A5.91 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?

C₆H₈O₇

Explanation:

1) Calculate the mass of carbon (C)

mass of C = % of C × mass of sample / 100mass 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 massnumber 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 numbernumber 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 numbernumber of moles of O = 1.2966 × 10²³ / 6.022 × 10²³ = 0.21531 moles

5) Find the mole ratios:

Summary of moles:

C: 0.184 molH: 0.24626 molO: 0.21531 mol

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

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

Multiply by 3 to round to integer numbers:

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

Multiply by 2 to round to integer numbers:

C:  3 × 2 = 6H: 4 × 2 = 8O: 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.