed{O \approx 4}`

Thus, the minimum or empirical formula found for the compound will be:

`\boxed{\boxed{Fe_1S_1O_4\:\:\:or\:\:\:FeSO_4}}\Longleftarrow(Empirical\:Formula)\end{array}}\qquad\checkmark`

Answer:

FeSO4 - Iron (II) Sulfate

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I Hope this helps, greetings ... Dexteright02! =)