Question

Calculate the mass of tetraborane (B4 H10) that contains a billion (1.0 x 10 *9) boron atoms. Be sure your answer has a unit symbol if necessary, and round it to significant digigits.

Answer 1

Answer:

a billion boron atoms are contained in a mass m= 2.21*10⁻¹⁴ gr of tetraborane

Explanation:

Since there are an Avogadro number (N= 6.023*10²³ molecules/mole) of molecules of tetraborane in a mole of tetraborane , and per each molecule there are 4 atoms of B . Then

number of moles n = a/(4*N)

where a= number of atoms

the corresponding mass of tetraborane m is

m= n*M

where M= molecular weight of tetraborane = 53.32 g/mol

then

m= n*M = a*M/(4*N)

replacing values

m= a*M/(4*N)= 1.0*10⁹ atoms of B * 53.32 gr/mole /( 4 atoms of B/molecule* 6.023*10²³ molecules/mole ) = 2.21*10⁻¹⁴ gr

m= 2.21*10⁻¹⁴ gr

Answer 2

The mass of tetraborane (B₄H₁₀) that contains a billion (1×10⁹) boron atoms is 2.21×10¯¹⁴ g

We'll begin by obtaining the number of mole of boron in 1 mole of B₄H₁₀. This can be obtained as follow:

1 mole of B₄H₁₀ = 53.32 g

Thus,

53.32 g of B₄H₁₀ contains 4 moles of B.

• Next, we shall determine the number of atoms in 4 moles of B.

From Avogadro's hypothesis,

1 mole of B = 6.02×10²³ atoms

Therefore,

4 moles of B = 4 × 6.02×10²³

4 moles of B = 2.408×10²⁴ atoms

Thus, we can say that:

2.408×10²⁴ atoms of boron is present in 53.32 g of B₄H₁₀.

• Finally, we shall determine the mass of B₄H₁₀ that contains 1×10⁹ atoms of boron. This can be obtained as follow:

2.408×10²⁴ atoms of boron is present in 53.32 g of B₄H₁₀.

Therefore,

1×10⁹ atoms of boron will be present in = $\frac{1*10^{9} * 53.32}{2.408*10^{24}}$ = 2.21×10¯¹⁴ g of B₄H₁₀

Therefore, the mass of tetraborane (B₄H₁₀) that contains a billion (1×10⁹) boron atoms is 2.21×10¯¹⁴ g

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