Question

2) A 0.77 mg sample of nitrogen gas reacts with chlorine gas to form 6.61 mg of a nitrogen

Answer 1

Answer:

NCl₃

Explanation:

From the question given above, the following data were obtained:

Mass of nitrogen (N) = 0.77 mg

Mass of chlorine (Cl) = 6.61 mg

Empirical formula =?

The empirical formula of the compound can be obtained as follow:

N = 0.77 mg

Cl = 6.61 mg

Divide by their molar mass

N = 0.77 / 14 = 0.055

Cl = 6.61 / 35.5 = 0.186

Divide by the smallest

N = 0.055 / 0.055 = 1

Cl = 0.186 /0.055 = 3

Therefore, the empirical formula of the compound is NCl₃