Boron forms a gaseous compound with hydrogen with empirical formula BH3. a 1.05L sample of this compound at a pressure of 88 mm Hg at 295K has a mass of .138g. what is the molecular formula of the compound?

Answer 1

Use PV = nRT to solve for n = number of moles BH3, then n = grams/molar mass. You know n and grams, solve for molar mass.
X*empirical formula mass = molar mass
Solve for X and substitute into
(BH3)_x so the molecular formula will be B_xH_3x