Chemistry

A compound containing boron and hydrogen has been synthesized. You perform two experiments to determine the molecular formula of the compound. A 0.492 g sample of the compound is burned in pure oxygen to produce B2O3 and 0.540 g of water. In the second experiment, when a 63.1 mg sample of the compound was placed in a 120.0 mL flask at 23 degrees C, the pressure was 98.6 mmHg. Determine the molecular formula of the compound.

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asked by Anna
  1. %H in the BxHy = (g H/mass sample)*100
    g H atoms (0.540 x 2/18)= about 0.0600
    and %H atoms = (0.06/0.492)*100 = about 12% but you should be more accurate than that on all calculations that follow. Since the only other element is B, then B must be 100%-12% = about 88%

    Now take 100 g sample that gives you about
    88 g B
    12 g H
    Convert to mols.
    88/atomic mass B = about 8
    12/atomic mass H = about 12
    Determine ratio which you can see is 8/8 = 1
    12/8 = 1.5
    so ratio is B2H3 and empirical formula mass is about 24.6

    Now what's the molar mass.
    PV = nRT
    n = PV/RT and I obtained approx 0.00064 (again you should confirm all of these more accurately).
    mols = g/molar mass and molar mass = g/mols = 0.0631/0.00064 = about 98

    Molar mass is about 98; empirical mass is about 24.6 so 98/24.6 = 3.98 which rounds to 4. I suspect when you do the math more accurately it will be much closer to 4.00. So the formula is
    (B2H3)4 or B8H12.

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