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chubhunter [2.5K]
2 years ago
8

A 5.91 g unknown sample analyzed by elemental analysis revealed a composition of 37.51 % C. In addition, it was determined that

the sample contains 1.4830 × 1023 hydrogen atoms and 1.2966 × 1023 oxygen atoms. What is the empirical formula?
Chemistry
1 answer:
SSSSS [86.1K]2 years ago
7 0

Answer:

  • C₆H₈O₇

Explanation:

<u>1) Calculate the mass of carbon (C)</u>

  • mass of C = % of C × mass of sample / 100

  • mass of C = 37.51% × 5.91 g / 100 = 2.21 g

<u>2) Calculate the number of moles of C</u>

  • number of moles = mass in grams / molar mass

  • number of moles of C = 2.21 g / 12.01 g/mol = 0.184 moles

<u>3) Calculate the number of moles of hydrogen atoms, H:</u>

  • number of moles = number of atoms / Avogadro's number

  • number of moles of H = 1.4830 × 10²³ / 6.022 × 10²³ = 0.24626 moles

<u>4) Calculate the number of moles of oxygen atoms, O:</u>

  • number of moles = number of atoms / Avogadro's number

  • number of moles of O = 1.2966 × 10²³ / 6.022 × 10²³ = 0.21531 moles

<u>5) Find the </u><u>mole ratios</u><u>:</u>

Summary of moles:

  • C: 0.184 mol
  • H: 0.24626 mol
  • O: 0.21531 mol

Divide every amount by the smallest number, which is 0.184:

  • C: 0.184 / 0.184 = 1
  • H: 0.24626 / 0.184 = 1.34
  • O: 0.21531 / 0.184 = 1.17

Multiply by 3 to round to integer numbers:

  • C:  1 × 3 = 3
  • H: 1.34 × 3 = 4.02 ≈ 4
  • O: 1.17 × 3 = 3.51

Multiply by 2 to round to integer numbers:

  • C:  3 × 2 = 6
  • H: 4 × 2 = 8
  • O: 3.51 × 2 ≈ 7

Use the mole ratios as superscripts to write the empirical formula

  • C₆H₈O₇ ← answer

Just as a reference, you can search in internet and find that one compound with that empirical formula is citric acid.

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