1. Find the empirical formula:

* Assume a 100 g sample: This makes the percentages directly into grams.

* Convert grams to moles:

* Moles of Boron (B): 78.14 g / 10.81 g/mol (molar mass of B) = 7.23 mol

* Moles of Hydrogen (H): 21.86 g / 1.01 g/mol (molar mass of H) = 21.64 mol

* Find the simplest whole-number ratio: Divide both mole values by the smaller one (7.23 mol):

* B: 7.23 mol / 7.23 mol = 1

* H: 21.64 mol / 7.23 mol ≈ 3

* Empirical formula: BH₃

2. Determine the molar mass:

* Use the ideal gas law (PV = nRT):

* P = 1.12 atm

* V = 74.3 mL = 0.0743 L

* R = 0.0821 L⋅atm/(mol⋅K)

* T = 27°C + 273.15 = 300.15 K

* Solve for n (number of moles):

* n = (PV) / (RT) = (1.12 atm * 0.0743 L) / (0.0821 L⋅atm/(mol⋅K) * 300.15 K) ≈ 0.0034 mol

* Calculate molar mass:

* Molar mass = mass / moles = 0.0934 g / 0.0034 mol ≈ 27.5 g/mol

3. Determine the molecular formula:

* Find the ratio of the molar mass to the empirical formula mass:

* Empirical formula mass of BH₃ = 10.81 g/mol + 3(1.01 g/mol) = 13.84 g/mol

* Ratio = 27.5 g/mol / 13.84 g/mol ≈ 2

* Multiply the subscripts in the empirical formula by the ratio:

* Molecular formula: (BH₃)₂ = B₂H₆

Therefore, the molecular formula of the gas compound is B₂H₆.