Question

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.

Answer 1

The molecular formula of this compound is C15H20O10N5

Step-by-step 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