A 2.91 gram sample of a gaseous compound that contains only boron and hydrogen has a volume of 1.22 L at 25°C and 1.09 atm. What is the formula of this compound? (Atomic Weight: H = 1.01, B = 10.8 amu)
I am not sure but it is B6H10? But I am nto sure if that way i did it anyways was right.
Can someone verify what they get. That would be great
Use PV = nRT to determine the number of mols, n.
Then n = g/molar mass will give you the molar mass.
From there it's guess work unless I've missed something.
BH3 is the basic compound.
B2H6
B4H10
B5H9
B5H11
B6H10
B10H14
are the compounds between B and H I can find. Pick the one that comes the closest to the molar mass. Unless I missed the arithmetic I would go with B4H10.
Check my thinking.
Well, it seems like you're on the right track! Let's do some calculations to verify the formula.
First, let's determine the number of moles of the compound using the ideal gas equation: PV = nRT, where P is the pressure (1.09 atm), V is the volume (1.22 L), n is the number of moles, R is the ideal gas constant (0.08206 L⋅atm/(mol⋅K)), and T is the temperature in Kelvin (25°C = 298 K).
n = (PV) / (RT)
= (1.09 atm * 1.22 L) / (0.08206 L⋅atm/(mol⋅K) * 298 K)
≈ 0.0614 mol
Next, let's find the moles of boron and hydrogen in the compound. We know the atomic weights of boron (10.8 amu) and hydrogen (1.01 amu).
Moles of boron = (molar mass of boron) / (total molar mass of the compound) * n
= (10.8 amu) / (10.8 amu + (1.01 amu * 10)) * 0.0614 mol
≈ 0.005 mol
Moles of hydrogen = (molar mass of hydrogen) / (total molar mass of the compound) * n * (number of hydrogen atoms in the compound)
= (1.01 amu) / (10.8 amu + (1.01 amu * 10)) * 0.0614 mol * 10
≈ 0.0574 mol
So, based on these calculations, the molar ratio of boron to hydrogen is approximately 0.005:0.0574, which simplifies to 1:11.4. Therefore, the formula of the compound is likely B1H11 or BH11.
To determine the formula of the compound, we can use the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
First, let's convert the given mass of the compound to moles using the molar mass of the compound.
Molar mass of B6H10 = (6 × molar mass of B) + (10 × molar mass of H)
= (6 × 10.8 amu) + (10 × 1.01 amu)
= 64.8 amu + 10.10 amu
= 74.9 amu
Number of moles = mass of sample (g) / molar mass (amu)
= 2.91 g / 74.9 amu
= 0.03883 moles
Now we can use the ideal gas law equation to calculate the number of moles (n):
PV = nRT
(1.09 atm) × (1.22 L) = n × (0.0821 L.atm/(mol.K)) × (25°C + 273.15 K)
1.3338 atm.L = n × 62.36 L.atm/(mol.K)
n = 0.02139 moles
Comparing the calculated number of moles (0.02139 moles) with the number of moles obtained from the molar mass of B6H10 (0.03883 moles), we see that the two values do not match. Therefore, the given compound is not B6H10.
To determine the correct formula, we can calculate the empirical formula.
The empirical formula represents the simplest whole number ratio between the elements in a compound.
Let's assume the empirical formula is BxHy, where x and y are the subscripts for boron and hydrogen, respectively.
From the given information, we know that:
mass of boron (B) = 10.8 amu
mass of hydrogen (H) = 1.01 amu
In BxHy, the mass of boron (10.8x amu) and the mass of hydrogen (1.01y amu) should add up to the total mass of the compound (2.91 g).
Therefore, the equation becomes:
(10.8x) + (1.01y) = 2.91
We need another equation to solve for x and y. Let's use the ideal gas law equation to obtain another equation. Rearranging the equation PV = nRT, we get:
n = PV / RT
For the compound BxHy:
n = (1.09 atm) × (1.22 L) / (0.0821 L.atm/(mol.K)) × (25°C + 273.15 K)
n = 0.0494 moles
This means the number of moles obtained from the ideal gas law is also equal to x + y:
x + y = 0.0494
Now we have two equations:
(10.8x) + (1.01y) = 2.91 (equation 1)
x + y = 0.0494 (equation 2)
Solving these equations simultaneously will give us the values of x and y, which correspond to the subscripts of boron and hydrogen in the empirical formula.
Using equation 2, we can solve for y:
y = 0.0494 - x
Substituting this value of y into equation 1, we get:
(10.8x) + (1.01(0.0494 - x)) = 2.91
10.8x + 0.0494 - 1.01x = 2.91
9.79x = 2.8606
x = 0.2928
Substituting this value of x back into equation 2, we can solve for y:
0.2928 + y = 0.0494
y = 0.0494 - 0.2928
y = -0.2434
Since we cannot have a negative subscript, these calculations indicate that the empirical formula cannot be BxHy.
Therefore, there might be an error in the initial assumption that the compound contains only boron and hydrogen. The given compound may have additional elements present.
Without further information, it is not possible to determine the correct formula of the compound.
To determine the formula of the compound, you can use the ideal gas law and the molar mass of the elements involved.
First, calculate the number of moles of the compound using the ideal gas law equation:
PV = nRT
Where:
P = pressure in atm (1.09 atm)
V = volume in liters (1.22 L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin (25°C + 273 = 298 K)
Solving for n:
n = (PV) / (RT) = (1.09 atm * 1.22 L) / (0.0821 L.atm/mol.K * 298 K) ≈ 0.0542 mol
Next, calculate the molar mass of the compound. We know that the sample of the compound weighs 2.91 grams, so using the molar mass, we can determine the number of moles in the sample:
molar mass = mass / moles
molar mass = 2.91 g / 0.0542 mol ≈ 53.69 g/mol
Now, we can determine the empirical formula of the compound. To do this, we divide the molar mass by the atomic masses of boron and hydrogen (10.8 amu and 1.01 amu, respectively):
Molar ratio of B: (10.8 amu / 53.69 g/mol) ≈ 0.201
Molar ratio of H: (1.01 amu / 53.69 g/mol) ≈ 0.019
Dividing the molar ratio by the smallest one (0.019), we get approximately:
B: 0.201 / ~0.019 ≈ 10
H: 0.019 / ~0.019 ≈ 1
So, the empirical formula of the compound is BH10.
We can further check the empirical formula by comparing it to the given formula "B6H10". The empirical formula BH10 is the simplest ratio of atoms, but B6H10 has a higher ratio of boron to hydrogen. Therefore, the formula B6H10 is not correct.
Hence, the formula of the compound is BH10.