Lysine, an essential amino acid in the human body, contains C,H,O, and N. In one experiment, the complete combustion of 2.175g of lysine gave 3.94g CO2 and 1.89g H2O. In a separate experiment, 1.873 g of lysine gave 0.436 g NH3.

(a) Calculate the empirical formula of lysine

(b) The approximate molar mass of lysine is 150g. What is the moleclar formula of the compound?

For (a) I think I found the # of each element except H.

For element C would this be right:
3.94g CO2 / 12 g C = 0.328 C

We can only calculate the empirical formula, which is the simplest ratio between the number of atoms of the constituents. This ratio may or may not be a sub-multiple of the actual formula.

We will first decide on the processes.
The complete combustion means that the limiting reagent will be the carbon and hydrogen in lysine, while oxygen is in excess supply.

The second process is where nitrogen is the limiting reagent that produces ammonia.

We will calculate the percentage weight of C, H and N, while that of O will be calculated by difference from 100%.

Each percentage by weight will then be divided by the respective atomic mass to obtain the proportion of number of atoms. From there, we divide the HCF (approx.) of the numbers to obtain the ratio of number of atoms, in integers.

Element, mass, %mass, #atoms, integer #atoms
C: 3.94*12/44=1.075; 49.4%; 4.1167; 3
H: 1.89*2/18=0.2100; 9.655%; 9.579; 7
N: 0.436*14/17=0.3591; 19.17%; 1.3693; 1
O: -; (100-49.4-9.655-19.17)=21.78%; 1.3609; 1

From the last column, we conclude that the empirical formula is C3H7NO.

To calculate the empirical formula of lysine, we need to determine the ratio of each element present in the compound.

(a) Let's start with the calculation for carbon (C):

Mass of CO2 produced = 3.94 g
1 mole of CO2 contains 1 mole of carbon (C)
Molar mass of carbon (C) = 12 g/mol

Number of moles of carbon (C) = (3.94 g CO2) / (44.01 g/mol CO2) = 0.0895 mol CO2
Number of moles of carbon (C) = 0.0895 mol

Similarly, you can calculate the number of moles for hydrogen (H), oxygen (O), and nitrogen (N) using the given masses of CO2, H2O, and NH3.

For hydrogen (H):
Mass of H2O produced = 1.89 g
1 mole of H2O contains 2 moles of hydrogen (H)
Molar mass of hydrogen (H) = 1 g/mol

Number of moles of hydrogen (H) = (1.89 g H2O) / (18.015 g/mol H2O) = 0.1049 mol H2O
Number of moles of hydrogen (H) = 0.2098 mol

For oxygen (O):
The mass of oxygen can be calculated by subtracting the masses of carbon and hydrogen from the total mass of CO2 and H2O:

Mass of oxygen (O) = (3.94 g CO2 + 1.89 g H2O) - (0.328 g C + 0.2098 g H)
Mass of oxygen (O) = 4.4922 g - 0.5378 g
Mass of oxygen (O) = 3.9544 g

Molar mass of oxygen (O) = 16 g/mol

Number of moles of oxygen (O) = (3.9544 g O) / (16 g/mol) = 0.2465 mol

For nitrogen (N):
The mass of nitrogen can be calculated by subtracting the mass of carbon from the total mass of lysine and NH3:

Mass of nitrogen (N) = (1.873 g lysine) - (0.328 g C)
Mass of nitrogen (N) = 1.544 g

Molar mass of nitrogen (N) = 14 g/mol

Number of moles of nitrogen (N) = (1.544 g N) / (14 g/mol) = 0.1103 mol

Now, we have the number of moles for each element:

Carbon (C) = 0.0895 mol
Hydrogen (H) = 0.2098 mol
Oxygen (O) = 0.2465 mol
Nitrogen (N) = 0.1103 mol

The empirical formula of lysine can be determined by dividing the moles of each element by the smallest number of moles. In this case, the smallest number of moles is 0.0895 mol (from carbon).

Empirical formula:
C: 0.0895 mol / 0.0895 mol = 1
H: 0.2098 mol / 0.0895 mol = 2.34 (approx.)
O: 0.2465 mol / 0.0895 mol = 2.75 (approx.)
N: 0.1103 mol / 0.0895 mol = 1.23 (approx.)

Therefore, the empirical formula of lysine is C1H2.34O2.75N1.23.

(b) To determine the molecular formula of lysine, we need to know the molar mass of the compound. The approximate molar mass of lysine is given as 150 g.

Divide the molar mass of lysine by the molar mass of the empirical formula to find the number of empirical formula units present in the molecular formula:

Molar mass of lysine = 150 g/mol
Molar mass of empirical formula (C1H2.34O2.75N1.23) = (1 * 12) + (2.34 * 1) + (2.75 * 16) + (1.23 * 14) = 150.6 g/mol (approx.)

