Question

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?

Answer 1

• C₆H₈O₇

Step-by-step explanation:

1) Calculate the mass of carbon (C)

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

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

2) Calculate the number of moles of C

• number of moles = mass in grams / molar mass

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

3) Calculate the number of moles of hydrogen atoms, H:

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

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

4) Calculate the number of moles of oxygen atoms, O:

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

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

5) Find the mole ratios:

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.