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7 December, 02:02

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?

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  1. 7 December, 02:19
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    C₆H₈O₇

    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.
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