Hello, Here is a post lab question that I need answered from my \"Determination

Question

Hello,

Here is a post lab question that I need answered from my "Determination of Molar Mass by Freezing Point" post-lab.

Question:

**NOTE**

DOUBLE CHECK YOUR FINAL ANSWER AND MAKE SURE YOU SHOW ALL WORK, I WILL DOWNRATE IF THE QUESTION IS INCORRECT SO PLEASE MAKE SURE EVERYTHING LOOKS GOOD. ONLY HIGH RATING TO CORRECT ANSWERS THANKS!

ALSO ANSWER ALL SUBPARTS NOT JUST ONE THANKS!

7. A forensic chemist is given a white solid that is suspected of being pure cocaine (C17H21NO4, molar mass 303.35 g/mol). She dissolves 1.22 t 0.01 g of the solid in 15.60 t 0.01 gbenzene. The freezing point is lowered by 1.32 t 0.04 oC. a. What is the molar mass of the substance? b. Assuming that the percent uncertainty in the calculated molar mass is the percent uncer- tainty in the temperature change, calculate the uncertainty in the molar mass. c. Could the chemist unequivocally state that the substance is cocaine? For example, is the uncertainty small enough to distinguish cocaine from codeine (C18H21NO3, molar mass 299.36 g/mol)? d. Assuming that the absolute uncertainties in the measurements of temperature and mass remain unchanged, how could the chemist improve the precision of her results?

Explanation / Answer

7 .a ANSWER

Based on the freezing point depression,

molar mass of the solid

T = Kfpm or

m = T/ Kf

= 1.32o C / 5.12o C kg/mol

= 0.2578 mol/ kg benzene

molar mass of the solute be

=(0.2578 mol / kg benzene) x (15.6 g benzene)x(1 kg/ 1000 g) = 0.004022 mol compound

the molar mass is:

= 1.22 g / 0.004022 mol = 303.3 g/mol-------------ANSWER

This molar mass correlates well with that of cocaine, supporting identification

============

molar mass of C17H21NO4 is 17*12+21+14+4*16=303g
the concentration is 1.22/15.6=0.078g cocaine/g benzene or 78g cocaine/kg benzene
the molality is 78/303=0.258 moles/kg C6H6
the freezing point depression dt=5.12*0.258=1.32
So ,the substance could be cocaine the uncertainty =0.04/1.32=0.03=3%

===============

Tf = Kf · m x i

Tf, the freezing point depression, is defined as Tf (pure solvent) Tf (solution) = 1.32

Kf, the cryoscopic constant, = 5.12 for benzene
m is the molality of the solution = X
i = vant hoff factor 1

1.32 = 5.12 x X x 1
molality =1.32 / 5.12 = 0.258 moles / Kg of solution

we have 1.22 g of the solid in 15.60 g of benzene or 1.22 x 1000 / 15.6 g in a kg of solution

= 78.21 g / kg

so .258 moles = 78.21g 1 mole weighs 78.21 / .258 g = 303.14 g/ mole

molecular weight of C17H21NO4 is 303.35 g/mol

hence it looks like cocaine

=============================

molar mass of codeine = 12*12+21+14+3*16=227
3% of the mass of cocaine= 303*3/100=9.1 . so the limit of mass measured previously would be 303-9.1<M<303+9.1 or 293.9<M<312.1
so the substance is surely not codeine

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

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