A compound contains only C, H, and O. To analyze the compound, you determine its

Question

A compound contains only C, H, and O. To analyze the compound, you determine its empirical formula by
analysis for C and H, and its molar mass by freezing point depression. (a) What is the
empirical formula if a 0.01510 g sample gives 0.01102 g of H2O and 0.04484 g of CO2
on combustion? (b) If you dissolve 1.000 g of the powder in 100.000 g of camphor,
you find that the melting point of the solution is 177.05 EC. Pure camphor melts at
179.75 EC. The kfp for camphor is !40.0 deg/molal. What is the molar mass of the
white powder? (c) What is the molecular formula of the white powder?

Explanation / Answer

Calculate mass oc H in 0.01102g H2O 2.016/18.015*0.01102 = 0.0012332g Calculate mass C in 0.04484g CO2 12.011/44.0096* 0.04484 = 0.01224g Calculate mass O = 0.01510 - ( 0.0012332+0.01224) = 0.001629g divide each by respective atomic mass C = 0.01224g/12.011 = 0.0010191 H = 0.0012332g /1.008 = 0.0012234 O = 0.001629/15.999 = 0.000101819 Divide through by smallest: C = 10.0 H = 12.0 O = 1.0 Empirical formula = C10H12O Question b) you have dissolved 10g powder in 1kg camphor melting point lowered by 179.75 - 177.05 = 2.7°C Kf for camphor = 40°C/molal therefore you have a 2.7/40 = 0.0675 molal solution. 10g = 0.0675mol 1mol = 10/0.0675 = 148.148g/mol Empirical formula, C10H12O = 120+12+16 = 148g/mol The empirical formula is the molecular formula

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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