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Chemistry Combustion Sample Calculation

A 32.2 mg sample of the organic compound from problem 1 was treated to combustion analysis which yielded 59.2 mg CO2 and 12.3 mg of H2O. Analysis by low resolution mass spectrometry yielded a molar mass of 191 g/mol for the pure
compound. Further analysis showed that the compound contained only carbon, hydrogen, oxygen, and chlorine. What is the molecular formula of the compound?

I solved problem 1, which was "A 132.2 mg sample of a pure organic compound containing only chlorine as the halogen was treated in this manner. The
aqueous layer required 13.63 mL of 0.1015 M AgNO3 to reach the argentometric endpoint. What is the percentage chlorine by mass in the organic compound?" and got an answer of 37.1% Cl, which im pretty sure is right.

Here's what I did so far....

12.3mg H2O/18.01528mg x (1 O/1 H2O) = 0.6827mmol O
12.3mg H2O/18.01528mg x (2 H/ 1 H2O) = 1.3655mmol H
59.23mg CO2/44.0098mg x (1 C/1 CO2= 1.345mmol C

...divide them all by the lowest mole ratio= C2H2O

now i know there has to be a Cl in there somewhere, and there's also a stoicheometric ratio of 6 from (191mg/mmol)/(32.2mg)=6, bringing the formula to C12H12O6. this still doesn't have a Cl in it and its MW is already 252. I'm not sure if I'm supposed to factor in the O contribution from the CO2 as well, or how the Cl is factored into the equation and still getting a MW of 191...?

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Solution Preview

What you know:

You have 32.2 mg of material
You have 37.2 % Cl
You have an overall molar mass of 191
You get 59.2 mg CO2 and 12.3 H2O from combustion

37.2% of 191 = 71 / 35.5 (mass of Cl) ...

Solution Summary

Chemistry combustion sample calculations are examined in the solution.