Answer:
The empirical formula is C3H12N2Cl2
Explanation:
Step 1: Data given
When 1.00 g of X is dissolved in water and allowed to react with excess AgNO3, all the chlorine (present as chloride) precipitates and 1.95 g of AgCl is collected.
When 1.00 g of X undergoes complete combustion,0.900 g of CO2 and 0.735 g of H2O are formed.
Atomic mass of C = 12.01 g/mol
Atomic mass of H = 1.01 g/mol
Atomic mass of N = 14.01 g/mol
Atomic mass of Cl = 35.45 g/mol
Atomic mass of Ag = 107.87 g/mol
Step 2: Calculate moles of Cl
Moles AgCl = Mass AgCl / molar AgCl
Moles AgCl = 1.95 grams / 143.32 g/mol
Moles AgCl = 0.0136 moles
For 1 mol AgCl we have 1 mol Cl
Moles Cl = <u>0.0136 moles Cl</u>
Mass Cl = 0.0136 moles * 35.45 g/mol = <u>0.482 grams Cl</u>
Step 3: Calculate moles CO2
1 g reacts with O2 to produce 0.9 g CO2
Moles CO2 = 0.9 grams / 44.01 g/mol = 0.02045 moles
For 1 mol CO2 we have 1 mol C
Moles C = <u>0.02045 moles C</u>
Mass C = 0.02045 moles *12.01 g/mol = <u>0.2456 grams C</u>
Step 4: Calculate moles H
Moles H2O = 0.735 grams / 18.02 g/mol
Moles H2O = 0.0408 moles H2O
In 1 mol H2O we have 2 moles H
Moles H = 2*0.0408 = <u>0.0816 moles H</u>
Mass H = <u>0.0824 grams H</u>
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Step 5: Calculate mass N
Mass N = 1.00 grams - 0.482 grams Cl - 0.2456 grams C -0.0824 grams H
Mass N = 0.190 grams N
Step 6: Calculate moles N
Moles N = 0.190 grams N /14.01 g/mol
Moles N = <u>0.0136 moles N</u>
Step 7: Calculate mol ratio
We divide by the smallest amount of moles
C: 0.02045 moles / 0.0136 moles = 1.5
H: 0.0816 moles / 0.0136 moles = 6
N: 0.0136 moles / 0.0136 moles = 1
Cl: 0.0136 moles/ 0.0136 moles = 1
The empirical formula is C3H12N2Cl2