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5 January, 14:30

An oxide of nitrogen was found by elemental analysis tocontain

30.4% nitrogen and 69.6% oxygen. If 23.0 g of thisgas were found to

occupy 5.6 L at STP, what are the empirical andmolecular formulas

for this oxide of nitrogen?

+2
Answers (1)
  1. 5 January, 17:54
    0
    Emperical formula: NO2

    Molecular formula: N2O4

    Explanation:

    Emperical formula

    N. O

    % by mass. 30.4. 69.6

    mass fraction. 30.4/14 = 2.17

    69.6/12 = 4.35

    2.17/2.17 = 1 4.35/2.17 = 2

    Emperical formula is NO2

    Molecular formula

    (NO2) n = molecular weight (M) of the oxide

    M = mRT/PV

    At STP, P = 1 atm, T = 273K

    M = 23*0.082057*273/1*5.6 = 92g/mol

    (NO2) n = 92

    (14+32) n = 92

    46n = 92

    n = 92/46 = 2

    Molecular formula = (NO2) 2 = N2O4
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