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ChemistryChemistry
13 May, 03:30

A sample of a compound containing C, O, and silver (Ag) weighed 1.372 g. On analysis, it was found to contain 0.288 g O and 0.974 g Ag. The molar mass of the compound was determined to be 308.8 g/mol. What is the molecular formula of the compound?

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Answers (2)
  1. 13 May, 05:16
    0
    The molecular formula is Ag2C2O4

    Explanation:

    Step 1: Data given

    A compound contains C, O and Ag

    The mass of the compound is 1.372 grams

    Mass of O = 0.288 grams

    Mass of Ag = 0.974 grams

    Molar mass C = 12.01 g/mol

    Molar mass O = 16.0 g/mol

    Molar mass Ag = 107.87 g/mol

    The molar mass of the compound = 308.8 g/mol

    Step 2: Calculate moles

    Moles = mass / molar mass

    Moles O = 0.288 grams / 16.0 grams

    Moles O = 0.018 moles

    Moles Ag = 0.974 grams / 107.87 g/mol

    Moles Ag = 0.00903 moles

    Moles C = 0.11 grams / 12.01 g/mol

    Moles C = 0.00916 moles

    Step 3: Calculate the mol ratio

    We divide by the smallest amount of moles

    O: 0.018 / 0.00903 = 2

    C: 0.0.00916 / 0.00903 = 1

    Ag: 0.00903/0.00903 =

    The empirical formula is AgCO2

    The molar mass of this empirical formula is 151.88

    We have to multiply the empirical formula by n

    n = 308.8/151.88 = 2

    2 * (AgCO2) = Ag2C2O4

    The molecular formula is Ag2C2O4
  2. 13 May, 05:35
    0
    The molecular formula for the compound is Ag₂C₂O₄, silver oxalate.

    Explanation:

    This is an easy excersise, we relate the moles of the elements in the compound with the molar mass of the total compound.

    The sample weighs 1.372 g, with 0.288 g of O, 0.974 g of Ag and the rest, C

    Therefore, 1.372g - 0.288 g - 0.974g = 0.112 g are C

    We convert the mass to moles

    0.288 g / 16g/mol = 0.018 moles of O

    0.974 g / 107.87 g/mol = 0.00903 moles of Ag

    0.112 g / 12 g/mol = 0.00933 moles of C

    Now, the use of the molar mass of the compound. Let's prepare some rules of three:

    1.372 g of compound contain 0.018 moles of O, 0.00903 moles of Ag and 0.00933 moles of C

    Then, 308.8 g of compound (a mol) must contain:

    (308.8. 0.018) / 1.372 = 4 moles of O

    (308.8. 0.00903) / 1.372 = 2 moles of Ag

    (308.8. 0.00933) / 1.372 = 2 moles of C

    The molecular formula for the compound is Ag₂C₂O₄, silver oxalate.
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Home » Chemistry » A sample of a compound containing C, O, and silver (Ag) weighed 1.372 g. On analysis, it was found to contain 0.288 g O and 0.974 g Ag. The molar mass of the compound was determined to be 308.8 g/mol. What is the molecular formula of the compound?
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