Acompound is composed of 72% iron and 27.6% oxygen. what is the empirical formula?

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explanation:

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The empirical formula is Fe_{3}O_{4}.

Explanation:

Empirical formula is the easiest way of represent a chemical compound with representing the ratio of each element present in the compound. For finding the empirical formula, first we have to know the molar mass present in the given percentage of elements.

As we know that the molar mass of iron is 55.8 g/mol and oxygen in 16 g/mol, then dividing the percentage of elements given with their respective molar mass will give us the moles present in this composition.So,

             \frac{72}{56}=1.29 \text { moles of } F e

             \frac{27.6}{16}=1.725 \text { moles of } 0

Now the least among the two is 1.29, so we have to divide the moles of Fe and O with this to determine the ratio of each element. So, we are getting the ratio of 1:1.3, then if we multiply it by 3, we can get integer terms as 3:4. Thus, we obtained that Fe is 3 and O is  4. So the empirical formula will be Fe_{3}O_{4}.

The empirical formula is Fe_{3}O_{4}.

Explanation:

Empirical formula is the easiest way of represent a chemical compound with representing the ratio of each element present in the compound. For finding the empirical formula, first we have to know the molar mass present in the given percentage of elements.

As we know that the molar mass of iron is 55.8 g/mol and oxygen in 16 g/mol, then dividing the percentage of elements given with their respective molar mass will give us the moles present in this composition.So,

             \frac{72}{56}=1.29 \text { moles of } F e

             \frac{27.6}{16}=1.725 \text { moles of } 0

Now the least among the two is 1.29, so we have to divide the moles of Fe and O with this to determine the ratio of each element. So, we are getting the ratio of 1:1.3, then if we multiply it by 3, we can get integer terms as 3:4. Thus, we obtained that Fe is 3 and O is  4. So the empirical formula will be Fe_{3}O_{4}.



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