Barium and nitrogen form two binary compounds containing 90.745% and 93.634% barium, respectively. determine the empirical formulas of these two compounds.

BaN, Ba₃N₂

Explanation:

In order to determine the empirical formula of a compound, we have to follow a series of steps.

Step 1: Determine the percent composition

Step 2: Divide each percentage by the atomic mass of each element

Step 3: Divide all numbers by the smallest one

Step 4: If necessary, multiply by a number so the coefficients are whole

Compound 1

Step 1

Ba: 90.745%

N: 100% - 90.745% = 9.255%

Step 2

Ba: 90.745/137.33 = 0.66078

N: 9.255/14.01 = 0.6606

Step 3

Ba: 0.66078/0.6606 ≈ 1

N: 0.6606/0.6606 = 1

The empirical formula is BaN.

Compound 2

Step 1

Ba: 93.634%

N: 100% - 93.634% = 6.366%

Step 2

Ba: 93.634/137.33 = 0.68182

N: 6.366/14.01 = 0.4544

Step 3

Ba: 0.68182/0.4544 = 1.5

N: 0.4544/0.4544 = 1

Step 4

Ba: 1.5 × 2 = 3

N: 1 × 2 = 2

The empirical formula is Ba₃N₂.

answer: option (b) is the correct answer.

explanation:

a covalent bond is formed by sharing of electrons. whereas polarity arises due to difference in electronegativity of two atoms.

therefore, more is the electronegativity of an atom which combines with the titanium, more readily it will induce a partial positive and negative charge.

thus, out of the given options fluorine is most electronegative and has an incomplete octet. therefore, it will readily combine with titanium to share the electrons and complete its octet.  

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