 Charles Kingsley

2022-01-03

A 0.77-mg sample of nitrogen reacts with chlorine to form 6.61 mg of the chloride. Determine the empirical formula of nitrogen chloride. Lynne Trussell

Step 1
Mass of chlorine
Find the mass of chlorine.
$=\text{mass of compound}-\text{mass of nitrogen}$
$=6.61mg-0.77mg$

Step 2
Convert the masses of these elements to moles using molar masses. Then determine the number of moles of each element relative to the other elements. Divide all quantities by the smallest amount of moles.

Therefore, the empirical formula is $NC{l}_{3}$ Debbie Moore

Step 1
The empirical formula of a species is determined by knowing the mass of all the constituting atoms. From the mass of an atom, the corresponding number of moles is calculated. After that, taking the simplest integral ratio between the number of moles of atoms, the empirical formula is determined.
Nitrogen chloride contains nitrogen (N) and chlorine (Cl) as constituting atoms.
Let's assume the given reaction goes to completion. Hence, the mass (M) of Cl present in 6.61 mg of nitrogen chloride is calculated as shown below:
$M=\left(6.61-0.77\right)mg$
$=5.84mg$
Step 2
The molar mass of N and Cl are 14.01 and $35.45\frac{g}{mol}$, respectively. So, the number of mol of N $\left({n}_{N}\right)$ and $Cl\left({n}_{Cl}\right)$ present in 0.77 mg of N and 5.84 mg of Cl, respectively, are calculated as shown below:
${n}_{N}=\frac{0.77mg}{14.01\frac{g}{mol}}$

$=5.5×{10}^{-5}mol$
${n}_{Cl}=\frac{5.84mg}{35.45\frac{g}{mol}}$

$=1.6×{10}^{-4}mol$
Step 3
Hence, the simplest integral ratio between ${n}_{N}$ and ${n}_{Cl}$ is calculated as shown below:
${n}_{N}:{n}_{Cl}$
$=5.5×{10}^{-5}mol:1.6×{10}^{-4}mol$
$=1:\frac{1.6×{10}^{-4}}{5.5×{10}^{-5}}$
$=1:2.9$
$\approx 1:3$
As the simplest integral ratio between the number of moles of constituting atoms represent the corresponding empirical formula, therefore, the empirical formula of nitrogen chloride is $NC{l}_{3}$ karton

Step 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
$\frac{0,055}{0.055}=1$
$\frac{0.165}{0.055}=3$
Therefore, empirical formula is $NC{l}_{3}$

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