A 24.00 g sample contains 14.60 g Cl and 9.400 g B. What is the percent composition (by mass) of boron in this sample?

The percent composition of boron in this sample is 39.17 %.

The mass of chlorine = 14.60 g

The mass of Boron = 9.400g

Mass of the sample = 24.00 g

The percentage composition of boron can be calculated below:

% composition of Boron = 9.4 / 24 × 100

% composition of Boron = 940 / 24

% composition of Boron = 39.1666666667

% composition of Boron ≈ 39. 17%

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39.2 %

Explanation:

The following data were obtained from the question:

Mass of sample = 24 g

Mass of Cl = 14.6 g

Mass of B = 9.4 g

Percentage composition of boron =?

The percentage composition (by mass) of boron in the sample can be obtained as illustrated below:

Percentage composition of boron = mass of B /mass of sample × 100

Percentage composition of boron = 9.4/24 × 100

Percentage composition of boron = 39.2 %

Therefore, the percentage composition (by mass) of boron in the sample is 39.2 %

the electronic configuration of an atom with eight electrons is 1s²2s²2p⁴. the orbital diagrams are shown in the figure.

as per the hunds rule:

1. electrons tend to avoid being in the same orbital as per as practicable.

2. electrons in different orbitals of the same energy will have their spin parallel.

this can be illustrated from the multiplicity of the orbitals of the atom. as there are four electrons in the p orbital and p orbital has 3 sub-shells. the electrons can be fed in different ways as shown in the figure but the correct orbital diagram will be as per the hunds rule.

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