I would like to ask about a problem that I am stuck on:

An oligopeptide is made up of glycine, alanine, and valine. Hydrolysis of X in $\pu{500 mL}$ of $\pu{1 M}$ $\ce{H2SO4}$ solution yields solution Y, condensation of solution Y yields mixture Z containing dipeptides, tripeptides, tetrapeptides, pentapeptides and corresponding amino acids. Burn half of mixture Z with a sufficient amount of air, absorb the combustion product into the excess $\ce{Ba(OH)2}$ solution, and it is seen that the weight of the tank increases by $74.225$ grams, the weight of the solution decreases by $161.19$ grams and at the same time $139.608$ liters (measured at STP) of inert gas has escaped. Allow solution Y to completely react with V liters of heated $\pu{2 M}$ $\ce{KOH}$ solution (used 20% excess compared to the required amount), evaporate the solution after reaction, the solid mass has a value of $m$. Find $m$. (Given $\ce{H} = 1$, $\ce{C} = 12$, $\ce{N} = 14$, $\ce{O} = 16$, $\ce{S} = 32$, $\ce{K} = 39$, $\ce{Ba} = 137$)

So far, I have only worked out that there are $\pu{1.195 mol}$ of $\ce{CO2}$ and $\pu{1.2025 mol}$ of $\ce{H2O}$ in the combustion reaction by utilizing the law of conservation. The weight increase of the tank is the weight of $\pu{CO2}$ and $\pu{H2O}$ combined: $$m_{\text{increase}} = m_{\ce{CO2}} + m_{\ce{H2O}} = \pu{74.225 g}$$ And, the weight decrease of the solution is due to the formation of $\ce{BaCO3}$ $$m_{\text{decrease}} = m_{\ce{BaCO3}} - (m_{\ce{CO2}} + m_{\ce{H2O}}) = \pu{161.19 g}$$ By substitution, I am able to find that $m_{\ce{BaCO3}} = 235.415$ which leads to the amount $$n_{\ce{BaCO3}} = \frac{235.415}{197} = 1.195 \Rightarrow n_{\ce{CO2}} = \pu{1.195 mol}$$ (conservation of the element $\ce{C}$)
After that, finding the amount of water is fairly simple. However, after this step I am stuck and unable to progress through this question. I would like to receive some help to solve this question.

  • $\begingroup$ There is no solution Y left to react with KOH, as it was already condensed to make solution Z. so you have m = 112 * V g KOH. :-P $\endgroup$
    – Poutnik
    Sep 1, 2023 at 15:30

2 Answers 2


This is a fairly simple peptide hydrolysis problem. The inert gas is nitrogen: $$n_{N_2}=\frac{139.608}{22.4}=6.2325 \text{ mol}$$ We can easily see that this nitrogen also contains the nitrogen that was already in the air.
Let's consider mixture Z, we will reduce its composition to the following "compounds": $\ce{C_2H_3ON, CH_2, H_2O}$
Let $n_{\ce{C2H3ON}}=2x, n_{\ce{CH2}}=2y, n_{\ce{H_2O}}=2z$
Half of mixture Z will have: $n_{\ce{C2H3ON}}=x, n_{\ce{CH2}}=y, n_{\ce{H_2O}}=z$
Then, the escaped nitrogen will be: $$n_{\ce{N_2}}=n_{\ce{N_2}(air)}+n_{\ce{N_2}(rxn)}=n_{\ce{N_2}(air)}+0.5x=6.2325 \text{ mol}$$ Now, let's consider the combustion reaction, by looking at the elements, we can easily formulate the following system of equations:
$$ \left\{ \begin{array}{ll} 2x+y&=1.195 \text{ (C atoms)}\\ 3x+2y+2z&=2\cdot1.2025 \text{ (H atoms)}\\ x+z+0.5(6.2325-0.5x)&=1.195\cdot2+1.2025 \text{ (O atoms)} \end{array} \right. $$ Solving this simple system gives us $x=0.375, y= 0.445, z=0.195$
Then, the composition of mixture Z will be: $$ \left\{ \begin{array}{} \ce{C2H3ON}: 0.75 \text{ mol} \\ \ce{CH2}: 0.89 \text{ mol} \\ \ce{H2O}: 0.39 \text{ mol} \end{array} \right. $$ Mixture Z is the important part in solution Y as it is the one that'll react with the potassium hydroxide solution. Let Z fully react with the $\ce{KOH}$ solution: $$\ce{Z +KOH->Z' +H2O}$$ (Z' is the salt mixture)
By conservation of mass, we can easily formuate: $$m_{\ce{Z'}}=m_{\ce{Z}}+m_\ce{KOH}-m_\ce{H2O}$$ We know that: $n_\ce{KOH}=n_\ce{Z}=n_\ce{C2H3ON}=0.75;\\ n_\ce{H2O}=0.39$
And: $m_\ce{Z}=m_\ce{C2H3ON}+m_\ce{CH2}+m_\ce{H2O}=57\cdot0.75+14\cdot0.89+18\cdot0.39=62.23 \text{ grams}$
Substituting all of them into our original mass equation, we have: $$ m_\ce{Z'}=62.23+0.75\cdot56-0.39\cdot18=97.21 \text{ grams} $$ Remember that when solution Y reacts with the $\ce{KOH}$ solution, not only does the Z part reacts, but the "acidic" part also reacts with it. What this means, is that the $\ce{KOH}$ here not only serves a part in creating the salt mixture, it also "neutralizes" the original acid. Therefore, we have to take the existence of the sulfuric acid into account:
$$ n_{\ce{KOH}(\text{total})}=n_{\ce{KOH}(\ce{Z})}+n_{\ce{KOH}(\text{acid})} = 0.75 + 2n_{\ce{H2SO4}} =0.75+2\cdot(500\cdot10^{-3}\cdot1) =1.75 \text{ mol} $$ This then leads us to the amount of excess $\ce{KOH}$: $n_{\ce{KOH}\text{(xs)}}=1.75\cdot0.2=0.35 \text{ mol}$
So, the solid mass contains Z', $\text{0.5 mol } \ce{K2SO4}, \text{0.35 mol }\ce{KOH} $
Hence, $m=97.21+0.5\cdot174+0.35\cdot56=203.81 \text{ grams}$.

  • $\begingroup$ Thank you saviour !! $\endgroup$ Sep 4, 2023 at 12:57

To go forward, the amount of inert gas produced by the combustion must be taken into account. This inert gas is probably nitrogen $\ce{N2}$.

Burning these peptides produce $139.6$ L $\ce{N2}$, or $n$($\ce{N2}$) = $\frac{pV}{RT}$ = $\frac{101325·0.1396}{8.314·273}$ = $6.21$ mol $\ce{N2}$. This amount of $\ce{N2}$ was produced by one half of the hydrolyzed $\ce{Z}$ solution. The total amount of amino acid should have burned in producing twice as much, or $12.42$ moles $\ce{N2}$. As every amino acid has $1$ nitrogen atom, the total numbers of amino acids producing $12.42$ mole $\ce{N2}$ is the double, and is equal to $24.84$ mol amino-acid. As a conséquence, the initial peptide contained $24.84$ mol amino-acid.

There should be something wrong here, because it is not possible to burn $24.84$ mol of any amino-acid, and getting only $1.195$ mole $\ce{CO2}$. Where is the mistake ?

  • $\begingroup$ I think that the inert gas includes the nitrogen from the reaction and the nitrogen that was already in the air because mixture Z was burned in air which has roughly 80% nitrogen. $\endgroup$ Sep 2, 2023 at 9:14

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