Question

A 0.77 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?

Answer 1

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

$Mass of Nitrogen = \frac{0.77}{14} = 0.055$

$Mass of chlorine = \frac{5.84}{35.5} = 0.165$

Each value is divided by small number

$\frac{0.055}{0.055} =1 ; \frac{0.165}{0.055} = 3$

$N_{1} Cl_{3}$

Therefore, empirical formula is $NCl_{3}$

Answer 2

Answer:

The empirical formula of the nitrogen chloride is $2NCl_{3}$

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:

$N_{2}$ + 3$Cl_{2}$ ->  $2NCl_{3}$

Step 2: We calculate the number of moles that are used from $N_{2}$ and the number of moles that we get from  $2NCl_{3}$.

moles of $N_{2}$ = 0,77mg $N_{2}$ * 1g/(1000mg) * 1/(28 g $N_{2}$/mol $N_{2}$)

moles of $N_{2}$ = 2,75 * 10^(-5)

moles of  $2NCl_{3}$ = 6,615mg 2NCl3 * 1g/(1000mg) * 1/(28 g  $2NCl_{3}$/mol  $2NCl_{3}$)

moles of  $2NCl_{3}$ = 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 $N_{2}$ produces 2 moles of  $2NCl_{3}$.

2.75 * 10^(-5) moles of $N_{2}$ = (5.5*10^(-5) moles of  $2NCl_{3}$) / 2

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

Have a nice day!