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19 May, 12:34

A sample of a compound contains only the elements sodium, sulfur, and oxygen. it is found by analysis to contain 0.979 g na, 1.365 g s, and 1.021 g o. determine its empirical formula.

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  1. 19 May, 12:51
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    First, convert the masses into moles using the molar weights: 23 g Na/mol, 32.06 g S/mol and 16 g O/mol.

    Mol Na: 0.979/23 = 0.04256522

    Mol S: 1.365/32.06 = 0.04257642

    Mol O: 1.021/16 = 0.0638125

    The least value is mol Na. Thus, divide all mole compositions to that of mol Na:

    Mol Na: 0.04256522/0.04256522 = 1

    Mol S: 0.04257642/0.04256522 = 1

    Mol O: 0.0638125/0.04256522 = 1.5

    The ratios must be whole numbers. To make 1.5 a whole number, multiply all ratios with 2 to keep it uniform.

    Mol Na: 1*2 = 2

    Mol S: 1*2 = 2

    Mol O: 1.5*2 = 3

    Thus, the empirical formula is Na₂S₂O₃.
  2. 19 May, 12:58
    0
    To determine the empirical formula for the compound that contains 0.979 g Na, 1.365 g S, and 1.021 g O, we convert these to mole units. The molar masses to be used are:

    Molar mass of Na = 23 g/mol

    Molar mass of S = 32 g/mol

    Molar mass of O = 16 g / mol

    The number of moles is obtained using the molar mass for each element.

    moles Na = 0.979 g Na / 23 g/mol Na = 0.04256

    moles S = 1.365 g Na / 32 g/mol Na = 0.04265

    moles O = 1.021 g O / 16 g/mol Na = 0.06326

    We then divide each with the smallest number of moles obtained.

    Na: 0.04256 / 0.04256 = 1

    S: 0.04265 / 0.04256 = 1.002 ≈ 1

    O: 0.06326 / 0.04256 = 1.49 ≈ 1.5

    We then have an empirical formula of NaSO₁.₅. However, chemical formulas must have only integers as subscripts, thus, we multiply each to 2. The empirical formula is then Na₂S₂O₃ also known as sodium thiosulfate.
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