Question

Calculate the number of hydrogen atoms in a 110.0 sample of tetraborane(B4H10) . Be sure your answer has a unit symbol if necessary, and round it to 4 significant digits.

Answer 1

To calculate the number of hydrogen atoms in a 110.0 g sample of tetraborane (B4H10), you must find the number of moles of the compound in the sample and then multiply that by Avogadro's number and the number of hydrogen atoms per mole. The final result is approximately 1.242 × 10² hydrogen atoms.

Step-by-step explanation:

To calculate the number of hydrogen atoms in a 110.0 g sample of tetraborane (B4H10), we first need to determine the molar mass of tetraborane. The molar mass of boron (B) is 10.81 g/mol and the molar mass of hydrogen (H) is 1.01 g/mol. Since tetraborane has 4 boron atoms and 10 hydrogen atoms, its molar mass is calculated as follows:

• (4 × 10.81 g/mol) + (10 × 1.01 g/mol) = 43.24 g/mol + 10.10 g/mol = 53.34 g/mol.

Next, we calculate the number of moles of tetraborane in the 110.0 g sample:

Number of moles = Mass of sample / Molar mass = 110.0 g / 53.34 g/mol ≈ 2.0616 moles.

Now, since each mole of tetraborane has 10 hydrogen atoms, the total number of hydrogen atoms can be found by multiplying the number of moles of tetraborane by Avogadro's number (6.022 × 1023 atoms/mol), and then by 10 (the number of hydrogen atoms per mole of tetraborane):

A number of hydrogen atoms = 2.0616 moles × 6.022 × 1023 atoms/mol × 10 ≈ 1.2416 × 1025 atoms.

Rounded to four significant digits, the number of hydrogen atoms is 1.242 × 1025 atoms.

Answer 2

`1.242 * 10^(25)\text{ atoms H}`

Step-by-step explanation:

You must convert the mass of B₄H₁₀ to moles of B₄H₁₀, then to molecules of B₄H₁₀, and finally to atoms of H.

1. Moles of B₄H₁₀

`\text{Moles of B$_(4)$H}_(10) = \text{110.0 g B$_(4)$H}_(10) * \frac{\text{1 mol B$_(4)$H}_(10)}{\text{53.32 g B$_(4)$H}_(10)} = \text{2.063 mol B$_(4)$H}_(10)`

2. Molecules of B₄H₁₀

`\text{No. of molecules} = \text{2.063 mol B$_(4)$H}_(10) * \frac{6.022 * 10^(23)\text{ molecules B$_(4)$H}_(10)}{\text{1 mol B$_(4)$H}_(10)}\\\\=1.242 * 10^(24)\text{ molecules B$_(4)$H}_(10)`

3. Atoms of H

`\text{Atoms of H} = 1.242 * 10^(24)\text{ molecules B$_(4)$H}_(10) * \frac{\text{10 atoms H}}{\text{1 molecule B$_(4)$H}_(10)}\\\\= \mathbf{1.242 * 10^(25)}\textbf{ atoms H}`