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Chapter 8: Problem 26

A sample containing the amino acid alanine, \(\mathrm{CH}_{3} \mathrm{CH}\left(\mathrm{NH}_{2}\right) \mathrm{COOH},\) plus inert matter is analyzed by the Kjeldahl method. A \(2.00-\mathrm{g}\) sample is digested, the \(\mathrm{NH}_{3}\) is distilled and collected in \(50.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4},\) and a volume of \(9.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\) is required for back-titration. Calculate the percent alanine in the sample.

Short Answer

Expert verified
The percent of alanine in the sample is 58.8%.

Step by step solution

01

Calculate moles of H2SO4 initially added

We start by calculating the moles of \(\mathrm{H}_2\mathrm{SO}_4\) initially added to the solution. Using the formula \(\text{{moles}} = \text{{concentration}}\times\text{{volume}},\) the concentration of \(\mathrm{H}_2\mathrm{SO}_4\) is \(0.150\,\mathrm{M}\) and the volume is \(50.0\,\mathrm{mL}\) or \(0.050\,\mathrm{L}\). Thus, the moles of \(\mathrm{H}_2\mathrm{SO}_4\)are:\[\text{{moles of }} \mathrm{H}_2\mathrm{SO}_4 = 0.150\,\mathrm{M} \times 0.050\,\mathrm{L} = 0.0075\,\text{{mol}}.\]
02

Calculate moles of NaOH used in back-titration

Next, we calculate the moles of \(\mathrm{NaOH}\) used for the back-titration bythe same formula. The concentration of \(\mathrm{NaOH}\) is \(0.100\,\mathrm{M}\) and the volumeis \(9.0\,\mathrm{mL}\) or \(0.009\,\mathrm{L}\). Thus, the moles of \\(\mathrm{NaOH}\)are:\[\text{{moles of }} \mathrm{NaOH} = 0.100\,\mathrm{M} \times 0.009\,\mathrm{L} = 0.0009\,\text{{mol}}.\]
03

Calculate moles of H2SO4 reacted with ammonia

The moles of \(\mathrm{NaOH}\) used in the back-titrationneutralize the excess \(\mathrm{H}_2\mathrm{SO}_4\). Therefore, to find out how much \(\mathrm{H}_2\mathrm{SO}_4\)reacted with the \(\mathrm{NH}_3\), we subtract the moles of \(\mathrm{NaOH}\) from the initial molesof \(\mathrm{H}_2\mathrm{SO}_4\):\[\text{{moles of }} \mathrm{H}_2\mathrm{SO}_4 \text{{ that reacted}} = 0.0075\,\text{{mol}} - 0.0009\,\text{{mol}} = 0.0066\,\text{{mol}}.\]
04

Relate moles of reacted H2SO4 to moles of NH3

Since each molecule of \(\mathrm{H}_2\mathrm{SO}_4\) can react with 2 molecules of \(\mathrm{NH}_3\), we find that the moles of \(\mathrm{NH}_3\)is twice the moles of \(\mathrm{H}_2\mathrm{SO}_4\)that reacted:\[\text{{moles of}}\,\mathrm{NH}_3 = 2 imes 0.0066\,\text{{mol}} = 0.0132\,\text{{mol}}.\]
05

Calculate mass of alanine from moles of NH3

The formula for alanine is \(\mathrm{C}_3\mathrm{H}_7\mathrm{NO}_2\),and its molar mass is approximately \(89.09\,\mathrm{g/mol}\). Each mole of \(\mathrm{NH}_3\) originates from \(1\,\text{{mol}}\,\text{{of}}\,\mathrm{C}_3\mathrm{H}_7\mathrm{NO}_2\).Thus, the mass of alanineis calculated as follows:\[\text{{mass of alanine}} = 0.0132\,\text{{mol}} \times 89.09\,\mathrm{g/mol} = 1.176\,\text{{g}}.\]
06

Calculate percent composition of alanine in the sample

Lastly, we determine the percentage of alanine in the original sample. The originalmass of the sample was \(2.00\,\mathrm{g}\) and we found \(1.176\,\mathrm{g}\) of alanine. The calculation is:\[\text{{percent alanine}} = \left(\frac{1.176\,\text{{g}}}{2.00\,\text{{g}}}\right) \times 100\%= 58.8\%.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Percent Composition
When we talk about percent composition, we're discussing what fraction of a mixture or compound a particular component makes up, expressed in percentage form. Imagine you have a fruit juice that is partly apple juice and partly other ingredients. If you know the mass of apple juice in the mix, you can determine what percentage of the juice is actually apple.
In the context of chemistry, like in the Kjeldahl method exercise, calculating the percent composition helps us understand how much of the sample is made up of a specific substance, such as alanine in this case.
  • Begin by finding the mass of the substance of interest in the entire mixture.
  • Next, take this mass and divide it by the total mass of your sample.
  • Multiply the result by 100 to convert the fraction to a percentage.
So, if your 2.00 g sample has 1.176 g of alanine, the percent of alanine is \[\left(\frac{1.176\,\text{g}}{2.00\,\text{g}}\right) \times 100\% = 58.8\%.\] This tells you that 58.8% of your mixture is alanine.
Titration
Titration is a scientific method used to determine the concentration of a particular substance in a solution. It's kind of like figuring out the exact amount of sugar you need in a cup of coffee to make it taste just right.
In the Kjeldahl method, titration plays a key role in analyzing nitrogen in compounds. Here's how it works:
  • A known volume of a reactant (e.g., an acid or base) with a known concentration is slowly added to a solution with an unknown concentration.
  • The goal is to reach an endpoint, which is when the reaction is exactly complete.
  • Indicators or pH meters are often used to detect this endpoint.
In our example, after capturing the ammonia, sulfuric acid (\(\text{H}_2\text{SO}_4\)) is used to bind the ammonia, and then sodium hydroxide (\(\text{NaOH}\)) is added in a back-titration to determine the amount of excess, unreacted acid. This helps us establish how much ammonia reacted, providing data needed to calculate the percent composition of alanine.
Mole Calculations
Understanding moles is like counting substances in a recipe using groups instead of single pieces. In chemistry, a mole is a way to count molecules or atoms.
Think of it as a chemist's dozen, where one mole is equal to Avogadro's number, approximately \(6.022 \times 10^{23}\) particles.
In the exercise, mole calculations are crucial for figuring out how much \(\text{NH}_3\) came from the alanine:
  • First, we calculate moles of substances like \(\text{H}_2\text{SO}_4\) based on their concentration (moles per liter) and volume (liters).
  • Moles of \(\text{NH}_3\) are determined using its relationship with \(\text{H}_2\text{SO}_4\), with each mole of \(\text{H}_2\text{SO}_4\) reacting with 2 moles of \(\text{NH}_3\).
  • Finally, by knowing moles of \(\text{NH}_3\), we can calculate moles and hence the mass of alanine, using its molar mass鈥攖his mass is necessary to determine its percent composition in the sample.
Thus, accurate mole calculations ensure you can unravel the puzzle of percentage composition in complex mixtures.

