A quantity of 6.882 g of an organic compound is dissolved in water to make 276.0 mL of solution. The solution has an osmotic pressure of 1.43 atm at 27°C. The analysis of this compound shows that it contains 41.8 percent C, 4.7 percent H, 37.3 percent O, and 16.3 percent N. Calculate the molecular formula of the compound.

The molecular formula of this compound is C15H20O10N5

Explanation:

Step 1: Data given

Mass of the compound = 6.882 grams

Volume of solution = 276.0 mL = 0.276 L

Osmotic pressure = 1.43 atm

Temperature = 27.0°C

The compound contains:

⇒41.8 % C

⇒ 4.7 % H

⇒ 37.3 % O

⇒ 16.3 % N

Step 2: Calculate mass

C = 41.8 % = 2.876 grams

H = 4.7 % = 0.322 grams

O = 37.3 % = 2.566 grams

N = 16.3 % = 1.121 grams

Moles = mass / molar mass

Moles C = 2.876 grams / 12 g/mol

Moles C = 0.2397 moles

Moles H = 0.322 grams / 1.01 g/mol

Moles H = 0.3188 moles

Moles O = 2.566 grams / 16 g/mol

Moles O = 0.1604 moles

Moles N = 1.121 grams / 14 g/mol

Moles N = 0.0801 moles

Step 3: Calculate mole ratio

We divide by the smallest number of moles

C: 0.2397 / 0.0801 = 3

H: 0.3188/0.0801 = 4

O = 0.1604 / 0.0801 = 2

N = 0.0801 / 0.0801 = 1

The empirical formula = C3H4O2N with molecular mass of 86.04 g/mol

Step 4: Calculate the molarity

Osmotic pressure = MRT

1.43 atm = M (0.08206 L*atm/mol*K) (300K)

M = 0.0581 mol/L

Step 5: Calculate molar mass of the compound

Since the solution contained 6.882 grams in a volume of 276 mL

6.882 grams/276 mL * 1000 mL/L = 24.9 g/L

Now, if you just take the ratio of those two calculations, you have:

24.9 g/L / 0.0581 mol/L = 429 g/mol

Step 6: Calculate the molecular formula

This means the molar mass of the compound is 429 g/mol

To find the molecular formula we have to multiply the empirical formula by n

n = 429 / 86 g/mol = 5

5 * (C3H4O2N) = C15H20O10N5

The molecular formula of this compound is C15H20O10N5