An analysis shows that 2.97 g of iron metal combines with oxygen to form 4.25 g of an oxide of iron. What is the empirical formula of the compound

Respuesta :

The empirical formula of the compound obtained when 2.97 g of iron combined with oxygen to produce 4.25 g of iron oxide is Fe₂O₃

  • We'll begin by calculating the mass of oxygen in compound.

Mass of Fe = 2.97 g

Mass of compound = 4.25 g

Mass of O =?

Mass of O = (Mass of compound) – (Mass of Fe)

Mass of O = 4.25 – 2.97

Mass of O = 1.28 g

  • Finally, we shall determine the empirical formula of the compound. This can be obtained as follow:

Fe = 2.97 g

O = 1.28

Divide by their molar mass

Fe = 2.97 / 56 = 0.053

O = 1.28 / 16 = 0.08

Divide by the smallest

Fe = 0.053 / 0.053 = 1

O = 0.08 / 0.053 = 1.5

Multiply by 2 to express in whole number

Fe = 1 × 2 = 2

O = 1.5 × 2 = 3

Therefore, the empirical formula of the compound is Fe₂O₃

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Considering the definition of empirical formula,  

The empirical formula is the simplest expression to represent a chemical compound, which indicates the elements that are present and the minimum proportion in whole numbers that exist between its atoms, that is, the subscripts of chemical formulas are reduced to the most integers. small as possible.

In this case, in first place you know:

  • Mass of iron= 2.97 g
  • Mass of compound= 4.25 g
  • Mass of oxygen= (Mass of compound) – (Mass of Fe)= 4.25 g - 2.97 g= 1.28 g

Then, you need to express the mass into percentage as follow:

  • Iron (Fe): [tex]\frac{2.97 grams}{4.25 grams}x100=[/tex] 69.88%
  • Oxygen (O): [tex]\frac{1.28 grams}{4.25 grams}x100=[/tex]30.12%

Assuming a 100 grams sample, the percentages match the grams in the sample. So you have 69.88 grams of iron and 30.12 grams of oxygen.

Then it is possible to calculate the number of moles of each atom in the molecule, taking into account the corresponding molar mass:

  • Iron (Fe): [tex]\frac{69.88 grams}{55.85\frac{grams}{mole} }=1.25 moles[/tex]  
  • Oxygen (O): [tex]\frac{30.12 grams}{16 \frac{grams}{mole} }=1.88 moles[/tex]

The empirical formula must be expressed using whole number relationships, for this the numbers of moles are divided by the smallest result of those obtained. In this case:

  • Iron (Fe): [tex]\frac{1.25 moles}{1.25 moles}=1[/tex]  
  • Oxygen (O): [tex]\frac{1.88 moles}{1.25 moles}= 1.5[/tex]

Finally all coefficients are multiplied by a whole number to ensure that the smallest proportion of whole number subscripts is obtained. In this case, it is multiplied by 2:

  • Iron (Fe): 1×2= 2  
  • Oxygen (O): 1.5×2= 3

Therefore the Fe : O mole ratio is 2:3

Finally, the empirical formula is Fe₂O₃.

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