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A sample of pure compound contains 2.04 ...

A sample of pure compound contains 2.04 g of sodium, `2.65xx10^(22)` atoms of carbon and 0.132 mol of oxygen atoms. Its empirical formula is

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`Na_2CO_3`
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Knowledge Check

  • A sample of pure compound contains 1.15 g of sodium, 3.01xx10^(22) atoms of carbon and 0.1 mol of oxygen atom. Its empirical formula is

    A
    `Na_(2)CO_(3)`
    B
    `NaCO_(2)`
    C
    `Na_(2)CO`
    D
    `Na_(2)CO_(2)`

  • A given sample of pure compound contains 9.81 g of Zn, 1.8xx10^(23) atoms of chromium, and 0.60 mol of oxygen atoms. What is the simplest formula?

    A
    (a)`ZnCr_(2)O_(7)`
    B
    (b)`ZnCr_(2)O_(4)`
    C
    (c )`ZnCrO_(4)`
    D
    (d)`ZnCrO_(6)`

  • Empirical formula is the simplest formula of the compound which gives the atomic ratio of various elements present in one molecule of the compound. However, the molecular formula of the compound gives the number of atoms of various elements present in one molecule of the compound. Molecular formula=(Empirical formula) xxn n=("Molecular mass")/("Empirical formula mass") A compound may have same empirical and molecular formulae Both these formulae are calculated by using percentage composition of constituent elements. A compound of Na, C and O contains 0.0887 mol Na, 0.132 mol O and 2.65xx10^(22) atoms of carbon . the empirical formula of the compound is:

    A
    `NaCO`
    B
    `Na_(3)C_(5)O_(2)`
    C
    `Na_(2)CO_(3)`
    D
    `Na_(0.0887)C_(2.65xx10^(22))O_(0.132)`

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