Answer:
The empirical formula of the compound is C8H14ClN5
C = 44.5%
H = 6.6 %
Cl = 16.4%
N = 32.5%
Explanation:
An herbicide is found to contain only C, H, N, and Cl. The complete combustion of a 100.0-mg sample of the herbicide in excess oxygen produces 83.16 mL of CO2 and 73.30 mL of H2O vapor at STP.
A separate analysis shows that the sample also contains 16.44 mg of Cl. Calculate its empirical formula. Determine the percentage of the composition of the substance.
Step 1: Data given
Mass of the sample = 100.0 mg
Volume of CO2 = 83.16 mL
Volule of H2O = 73.30 mL
A separate analysis shows that the sample also contains 16.44 mg of Cl.
Step 2: The balanced equation
CwHxNyClz + O2 ⇒ CO2 + H2O
Step 3: Calculate moles of CO2
Moles CO2 = (p*V)/(R*T)
Moles CO2 = (1*0.08316)/(0.08206*273)
Moles CO2 = 0.003712 moles
Step 4: Calculate moles of H2O
Moles H2O = (1*0.07330)/(0.08206*273)
Moles H2O = 0.003272
Step 5: Calculate moles of Cl
Moles Cl = mass Cl / molar mass CL
Moles Cl = 0.01644 grams / 35.45 g/mol
Moles Cl = 4.64 *10^-4 moles
Step 6: Calculate moles C
CO2 contains 1 mol C
For 0.003712 mol of CO2 we have 0.003712 mol of C
Step 7: Calculate moles of H
H2O contains 2 mole of H
For 0.00372 moles of H2O we have 2*0.003272 = 0.006544 mol H
Step 8: Calculate mass of C and H
0.003712 mol C *12.01g/mol = 0.04458g C
0.006545 mol H *1.01g/mol = 0.006610g H
Step 9: Calculate mass of N
0.1 g - (0.04458g C+ 0.006610g H+0.01644g Cl)= 0.03237g N
moles N = 0.03237 / 14.00 g/mol = 0.002312
Step 10: Calculate mol ratio
C: 0.003712 / 0.000464 = 8
H: 0.006545/ 0.000464 = 14
Cl: 0.000464 / 0.000464 = 1
N: 0.002312 / 0.000464 = 5
The empirical formula of the compound is C8H14ClN5
The molar mass is 215.59 g/mol
Step 11: Determine the percentage of the composition of the substance.
C = ((8*12)/215.59) *100% = 44.5%
H = ((14*1.01)/215.59)*100% = 6.6 %
Cl = (35.45 / 215.59)*100% = 16.4%
N = ((5*14)/215.59) * 100% = 32.5%
