Lysine, an essential amino acid in the human body, contains C, H, O, and N . In one experiment,

Lysine, an essential amino acid in the human body, contains...

Lysine, an essential amino acid in the human body, contains $\mathrm{C}, \mathrm{H}, \mathrm{O},$ and $\mathrm{N} .$ In one experiment, the complete combustion of $2.175 \mathrm{g}$ of lysine gave $3.94 \mathrm{g}$ $\mathrm{CO}_{2}$ and $1.89\ \mathrm{g}\ \mathrm{H}_{2} \mathrm{O} .$ In a separate experiment, $1.873 \mathrm{g}$ of lysine gave $0.436 \mathrm{g} \mathrm{NH}_{3} .$ (a) Calculate the empirical formula of lysine. (b) The approximate molar mass of lysine is $150 \mathrm{g} .$ What is the molecular formula of the compound?

Key Concepts

Molecular Formula Determination

Molecular formula determination uses the empirical formula in conjunction with the compound’s molar mass to establish the actual number of atoms of each element present in the molecule. By comparing the empirical formula mass to the experimentally determined molar mass, one can calculate a multiplication factor that scales the empirical formula to yield the molecular formula, providing the complete formula of the compound.

Nitrogen Determination in Combustion Analysis

When a compound contains nitrogen, additional steps must be taken during combustion analysis because nitrogen does not typically form an easy-to-analyze oxide like carbon or an element like hydrogen. Instead, nitrogen may be converted into a compound such as ammonia or nitrogen oxide, requiring separate quantification to accurately determine the nitrogen content in the original molecule.

Empirical Formula Determination

Empirical formula determination involves calculating the simplest whole number ratio of the atoms in a compound based on experimental data. After converting the masses of the constituent elements to moles, the ratio is simplified to the lowest terms, providing the empirical formula, which may be different from the molecular formula but reflects the fundamental composition of the compound.

Combustion Analysis

Combustion analysis is a technique used to determine the elemental composition of a compound by burning it in excess oxygen. From the masses of the combustion products, such as carbon dioxide and water, one can deduce the amounts of carbon and hydrogen, respectively, that were present in the original compound. This method is fundamental in analytical chemistry for determining the qualitative and quantitative aspects of the compound's makeup.

Transcript

00:03 Given information for part a, 2 .175 gram of lysine, gave 3 .94 gram of co2 and 1 .89 gram of h2.

00:56 Present in 3 .94 gram is equal to 3 .94 divided by 444 mole is equal to 0 .0895 if all the carbon has been converted into carbon dioxide then the number of moles of carbon present in lysine will also be the number of more of carbon present in lysine will also be 0 .0895 therefore, mass of carbon in lysine and therefore mass of carbon in lysine is equal to 0 .0895 multiply 12 is equal to 1 .074 gram.

03:22 Mass percentage of carbon in 2 .175 gram of lysine is equal to we have 1 .074 divided by 2 .175 multiply 100 is equal to 49 .9.

04:11 37 % and number of malls of water present in 1 .89 gram is equal to 1 .89 divided by 18 mall is equal to 0 .10.

05:07 5 mole.

05:18 Each water molecules contains 2 hydrogens.

05:23 Then the number of mole of hydrogen present in lysine will be number of moles of, numbers of molecules of hydrogen present in lysine will be to multiply 0.

06:02 105 mall is equal to 0 .211 mole.

06:15 Therefore, mass of hydrogen in lysine is equal to 0 .0 .21 multiply 1 is equal to 0 .21 gram.

06:53 And mass percentage of hydrogen in 2 .175 gram lysine is equal to 0 .1 .21 divided by 2 .175 multiply 100 is equal to 9 .6.

07:44 Percent mass percent of hydrogen present in lysine is 9 .6 percent and next is 1 .873 gram of lysine gave 0 .436 gram of nh3 so number of of ammonia present in 0 .436 divided by 17 mall is equal to 0 .0256 mall.

09:14 If all the nitrogen has been converted into ammonia, then the number of moles of nitrogen present, then the number of mores of nitrogen present.

09:28 Number of moles of nitrogen present in lysine will be 0 .0256 mole.

10:04 Therefore, mass of nitrogen in lysine is equal to 0 .0256 mole.

10:25 Multiply 14 is equal to 0 .358 gram mass of nitrogen present in lysine is 0 .358 gram and mass percentage of nitrogen nitrogen in 1 .873 gram of lysine is equal to 0 .358 divided by 1 .8, 1 .873 multiply 100 is equal to 19 .17%...

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