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.

%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.

To determine the molecular formula of the compound containing boron and hydrogen, we need to analyze the information from the experiments.

Let's start by calculating the number of moles of B2O3 and water produced in the first experiment.

1. Calculate the number of moles of B2O3:

Mass of B2O3 = Mass of compound burned - Mass of water produced
Mass of B2O3 = 0.492 g - 0.540 g
Mass of B2O3 = -0.048 g (mass of B2O3 cannot be negative, there seems to be an error in the given values)

Please provide the correct value for the mass of the compound burned.

Note: The given data seems to be inconsistent because the mass of the compound burned should be greater than the mass of the water produced.

To determine the molecular formula of the compound, we need to use the information from both experiments.

First Experiment:
In the first experiment, the compound is burned in pure oxygen to produce B2O3 and water. From the given data, we know that 0.492 g of the compound produces 0.540 g of water.

1. Calculate the moles of water produced:
Using the molar mass of water (H2O), which is 18.015 g/mol, we can calculate the number of moles of water produced:
moles of water = mass of water / molar mass of water
moles of water = 0.540 g / 18.015 g/mol

2. Calculate the moles of boron trioxide (B2O3) produced:
Since the molecular formula of boron trioxide is B2O3, we can conclude that the same number of moles of boron (B) reacts with the water to produce B2O3. Therefore, the moles of boron trioxide produced are equal to the moles of water produced.

3. Calculate the moles of boron:
Since B2O3 has a molecular weight of 69.62 g/mol, the number of moles of boron is calculated as follows:
moles of boron = moles of B2O3 = moles of water

So, we have found the number of moles of boron in the compound from the first experiment.

Second Experiment:
In the second experiment, we can use the ideal gas law to find the number of moles of the compound.

1. Convert the mass of the sample to grams:
63.1 mg = 0.0631 g

2. Convert the volume of the flask to liters:
120.0 mL = 120.0 / 1000 = 0.1200 L

3. Convert the temperature to Kelvin:
23 degrees Celsius = 23 + 273 = 296 Kelvin

4. Convert the pressure to atmospheres:
98.6 mmHg = 98.6 / 760 = 0.1297 atm

5. Use the ideal gas law equation: PV = nRT
n = (PV) / (RT)
where P is the pressure, V is the volume, R is the ideal gas constant (0.0821 L atm/mol K), and T is the temperature in Kelvin.

Now, substitute the given values into the equation:
n = (0.1297 atm * 0.1200 L) / (0.0821 L atm/mol K * 296 K)

Once we calculate the value of "n" from the second experiment, we will have the number of moles of the compound.

Comparison and Determining Molecular Formula:
Now, compare the number of moles of boron from the first experiment with the number of moles of the compound from the second experiment.

The ratio of moles of boron to moles of the compound will give us the subscripts needed for the molecular formula. If the ratio is not a whole number, multiply both subscripts by a constant to get the smallest whole numbers.

Let's say the number of moles of boron from the first experiment is x, and the number of moles of the compound from the second experiment is y.

If the ratio is x:y, the molecular formula would be BxHy.

Therefore, the molecular formula of the compound can be determined by comparing the number of moles of boron from the first experiment with the number of moles of the compound from the second experiment.