[Carolyng]

Hello, I just need to know what I'm missing with a question;

 In making candy, a certain recipe calls for heating an aqueous sucrose solution to the "softball" stage. Which has a boiling point of 235-240 F. Wat is the range of mass percentages of the solution of sugar (C12H22O12) that boils at those two temperature?

What I've done so far;

 -Converted the temps to; 112.778-115.556 C
- used Change Tf=KfM that =12.77m and 30.39m respectfully

-calculated molality
-12.77x(342.3)/mol(1kg)/1000g=8.29
-30.39(342.3)/mol(1kg/1000g)=104.02

-Calculate mass %
-8.29/8.29+1(100)= 89.23%
-30.39/30.39+1(100)=91.04%

The books answer gives; 81.8% and 84.6% respectively.  I went over every step and can't decide what I'm doing wrong. Thank you. 

[mjc123]

-Converted the temps to; 112.778-115.556 C
- used Change Tf=KfM that =12.77m and 30.39m respectfully

Where do you get 12.77m from? (And you mean "respectively")

-calculated molality
-12.77x(342.3)/mol(1kg)/1000g=8.29
-30.39(342.3)/mol(1kg/1000g)=104.02

Not molality (you already calculated that, if wrongly); you mean g solute/g solvent; I don't think there's a name for that. And both calculations are wrong.

-Calculate mass %
-8.29/8.29+1(100)= 89.23%
-30.39/30.39+1(100)=91.04%

Why use g/g for one and molality for the other?
Your approach is correct, but you are just being very careless with the calculations.
Besides, I doubt if boiling point varies linearly with molality at such high concentrations.