Historically, freezing point depression measurements were important in chemistry

Question

Historically, freezing point depression measurements were important in chemistry as a way to determine Molar Mass, which along with other data could then be used to determine molecular formulas. This problem illustrates that use. It also serves as a reminder about calculations from Chem 221 that you need to remember how to do. A compound is composed of the elements C, H and O. The mass % carbon is 68.9% and hydrogen is 4.96%. When 4.63 g of the compound was dissolved in 100 g of benzene, the solution had a freezing point of 3.56 degree C. The normal freezing point of benzene is 5.50 degree C and its K_f = 5.12 degree C/m. What is the molecular formula of the compound?

Explanation / Answer

Empirical formula

moles of C = 68.9/12 = 5.741 mol

moles of H = 4.96/1 = 4.96 mol

moles of O = (100 - (68.9 + 4.96))/16 = 1.634 mol

Divide by smallest factor

C = 5.741/1.634 = 3

H = 4.96/1.634 = 3

O = 1.634/1.634 = 1

Emprical formula = C3H3O

Empirical formula mass = 3 x 12 + 3 x 1 + 16 = 55

From freezing point,

molality = (3.56 - 5.50)-5.12 x 1 = 0.38 m

molar mass = 4.63/0.38 x 0.1 = 121.84

Factor = 121.84/55 = 2

Molacular formula of compound = 2 x C3H3O = C6H6O2

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.