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Most popular questions from this chapter

The Kjeldahl procedure was used to analyze \(500 \mu \mathrm{L}\) of a solution containing \(50.0 \mathrm{mg}\) protein \(/ \mathrm{mL}\) solution. The liberated \(\mathrm{NH}_{3}\) was collected in \(5.00 \mathrm{~mL}\) of \(0.0300 \mathrm{M} \mathrm{HCl}\). The remaining acid required \(10.00 \mathrm{~mL} 0.010 \mathrm{M} \mathrm{NaOH}\) for complete titration. What is the weight percent of nitrogen in the protein?

Professor Alexander Scheeline, University of Illinois, has developed a spreadsheet Universal Acid Titrator (see the book's website, 8.52 Universal Acid Titrator.xlsx). You can type in the \(\mathrm{p} K_{a}\) value(s) for any acid (up to a decaprotic acid!) in row \(11,\) or \(\mathrm{p} K_{a}\) value(s) in row \(9,\) and generate a titration plot (pH vs. Volume base) for titration with strong, monoprotic base like \(\mathrm{NaOH}\). Delight Professor Scheeline by playing with the spreadsheet and generating different titration curves. (We recommend that when the higher K's are irrelevant, directly enter 0 for these \(K\) values in row \(11 .\) ) Rather than computing the \(\mathrm{pH}\) for a given amount of base added, this approach, also based on charge balance, computes how much base you have to add to get to a given pH. Here are the following challenge questions: (a) Why is it so much easier to directly compute the needed base amounts without involving solver or Goal Seek rather than computing the \(\mathrm{pH}\) for a given amount of base added? (b) Especially at low pH values why are some of the calculated base volumes negative? (They are never plotted.) (c) For any acid of your choice (or your creation), calculate the \(\mathrm{pH}\) when no base has been added \((\mathrm{V}=0)\) by using detailed calculation or Goal Seek or Solver. Then use the current spreadsheet to calculate how much base you will need to get to this pH. See if you indeed get zero. (d) For the more adventurous: After computing the titration curve with this spreadsheet (that ignores activity corrections), do a successive approximations calculation at each \(\mathrm{pH}\) that takes ionic strength and activity corrections into account, and compute a revised titration curve. Compare the activity corrected and uncorrected curves: where does activity correction matter most?

The \(\mathrm{p} K_{a}\) of finalic acid is 6.50 (a) What is the \(\mathrm{pH}\) of an equimolar solution of finalic acid and sodium finalate? Assume that both solutions are at \(0.1 M\) (b) Would you expect the buffer capacity to be higher or lower if you decrease the concentration of each constituent to \(0.01 M ?\) Why? (c) What is the effect of water dissociation on the \(\mathrm{pH}\) of these solutions? (d) Only considering the \(\mathrm{pH}\) of the solution, would it be safe to drink a \(0.1 \mathrm{M}\) solution of finalic acid? How about a \(0.01 M\) solution?

Sodium hydroxide and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) will titrate together to a phenolphthalein end point \(\left(\mathrm{OH}^{-} \rightarrow\right.\) \(\left.\mathrm{H}_{2} \mathrm{O} ; \mathrm{CO}_{3}^{2-} \rightarrow \mathrm{HCO}_{3}^{-}\right) .\) A mixture of \(\mathrm{NaOH}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is titrated with \(0.250 \mathrm{M} \mathrm{HCl}\) requiring \(26.2 \mathrm{~mL}\) for the phenolphthalein end point and an additional \(15.2 \mathrm{~mL}\) to reach the modified methyl orange end point. How many milligrams \(\mathrm{NaOH}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) are in the mixture?

(a) A \(0.9872 \mathrm{~g}\) sample of unknown potassium hydrogen phthalate (KHP) required \(28.23 \mathrm{~mL}\) of \(0.1037 M \mathrm{NaOH}\) for neutralization. What is the percentage of \(\mathrm{KHP}\) in the sample?
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