Chemistry, 26.06.2019 11:00 bermudezs732

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-.

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Answer from: elenadelfosse

The empirical formula of the compound is Fe_1S_1O_4

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= $\frac{36.76 g}{55.84 g/mol}=0.658 mol$

Moles of sulfur= $\frac{21.11 g}{32.06 g/mol}=0.658 mol$

Moles of oxygen= $\frac{42.13 g}{16 g/mol}=2.633 mol$

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 = $\frac{0.658 mol}{0.658 mol}=1$

For sulfur element = $\frac{0.658 mol}{0.658 mol}=1$

For oxygen element = $\frac{2.633 mol}{0.658 mol}=4.001\approx 4$

So, the empirical formula of the compound is Fe_1S_1O_4

Answer from: kitttimothy55

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:

$Fe: \dfrac{36.76\:\diagup\!\!\!\!\!g}{55.84\:\diagup\!\!\!\!\!g/mol} \approx 0.658\:mol$

$S: \dfrac{21.11\:\diagup\!\!\!\!\!g}{32.06\:\diagup\!\!\!\!\!g/mol} \approx 0.658\:mol$

$O: \dfrac{42.13\:\diagup\!\!\!\!\!g}{16\:\diagup\!\!\!\!\!g/mol} \approx 2.633\:mol$

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:

$Fe: \dfrac{0.658}{0.658}\to\:\:\boxed{Fe = 1}$

$S: \dfrac{0.658}{0.658}\to\:\:\boxed{S = 1}$

$O: \dfrac{2.633}{0.658}\to\:\:\boxed{O \approx 4}$

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

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

FeSO4 - Iron (II) Sulfate

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

Answer from: Quest

i believe the answer is true.

Answer from: Quest

Ithink that the answer is b

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