Can anyone help me solve this?

  • Thread starter PhysicsinCalifornia
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In summary, Dimethylglyoxime (DMG) is a compound with a molecular mass of 116.12 u. In a combustion analysis, a 2.464-g sample of DMG yields 3.735 g CO2 and 1.530 g H2O. In a separate experiment, the nitrogen in a 1.868 g sample of DMG is converted to NH3(g) and neutralized by passing it into 50.00 mL of 0.3600 M H2SO4(aq). The excess H2SO4(aq) then requires 18.63 mL of 0.2050 M NaOH(aq) for neutralization. Using these data, the most plausible
  • #1
PhysicsinCalifornia
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Dimethylglyoxime (DMG) is a carbon-hydrogen-nitrogen-oxygen compound with a molecular mass of 116.12 u. In a combustion analysis, a 2.464-g sample of DMG yields 3.735 g CO2 and 1.530 g H2O. In a separate experiment, the nitrogen in a 1.868 g sample of DMG is converted to NH3(g), and the NH3 is neutralized by passing it into 50.00 mL of 0.3600 M H2SO4(aq). After neutralization of the NH3, the excess H2SO4(aq) requires 18.63 mL of 0.2050 M NaOH(aq) for its neutralization. Using these data, determine for dimethylglyoxime the most plausible Lewis structure

I don't understand the neutralization part. Can anyone give me clues on how to solve this?
 
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  • #2
This is probably Kjeldahl's analysis for nitrogen content. The NH3 is converted to NH4+, you know the total number moles of H2SO4 that was initially present, it was consumed with the reaction with NH3, the rest that had not reacted is titrated with the NaOH. From this you should be able to deduce the moles of NH3, and thus nitrogen, was present.
 
  • #3
from the final neutralisation between NaOH and H2SO4, you should know homework many moles of H2SO4(excess) reacted with NaOH by using M1V1=M2V2 equation. then you deduct this number of moles of H2SO4 with the initial H2SO4, you will get the number of moles H2SO4 that reacted with NH3. From this you can tell the molesof NH3 that exist in the DMG bond. Thats all i know.
 

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