Hello! Need a little help for a chem MCQ :/
In an experiment, one such compound, X, containing nitrogen, oxygen and hydrogen was extracted from a fertiliser. Upon burning 26.6g of compound X, 14.7g of dinitrogen oxide, N2O and 12.0g of water were obtained as the only products. What is the empirical formula of Compound X?
A NH2O
B N2H2O2
C N2H4O3
D NH4O3
Originally posted by HAHAHjoo:Hello! Need a little help for a chem MCQ :/
In an experiment, one such compound, X, containing nitrogen, oxygen and hydrogen was extracted from a fertiliser. Upon burning 26.6g of compound X, 14.7g of dinitrogen oxide, N2O and 12.0g of water were obtained as the only products. What is the empirical formula of Compound X?
A NH2O
B N2H2O2
C N2H4O3
D NH4O3
Moles of N in N2O generated = moles of N in sample of X.
Moles of H in H2O generated = moles of H in sample of X.
Find sample mass of N and H in sample of X.
Total sample mass of X (ie. 26.6g) - (sample mass of N and H) = sample mass of O
From sample mass of O, find moles of O in sample of X.
From ratio of moles of N, H and O, determine empirical formula of X.
A 0.10 mol dm–3 solution of NH2CH2CH2NH2 has a pH of 11.3. When 30 cm3 of 0.10 mol dm–3 HCl is added to 10 cm3 of a 0.10 mol dm–3 solution of NH2CH2CH2NH2, the final pH is 1.6.
Using the following axes, sketch the pH changes that occur during this addition of
HCl(aq).
ANS: Since it is a diprotic base there will be two curves. But how to calculate the end-point?
Originally posted by hoay:A 0.10 mol dm–3 solution of NH2CH2CH2NH2 has a pH of 11.3. When 30 cm3 of 0.10 mol dm–3 HCl is added to 10 cm3 of a 0.10 mol dm–3 solution of NH2CH2CH2NH2, the final pH is 1.6.
Using the following axes, sketch the pH changes that occur during this addition of
HCl(aq).ANS: Since it is a diprotic base there will be two curves. But how to calculate the end-point?
Data given only allows the student to calculate Kb1, but not Kb2. If Kb2 is not given, the pH for the 1st and 2nd equivalence points cannot be determined.
The final pH of 1.6 is due primarily to the excess HCl, and not due to the acidity of N2H6 2+, since the former acid is much stronger than the latter.
Since the question says "sketch" and since Kb2 is not given, the student will not be required to determine the exact pH of the 1st and 2nd equivalence points.
If Kb2 was given, then we can calculate both Ka values (ie. of NH2CH2CH2NH3+ and +H3NCH2CH2NH3+, aka protonated ethylenediamine and diprotonated ethylenediamine).
Only if we have both Kb (or therefore, both Ka) values, then we can calculate the pH of the 1st and 2nd equivalence points as follows :
1st equivalence point : only 1 amphiprotic species present, use pH = 1/2 (pKa1 + pKa2)
2nd equivalence point : only 1 acidic species present, use Ka of diprotonated ethylenediamine.
Amines are basic since the lone pair of nitrogen is available for donation. Amides are not neutral because the lone pair of N is delocalized over the C and O atoms.
What about Nitriles e.g CH3-CN.?? the lone pair is present....Are they basic?
secondly, i am curious to know is basicity has anything to do with hybridization? Because in methylamine the C is sp3 hybridized, in amide it sp2 and in -CN it sp hybridized.
Originally posted by hoay:Amines are basic since the lone pair of nitrogen is available for donation. Amides are not neutral because the lone pair of N is delocalized over the C and O atoms.
What about Nitriles e.g CH3-CN.?? the lone pair is present....Are they basic?
secondly, i am curious to know is basicity has anything to do with hybridization? Because in methylamine the C is sp3 hybridized, in amide it sp2 and in -CN it sp hybridized.
You're correct. There are two distinct factors at work here : resonance, and % s-orbital character.
Resonance
If the lone pair is delocalized by resonance (through the sideways overlapping of p orbitals), such as in the case of amides and phenylamine, the lone pair will accordingly be less available to accept a proton.
Percentage s-orbital character
Since s-orbitals are located closer to the nucleus compared to p-orbitals, when a lone pair occupies an orbital with greater % s-orbital character (eg. sp hybridized orbital in nitriles > sp2 hybridized orbital in amides > sp3 hybridized orbital in amines), the lone pair is accordingly more strongly held back electrostatically by the positively charged nucleus, and consequently has a reduced basicity as well as nucleophilicity.
Amides are non-basic for both reasons, while nitriles are non-basic due to the % s-orbital character factor.
Based on the factor of % s-orbital character, nitriles are even less basic compared to amides. However, amides have the additional factor of resonance, and thus both amides and nitriles may be considered approximately equally non-basic overall.