Asked by mel
compund containing only C, H, and N yeilds the following data-
1. complete combustion of 34.3 mg of the compund producedmg of CO2 and 40.3 mg of H2O
2. a 65.2mg sample was analyzed by the dumas mthod, giving 35.6 mL of N2 at 740 torr and 25C
3. effusion rate as a gas was 24.6mL/min. effusion rate of argon gas under same conditions is 26.4 mL/min
what is the molecular formula of compound?
1. complete combustion of 34.3 mg of the compund producedmg of CO2 and 40.3 mg of H2O
2. a 65.2mg sample was analyzed by the dumas mthod, giving 35.6 mL of N2 at 740 torr and 25C
3. effusion rate as a gas was 24.6mL/min. effusion rate of argon gas under same conditions is 26.4 mL/min
what is the molecular formula of compound?
Answers
Answered by
DrBob222
You didn't type all the data.
1. I would first convert mg CO2 and mg H2O to percent CO2 and percent H2O in the 34.3 mg sample. I'm a little confused by the statement that "complete" combustion produced....... ; otherwise, the difference of 100% - %CO2 - %H2O = %N.
Convert percents so obtained to mols C, H, and N and obtain the ratio. This should give you the empirical formula.
3. (rate cmpd/rate argon)= sqrt [(molar mass argon/molar mass cmpd)]
1. I would first convert mg CO2 and mg H2O to percent CO2 and percent H2O in the 34.3 mg sample. I'm a little confused by the statement that "complete" combustion produced....... ; otherwise, the difference of 100% - %CO2 - %H2O = %N.
Convert percents so obtained to mols C, H, and N and obtain the ratio. This should give you the empirical formula.
3. (rate cmpd/rate argon)= sqrt [(molar mass argon/molar mass cmpd)]
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