0.596g of a gaseous compound containing only boron and hydrogen occupies 484 mL at NTP. When the compound was ignited in excess of oxygen, all its hydrogen was recovered as 1.17g of `H_(2)O` and all the boron was present as `B_(2)O_(3)`. Find the molecular formula of the compound (B=10.8, H=1)

Explanation

To find the molecular formula of the compound containing boron and hydrogen, we will follow these steps:

### Step 1: Calculate the number of moles of the gaseous compound
Given:
- Mass of the compound = 0.596 g
- Volume of the compound at NTP = 484 mL = 0.484 L
- Molar volume at NTP = 22.4 L/mol

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