Final answer:
To determine the empirical formula, we find the ratio of the number of moles of each element in the compound. The empirical formula for the compound is BH3. To determine the molecular formula, we divide the molar mass of the compound by the molar mass of BH3. The molecular formula of the compound is B3H9.
Step-by-step explanation:
The empirical formula of a compound is the simplest, most reduced ratio of the elements present in the compound. To determine the empirical formula, we must find the ratio of the number of moles of each element in the compound. Given that the compound is composed of boron and hydrogen, and that 0.025 g of the compound produces 0.063 g of B2O3, we can determine the moles of each element.
First, we calculate the moles of B2O3 produced by dividing its mass by its molar mass. The molar mass of B2O3 is 69.62 g/mol (2 x 10.81 for B + 3 x 16.00 for O). Therefore, the moles of B2O3 produced is 0.063 g / 69.62 g/mol = 0.000904 mol.
Since 1 mol of B2O3 contains 2 mol of B, we can determine the moles of B by multiplying the moles of B2O3 by 2. Therefore, the moles of B is 0.000904 mol * 2 = 0.001808 mol.
Similarly, since 1 mol of B2O3 contains 3 mol of O, we can determine the moles of O by multiplying the moles of B2O3 by 3. Therefore, the moles of O is 0.000904 mol * 3 = 0.002712 mol.
Next, we calculate the moles of H by subtracting the moles of B from the total moles of the compound. The moles of H is 0.002712 mol - 0.001808 mol = 0.000904 mol.
Finally, we divide the number of moles of each element by the smallest number of moles to obtain the simplest whole-number ratio. In this case, the smallest number of moles is 0.000904 mol. Dividing the moles of each element by 0.000904 mol gives us the ratio: B: 2 / 0.000904 = 2.21, H: 0.000904 / 0.000904 = 1.
Since we cannot have a non-integer ratio, we round the ratio to the nearest whole number to obtain the empirical formula. The empirical formula of the compound is therefore BH3.
To determine the molecular formula, we need to know the molar mass of the compound. Given that the molar mass is approximately 28 amu, we can compare it to the empirical formula. The molar mass of BH3 is approximately 11 amu (1 x 10.81 for B + 3 x 1.01 for H). Since the molar mass of the compound is larger than the molar mass of BH3, the molecular formula must have more than one BH3 unit.
To find the molecular formula, we divide the molar mass of the compound by the molar mass of BH3. The result is 28 amu / 11 amu = 2.55.
Rounding this ratio to the nearest whole number gives us the number of BH3 units in the compound, which is 3. Therefore, the molecular formula of the compound is B3H9.