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?

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

Once the empirical formula is established, the molecular formula can be determined by comparing the empirical formula mass with the compound’s actual average molar mass. The molecular formula is an integer multiple of the empirical formula, ensuring that the calculated molar mass aligns with the experimental molar mass.

Combustion Analysis

This is an analytical technique used to determine the elemental composition of a compound by burning it completely in oxygen. By measuring the masses of the combustion products (such as CO2 and H2O), the amounts of carbon and hydrogen can be deduced. Often, additional measurements (like ammonia production) are used to determine the amounts of other elements (such as nitrogen), making combustion analysis a powerful method for elemental analysis.

Empirical Formula Determination

The empirical formula represents the simplest whole-number ratio of atoms in a compound. It is determined by converting the masses of each element to moles and then normalizing these amounts to reveal the smallest ratio between the different elements in the substance.

Stoichiometry and Mass-to-Mole Conversions

Stoichiometry involves converting measured masses of substances into moles using their molar masses. This conversion is essential in determining the proportions of elements present in the compound, which is a crucial step in obtaining both the empirical and molecular formulas.

Transcript

00:01 In this question i want to determine the empirical formula of the given amino.

00:06 So what we are going to do, the first task is to determine the carbon.

00:10 Remember, we know this to be an amino acid, so it contains h, c, n, and oxygen.

00:16 So the source of oxygen, rather, the way in which we are going to be calculating these.

00:21 First of all look at carbon.

00:23 We're going to calculate the amount of carbon using co2 and we're going to look into nitrogen using information of carbon.

00:31 Of ammonia and we are going to look into hydrogen using h2o.

00:36 So once we have determined hc and o, we can then use, rather hc and n, we can then use the difference because we know 100 % should be equal to all of these.

00:48 So once we say we get all of these rather the hc and n, we can say a hundred percent minus the sum of these to determine that of oxygen.

00:59 Now first of all, let's just calculate.

01:01 The mass of carbon in co2.

01:03 We're looking into the mass of carbon.

01:07 That is mass of carbon, this is going to be equal to the number of moles divided by the molar mass.

01:12 Remember, number of moles is equal to mass divided by the molar mass.

01:18 So if we are looking for the mass, this is going to be the number of moles multiplied by the molar mass.

01:23 We are getting these number of moles from the given compounds.

01:27 That is, for carbon, we are going to be looking into the number of moles of carbon dioxide which is going to be equal to 3 .94 divided by the molar mass of carbon dioxide which is 44 these are the number of moles and since our carbon and co2 are in the ratio of 1 is to 1 the number of more of co2 is also equal to the number of carbon so these are the number of moles of carbon multiplied by the molar mass the molar mass of carbon which is 12 so the mass of carbon this is 1 .07 5 grams.

02:04 Now we're going to apply a similar approach.

02:08 Let's just deal with hydrogen.

02:09 The mass of hydrogen this is going to be equal to the mass of h2o divided by the molar mass of h2o multiplied by 2 because h2o the ratio of hydrogen to h2 we have 1 is 2.

02:28 So we multiply this by 2 and we multiply by the mr the model mass of hydrogen which is 1 so what we have here is 0 .0 0 .214 grams of hydrogen...

Please give Ace some feedback