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?
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 BxH3x
To determine the molecular formula of the compound, we need to find the molar mass of the compound with empirical formula BH3. Then, we can compare it to the molar mass of boron (B) to determine the molecular formula.
Let's start by calculating the molar mass of the empirical formula BH3. The molar mass of boron (B) is approximately 10.81 g/mol, and the molar mass of hydrogen (H) is approximately 1.01 g/mol.
The empirical formula BH3 contains one boron atom and three hydrogen atoms. Therefore, the molar mass of BH3 can be calculated as follows:
Molar mass of BH3 = (Atomic mass of B) + 3 x (Atomic mass of H)
= (10.81 g/mol) + 3 x (1.01 g/mol)
= 10.81 g/mol + 3.03 g/mol
= 13.84 g/mol
The molar mass of the empirical formula BH3 is approximately 13.84 g/mol.
Now, we can determine the number of empirical formula units present in the given mass of the sample. The mass of the sample is given as 0.138 g.
Number of empirical formula units = Mass of sample / Molar mass of empirical formula
= 0.138 g / 13.84 g/mol
≈ 0.00997 mol
Next, we can use the ideal gas law equation, PV = nRT, to calculate the number of moles of the gas in the sample. We are given the pressure (P) as 88 mm Hg and the volume (V) as 1.05 L. The gas constant (R) is 0.0821 L·atm/(mol·K), and the temperature (T) is 295 K.
PV = nRT
Solving for n (number of moles):
n = PV / RT
= (88 mm Hg) x (1 atm / 760 mm Hg) x (1.05 L) / (0.0821 L·atm/(mol·K) x 295 K)
≈ 0.00347 mol
The calculated number of moles of the gas from the ideal gas law is approximately 0.00347 mol.
Comparing the two values obtained, we see that the number of empirical formula units is greater than the number of moles obtained from the ideal gas law. This suggests that there are multiple empirical formula units in the molecular formula.
To find the molecular formula, we divide the number of empirical formula units by the number of moles obtained:
Number of empirical formula units / Number of moles = (0.00997 mol) / (0.00347 mol)
≈ 2.87
The number 2.87 is approximately 3. Therefore, the molecular formula of the compound is three times the empirical formula BH3, which is B3H9.
Hence, the molecular formula of the compound is B3H9.