2- A 0.60 sample an unknown organic acid found in muscle cells is burned in air and found to contain 0.24 grams of carbon, 0.040 grams of hy

Question

2- A 0.60 sample an unknown organic acid found in muscle cells is burned in air and found to contain 0.24 grams of carbon, 0.040 grams of hydrogen, with the rest being oxygen. If the molecular weight of the substance is 90 grams/n, what is the molecular formula

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Thu Nguyệt 6 months 2021-07-12T11:12:38+00:00 1 Answers 173 views 0

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    2021-07-12T11:14:04+00:00

    Answer:

    C₃H₆O₃

    Explanation:

    To solve this question we need to find, as first, the moles of each atom in order to find empirical formula (Simplest whole-number ratio of atoms present in a molecule).

    With the molar mass of the substance and the empirical formula we can find the molecular formula as follows:

    Moles C -Molar mass:12.0g/mol-

    0.24g * (1mol/12.0g) = 0.020 moles C

    Moles H = Mass H because molar mass = 1g/mol:

    0.040 moles H

    Moles O -Molar mass: 16g/mol-

    Mass O: 0.60g – 0.24g – 0.040g = 0.32g O

    0.32g O * (1mol/16g) = 0.020 moles O

    Ratio of atoms (Dividing in moles of C: Lower number of moles):

    C = 0.020 moles C / 0.020 moles C = 1

    H = 0.040 moles H / 0.020 moles C = 2

    O = 0.020 moles O / 0.020 moles C = 1

    Empirical formula:

    CH₂O.

    Molar mass CH2O:

    12g/mol + 2*1g/mol + 16g/mol = 30g/mol

    As molecular formula has a molar mass 3 times higher than empirical formula, the molecular formula is 3 times empirical formula:

    C₃H₆O₃

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