A0.77 mg sample of nitrogen reacts with chlorine to form 6.615 mg of the chloride. part a what is the empirical formula of the nitrogen chloride?

From the given:

Reacted mass of nitrogen = 0.77 mg

Mass of chlorine = 6.615 mg

Reacted mass of chlorine = 6.615 mg - 0.77 = 5.84 mg

Each value is divided by small number

Therefore, empirical formula is

The empirical formula of the nitrogen chloride is

Explanation:

Step 1: Form the formation equation of nitrogen chloride. The N and Cl are diatomic molecules, so they have subindices 2. We balance the chemical equation remaining:

+ 3 ->  

Step 2: We calculate the number of moles that are used from and the number of moles that we get from  .

moles of = 0,77mg * 1g/(1000mg) * 1/(28 g /mol )

moles of = 2,75 * 10^(-5)

moles of   = 6,615mg 2NCl3 * 1g/(1000mg) * 1/(28 g  /mol  )

moles of   = 5,5*10^(-5)

Step 3: We verify that the number of moles formed is correct using the reaction balance coefficients, taking into account that 1 mol of produces 2 moles of  .

2.75 * 10^(-5) moles of = (5.5*10^(-5) moles of  ) / 2

2,75 * 10^(-5) moles = 2,75 * 10^(-5) moles

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