A compound is made up of carbon, hydrogen, and oxygen atoms. It has 52.14% C and 34.73% O.

a. What is the empirical formula?

b. The molar mass of the compound is 138 g/mol. What is the molecular formula?



Answer :

Considering the definition of empirical and molecular formula, the empirical formula is C₂H₆O and the molecular formula is C₆H₁₈O₃.

Empirical formula

The empirical formula is the simplest expression to represent a chemical compound, regardless of its structure because indicates the elements that are present and the minimum proportion that exist between its atoms.

Molecular formula

The molecular formula is the chemical formula that indicates the number and type of different atoms present in the molecule, but it indicates the actual number of atoms that make up a molecule.

Molecular formula in this case

In this case, you know:

  • C: 52.14 %
  • O: 34.73  %
  • H: 13.13 % Note that the percentage must be 100%, and with carbon and oxygen there is 86.87%. So the rest is hydrogen.

Assuming a 100 grams sample, the percentages match the grams in the sample: you have 52.14 grams of C, 34.73 grams of O and 13.13 grams of H.

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

  • C: [tex]\frac{52.14 g}{12 \frac{g}{ymol} }[/tex]= 4.345 moles
  • O: [tex]\frac{34.73 g}{16 \frac{g}{ymol} }[/tex]= 2.17 moles
  • H: [tex]\frac{13.13 g}{1 \frac{g}{ymol} }[/tex]= 13.13 moles

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:

  • C: [tex]\frac{4.345 moles}{2.17 moles}[/tex]= 2
  • O: [tex]\frac{2.17 moles}{2.17 moles}[/tex]= 1
  • H: [tex]\frac{13.13 moles}{2.17 moles}[/tex]= 6

Therefore the C: H: O mole ratio is 2: 6: 1

Then, the empirical formula is C₂H₆O₁= C₂H₆O, with a empirical mass of 2×12 g/mol + 6×1 g/mol + 16 g/mol= 46 g/mol

The molecular formula can be calculated as MF= n(EF)

where:

  • MF= molecular formula
  • n=molecula mass÷ empirical mass
  • EF= empirical formula

In this case, the value n can be calculated:

n= 138 g/mol÷ 46 g/mol

Solving:

n= 3

Then, the molecular formula can be calculated as MF= 3×EF

Finally, the molecular formula is C₆H₁₈O₃.

In summary, the empirical formula is C₂H₆O and the molecular formula is C₆H₁₈O₃.

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