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
[tex]Mass of Nitrogen = \frac{0.77}{14} = 0.055[/tex]
[tex]Mass of chlorine = \frac{5.84}{35.5} = 0.165[/tex]
Each value is divided by small number
[tex]\frac{0.055}{0.055} =1 ; \frac{0.165}{0.055} = 3[/tex]
[tex]N_{1} Cl_{3}[/tex]
Therefore, empirical formula is [tex]NCl_{3}[/tex]
Answer:
The empirical formula of the nitrogen chloride is [tex]2NCl_{3}[/tex]
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:
[tex]N_{2}[/tex] + 3[tex]Cl_{2}[/tex] -> [tex]2NCl_{3}[/tex]
Step 2: We calculate the number of moles that are used from [tex]N_{2}[/tex] and the number of moles that we get from [tex]2NCl_{3}[/tex].
moles of [tex]N_{2}[/tex] = 0,77mg [tex]N_{2}[/tex] * 1g/(1000mg) * 1/(28 g [tex]N_{2}[/tex]/mol [tex]N_{2}[/tex])
moles of [tex]N_{2}[/tex] = 2,75 * 10^(-5)
moles of [tex]2NCl_{3}[/tex] = 6,615mg 2NCl3 * 1g/(1000mg) * 1/(28 g [tex]2NCl_{3}[/tex]/mol [tex]2NCl_{3}[/tex])
moles of [tex]2NCl_{3}[/tex] = 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 [tex]N_{2}[/tex] produces 2 moles of [tex]2NCl_{3}[/tex].
2.75 * 10^(-5) moles of [tex]N_{2}[/tex] = (5.5*10^(-5) moles of [tex]2NCl_{3}[/tex]) / 2
2,75 * 10^(-5) moles = 2,75 * 10^(-5) moles
Have a nice day!
