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.

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Lynne Trussell · Accepted Answer

Step 1 Mass of chlorine Find the mass of chlorine. =mass of compound−mass of nitrogen =6.61mg−0.77mg =5.84mg Cl 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. 0.77mg N1g N1mol N1000mg N14.0067g N =5.4974×10−5mol N 5.84mg Cl1g Cl1mol Cl1000mg Cl35.453g Cl =1.6473×10−4mol Cl 5.4974×10−5mol N5.4974×10−5mol=1 N 1.6473×10−4mol Cl5.4974×10−5mol=3 Cl Therefore, the empirical formula is NCl3

Debbie Moore · Accepted Answer

Step 1The 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&#039;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=(6.61−0.77)mg=5.84mgStep 2The molar mass of N and Cl are 14.01 and 35.45gmol, respectively. So, the number of mol of N (nN) and Cl(nCl) present in 0.77 mg of N and 5.84 mg of Cl, respectively, are calculated as shown below:nN=0.77mg14.01gmol=7.7×10−4g14.01gmol; (1mg=10−3g)=5.5×10−5molnCl=5.84mg35.45gmol=5.84×10−3g35.45gmol; (1mg=10−3g)=1.6×10−4molStep 3Hence, the simplest integral ratio between nN and nCl is calculated as shown below:nN:nCl=5.5×10−5mol:1.6×10−4mol=1:1.6×10−45.5×10−5=1:2.9≈1:3As 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 NCl3

karton · Accepted Answer

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 =0.7714=0.055 Mass of chlorine =5.8435.5=0.165 Each value is divided by small number 0,0550.055=1 0.1650.055=3 Therefore, empirical formula is NCl3

A 0.77-mg sample of nitrogen reacts with chlorine to form

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