Question

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.

Answer 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.