Heat of Vaporization Equation Question (I made a mistake somewhere in equation) - *10 Points* ?

Question

Hi, I have worked this following problem out, but it keeps coming up that the answer is "off by a multiple of ten". (I hate webassign!!)

My teacher has not responded back to me, and I need to know where I have gone wrong. Can someone please help??

My question:

The fluorocarbon compound C 2 Cl 3 F 3 has a normal boiling point of 47.6°C. The specific heats of C 2 Cl 3 F 3 ( l ) and C 2 Cl 3 F 3 ( g ) are 0.91 J/g-K and 0.67 J/g-K, respectively. The heat of vaporization for the compound is 27.49 kJ/mol. Calculate the heat required to convert 27.0 g of C 2 Cl 3 F 3 from a liquid at 8.00 ° C to a gas at 58.00 ° C .

My work:

27 x .91 x (47.6 - 8) = 972.972 = 973 (rounded)

(27)/(187.3762) x 27.49 x 10^3 = 3961.175 = 3961 (rounded)

27 x .67x (58.00 - 47.6) = 188.136 = 188 (rounded)

Heat required = 973+ 3961 +188 = 5122 J (Came up as "Your answer is off by a multiple of ten")

I even un-rounded the figures, and came up with 5122.283 and typed in 5122.3, and it came up incorrect.

Can someone please help me out on this? I will offer 10 best answer points to the best answer.

Thank you in advance!

Answer 1

Showing your units with your work may help you find solutions easier.

Find MW of compound before starting calculations for step 2.  

Change in C = change in K

Step 1:  heating liquid to boiling point

H1 = 27*(47.6 - 8.00)*0.91 J

Step 2: evaporation of liquid to gas

H2 = (27/MW)*27.49*1000 J       Hvap in J/mol need to convert g to mol

Step 3:  heating gas to final temperature

H3 = 27*(58.00-47.6)*0.67 J

Since our calculations agree, check the original problem.  The mass may be 2.7 g and not 27 g.