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.

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%

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation:

b.

the accuracy of the data.