Number of empirical formula units in the molecular formula:
150 g / 150.6 g/mol = 0.995 (approx.)

Since the number of empirical formula units is close to 1, the molecular formula of lysine should be very similar to the empirical formula.

Therefore, the molecular formula of lysine is approximately C1H2.34O2.75N1.23.

To calculate the empirical formula of lysine, you need to determine the moles of each element present in the compound. Here's how you can do it:

(a) Calculating the number of moles for each element:

1. Carbon (C):
To find the moles of carbon, you need to convert the mass of carbon dioxide (CO2) to moles of carbon using the molar mass of carbon:
0.328 g C / 12 g/mol C = 0.0273 mol C

2. Hydrogen (H):
The information given does not directly provide the mass of hydrogen. However, you know from the compound's molecular formula that lysine contains hydrogen. To determine the moles of hydrogen, you need to use the information provided in the combustion of lysine:
From the combustion, you know that 1.89 g of water (H2O) is formed from 2.175 g of lysine. The difference in mass between the initial lysine and the water formed is due to the combustion of hydrogen present in lysine:
2.175 g - 1.89 g = 0.285 g of hydrogen

Next, convert the mass of hydrogen to moles using the molar mass of hydrogen:
0.285 g H / 1 g/mol H = 0.285 mol H

3. Oxygen (O):
The mass of oxygen is not directly given, but you can calculate it by subtracting the mass of carbon (from the CO2) and the mass of hydrogen (from the water) from the total mass of lysine:
Total mass of lysine = mass of CO2 + mass of H2O + mass of N
Total mass of lysine = 3.94 g + 1.89 g + mass of N - (mass of C + mass of H)
Total mass of lysine = 3.94 g + 1.89 g + 1.873 g - (0.328 g + 0.285 g) = 7.297 g

Now, subtract the mass of carbon and hydrogen from the total mass of lysine to find the mass of oxygen:
Mass of oxygen = total mass of lysine - (mass of carbon + mass of hydrogen)
Mass of oxygen = 7.297 g - (0.328 g + 0.285 g) = 6.684 g

Convert the mass of oxygen to moles using the molar mass of oxygen:
6.684 g O / 16 g/mol O = 0.4185 mol O

4. Nitrogen (N):
The mass of nitrogen is given directly as 0.436 g NH3.
Convert the mass of nitrogen to moles using the molar mass of nitrogen:
0.436 g N / 14 g/mol N = 0.031 mol N

(b) Calculating the empirical formula:
Now that you have the number of moles for each element, divide each mole value by the smallest mole value to get the ratio of atoms:
Carbon (C): 0.0273 mol C / 0.0273 mol C = 1
Hydrogen (H): 0.285 mol H / 0.0273 mol C ≈ 10.45 (round to the nearest whole number) ≈ 10
Oxygen (O): 0.4185 mol O / 0.0273 mol C ≈ 15.35 (round to the nearest whole number) ≈ 15
Nitrogen (N): 0.031 mol N / 0.0273 mol C ≈ 1.14 (round to the nearest whole number) ≈ 1

The empirical formula for lysine is approximately C1H10N1O15.

To determine the molecular formula using the given molar mass of approximately 150 g/mol:

(c) Calculating the molecular formula:
The molar mass of the empirical formula (C1H10N1O15) can be calculated as follows:
Molar mass = (atomic mass of carbon × number of carbon atoms) + (atomic mass of hydrogen × number of hydrogen atoms) + (atomic mass of nitrogen × number of nitrogen atoms) + (atomic mass of oxygen × number of oxygen atoms)
Molar mass = (12 g/mol × 1) + (1 g/mol × 10) + (14 g/mol × 1) + (16 g/mol × 15)
Molar mass = 12 g/mol + 10 g/mol + 14 g/mol + 240 g/mol
Molar mass = 276 g/mol

To obtain the molecular formula, divide the given molar mass (150 g/mol) by the molar mass of the empirical formula (276 g/mol):
Molecular formula = (Empirical formula) × (150 g/mol ÷ 276 g/mol)

Now, multiply each subscript in the empirical formula by (150 g/mol ÷ 276 g/mol) to get the molecular formula:
Molecular formula = C1H10N1O15 × (150 g/mol ÷ 276 g/mol)
Molecular formula ≈ C0.3623H3.6232N0.3623O5.4348

Rounded to the nearest whole number, the molecular formula of lysine is C0.4H4N0.4O5.

Also when I found the # of N

I got it from 1.873g lysine and for the other elements C and O the lysine was 2.175g

would i have to somehow add those number of lysine to find the empirical formula?