Question

Info needed For reaction 2, the combustion of sucrose, the balanced equation is:
C₁₂H₂₂O₁₁ (s)     +    12 O₂ (g)      →      12 CO₂ (g)     +    11 H₂O (l)     
& table

Quantity	Value	Unit	MARKS
mass of sucrose	1.0255	g	1 Mark
Initial Temperature (Ti )	24.186	°C	1 Mark
Final Temperature (Tf )	25.726	°C	1 Mark
ΔT	Click or tap here to enter text.	°C	1 Mark

1. The reaction studied in part 2 of this laboratory shows an example of how bomb calorimetry can be used to estimate the caloric content of food. Sucrose which is also known as table sugar is a component of many foods. The energy content of sucrose has been measured in this experiment, but the value obtained is expressed in terms which only a scientist (such as you) can easily understand. In order to present the information in a format which the general public can understand, the result should be changed to a more general unit.
   1. Convert the value for the energy content of “sugar” (sucrose), measured in reaction 2, from units of kJ∙mol^-1 into units of Calories per gram (Cal∙g^-1).

Hints:

• use the molar mass of sucrose to convert moles to grams
• don’t forget to convert from kJ into J
• the calorie (spelled with a lower-case “c”) is defined as 1 cal = 4.184 J
• When referring to nutritional content the unit is actually the Calorie (spelled with an upper-case “C”) where 1 Cal = 1000 cal.

2. To check that your answer makes sense, look at the nutritional information on any packaged food containing a high sugar content (such as a cola drink, pancake syrup, jam, etc.). List the food item which you have chosen and quote the Calories and mass of “sugar”- sometimes listed as “carbohydrate”. Note that on nutritional labels, the calorie content is listed in terms of the Calorie (“upper-case C”) even if the label uses the “lower-case c”.

Calculate the caloric content (Calories per gram of “sugar”). Is the answer close to the value calculated in part a?

Answer 1

Here mass of water is missing. But below procedure will help you.

Suppose if mass of water is given as m grams. Then, heat abosorbed by water;

Q = m * s * Delta T

s is specific heat capacity of water and Delta T is rise in temperature.

So, Q in J = m * 4.184 * (25.726 - 24.186) = 6.443 m J

1 Cal = 4.184 J

Q in Cal = m * 1 * (25.726 - 24.186)

Now we can calculate moles of Sucrose = mass in grams/molar mass of sucrose (n) = 1.0255/342.3 = 0.002996 moles

----> Heat of combustion (Delta H) = - Q/n in J/mol

1000 J = 1 KJ

Delta H in KJ/mol = - Q / (n * 1000) KJ/mol

---->  

SO, Delta H in Cal/g = - Q in Cal / mass of sucrose = - Q in Cal / 1.0255 Cal/g ....Answer

You just need to put mass of water.

Let me know if any doubt.