Question

Calculate the number of boron in a 130.0 g sample of tetraborane (B4H10)Be sure your answer has a unit symbol if necessary, and rounded to four significant digits.

Answer 1

in a 130.0 g sample of tetraborane (B4H10), there are approximately 9.752 × 10²³ boron atoms.

Boron (B) atomic mass ≈ 10.81 g/mol

Hydrogen (H) atomic mass ≈ 1.01 g/mol

Molar mass of tetraborane (
`Br_(4) H_(10)`) = (4 × Boron atomic mass) + (10 × Hydrogen atomic mass)

Molar mass of tetraborane = (4 × 10.81 g/mol) + (10 × 1.01 g/mol)

Molar mass of tetraborane ≈ 43.24 g/mol + 10.1 g/mol

Molar mass of tetraborane ≈ 53.34 g/mol

Calculate the number of moles of tetraborane in 130.0 g:

Number of moles = Mass / Molar mass

Number of moles = 130.0 g / 53.34 g/mol

Number of moles ≈ 2.438 moles

Use the ratio of boron atoms per molecule of tetraborane:

From the chemical formula B4H10, there are 4 boron atoms per molecule of tetraborane.

Number of boron atoms = Number of moles of tetraborane × Number of boron atoms per mole

Number of boron atoms = 2.438 moles × 4

Number of boron atoms ≈ 9.752

Therefore, in a 130.0 g sample of tetraborane (B4H10), there are approximately 9.752 × 10²³ boron atoms.

Answer 2

First, we will need the atomic weight of the compounds:

B=10.8110 u

H=1.00784 u

Now, we will calculate the molecular weight of the molecule B4H10:

`\begin{gathered} B\colon\text{ 10.8110 x 4 = 43.244} \\ H\colon\text{ 1.00784 x 10 = 10.0784} \\ \text{Total weight = 43.244 + 10.0784 = 53.3224 g/mol} \end{gathered}`

Now that we have the molecular weight we proceed to calculate the number of molecules present in 130 g of sample:

`\text{Mol of B4H10 = }\frac{130\text{ g}}{53.3224\text{ g/mol}}\text{ = 2.4380 mol}`

So, we have 2.4380 mol of B4H4 in 130 g of sample, each molecule has 4 atoms of boron, which means that we will have:

2.4380 x 4 = 9.7520 atoms of B

The answer will be: 9.7520 atoms of B, here we have the answer with 4 significant digits.