One of the hydrides of boron has molar mass 122.22 g mol^-1 and is 11.55% by mass hydrogen. How do you determine its molecular formula?

1 Answer
Jun 10, 2017

Well, we assume that we have a #100*g# mass of the stuff........and we gets a molecular formula of #B_10H_14#

Explanation:

Well, we assume that we have a #100*g# mass of the stuff........and then proceed to calculate the #"empirical and molecular formulae"#.

And thus #"moles of boron"=((100-11.55)*g)/(10.81*g*mol^-1)=8.18*mol.#

And #"moles of hydrogen"=(11.55*g)/(1.00794*g*mol^-1)=11.46*mol.#

We divide thru by the lowest molar quantity to get an trial empirical formula of #BH_(1.4)#. But because, the empirical formula is by definition the SIMPLEST WHOLE NUMBER RATIO DEFINING CONSTITUENT ATOMS IN A SPECIES, we multiply the trial formula by #5# to gets #B_5H_7#.

And now we use the molecular mass to get the #"molecular formula"#, which is a whole number multiple of the #"empirical formula"#.

And thus we solve for #n# in the expression:

#nxx(5xx10.81+7xx1.00794)*g*mol^-1=122.22*g*mol^-1#

#n=(122.22*g*mol^-1)/((5xx10.81+7xx1.00794)*g*mol^-1)=2#

And thus our molecular formula is #2xxB_5H_7=B_10H_14#, so-called #"decaborane"#.

I have used this a couple of times, and funnily enuff it is said that the compounds has a chocolately odour. I cannot verify this odour because it smelled pretty foul.......