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Calculation of Energy and Mass in a Reduced-Calorie Soft Drink, Study notes of Chemistry

Solutions to two chemistry problems related to the energy and mass of a reduced-calorie soft drink. The first problem involves determining the equation for the complete combustion of sucrose and calculating the mass of tagatose required to produce a can of soft drink with the same sweetness as a standard one. The second problem calculates the energy obtained from the tagatose in the reduced-calorie soft drink.

What you will learn

  • What is the equation for the complete combustion of sucrose?
  • How much tagatose is required to produce a can of soft drink with the same sweetness as a standard one?
  • How much energy will a person obtain from a reduced-calorie can of soft drink?

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CHEM1101 2007-N-8 November 2007
โ€ข Normal table sugar is pure sucrose, C
12
H
22
O
11
(s). Give the equation for the complete
combustion of sucrose.
Marks
9
C
12
H
22
O
11
(s) + 12O
2
(g) ๎˜
๎˜๎˜
๎˜ 12CO
2
(g) + 11H
2
O(l)
A standard can (375 mL) of soft drink contains 795 kJ of energy. Assume the only
ingredients are water and sucrose and that the energy obtained from sucrose by the
body is the same as that obtained by combustion. Tagatose, C
6
H
12
O
6
, is a low-calorie
sweetener with a calorific value to humans of only 6.2 kJ g
โ€“1
. On a weight for weight
basis, it is 92% as sweet as sucrose. What mass of tagatose would be needed to
produce a 375 mL can of drink with the same sweetness as a standard soft drink?
Data: โˆ†
f
Hยฐ (C
12
H
22
O
11
(s)) = โ€“2221.7 kJ mol
โ€“1
โˆ†
f
Hยฐ (CO
2
(g)) = โ€“393.5 kJ mol
โ€“1
โˆ†
f
Hยฐ (H
2
O(l)) = โ€“285.8 kJ mol
โ€“1
The molar mass of sucrose, C
12
H
22
O
11
is:
M
sucrose
= 12 ร— 12.01 (C) + 22 ร— 1.008 (H) + 11 ร— 16.00 (O) = 342.296 g mol
-1
As
o o o
rxn f f
H m H (products) n H (reactants)
โˆ† = โˆ† โˆ’ โˆ†
โˆ† = โˆ† โˆ’ โˆ†โˆ† = โˆ† โˆ’ โˆ†
โˆ† = โˆ† โˆ’ โˆ†
โˆ‘ โˆ‘
โˆ‘ โˆ‘โˆ‘ โˆ‘
โˆ‘ โˆ‘ , the โˆ†
โˆ†โˆ†
โˆ†
comb
H
o
(sucrose)
is given by:
o o o
comb f f
o o
f 2 f 2
o o
f 12 22 11 f 2
1
H m H (products) n H (reactants)
(12 H (CO (g)) 11 H (H O(l)))
( H (C H O (g)) 12 H (O (g)))
(12 393.5 11 285.8) ( 2221.7 12 0) 5644.1kJ mol
โˆ’
โˆ’โˆ’
โˆ’
โˆ† = โˆ† โˆ’ โˆ†
โˆ† = โˆ† โˆ’ โˆ†โˆ† = โˆ† โˆ’ โˆ†
โˆ† = โˆ† โˆ’ โˆ†
โˆ‘ โˆ‘
โˆ‘ โˆ‘โˆ‘ โˆ‘
โˆ‘ โˆ‘
= โˆ† + โˆ†
= โˆ† + โˆ†= โˆ† + โˆ†
= โˆ† + โˆ†
โˆ’ โˆ† + โˆ†
โˆ’ โˆ† + โˆ†โˆ’ โˆ† + โˆ†
โˆ’ โˆ† + โˆ†
= ร— โˆ’ + ร— โˆ’ โˆ’ โˆ’ + ร— = โˆ’
= ร— โˆ’ + ร— โˆ’ โˆ’ โˆ’ + ร— = โˆ’= ร— โˆ’ + ร— โˆ’ โˆ’ โˆ’ + ร— = โˆ’
= ร— โˆ’ + ร— โˆ’ โˆ’ โˆ’ + ร— = โˆ’
As a standard can generates 795 kJ of energy, it must contain
1
795kJ
5644.1kJ mol
=
0.141 mol of sucrose. Using the molar mass from above, this corresponds to a
mass of:
mass = number of moles ร— molar mass = 0.141 ร— 342.296 = 48.2 g
As tagatose only has 92% of the sweetness of sucrose, more is required to match
the sweetness provided by sucrose:
mass of tagatose required = 48.2
52.4g
0.92 =
==
=
Answer: 52.4 g
ANSWER CONTINUES ON THE NEXT PAGE
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CHEM1101 2007-N-8 November 2007

  • Normal table sugar is pure sucrose, C 12

H

22

O

11

(s). Give the equation for the complete

combustion of sucrose.

Marks

C

12

H

22

O

11

(s) + 12O

2

(g)  

 12CO

2

(g) + 11H

2

O(l)

A standard can (375 mL) of soft drink contains 795 kJ of energy. Assume the only

ingredients are water and sucrose and that the energy obtained from sucrose by the

body is the same as that obtained by combustion. Tagatose, C

6

H

12

O

6

, is a low-calorie

sweetener with a calorific value to humans of only 6.2 kJ g

  • . On a weight for weight

basis, it is 92% as sweet as sucrose. What mass of tagatose would be needed to

produce a 375 mL can of drink with the same sweetness as a standard soft drink?

Data: โˆ†

f

H ยฐ (C

12

H

22

O

11

(s)) = โ€“2221.7 kJ mol

f

H ยฐ (CO

2

(g)) = โ€“393.5 kJ mol

f

H ยฐ (H

2

O(l)) = โ€“285.8 kJ mol

The molar mass of sucrose, C

12

H

22

O

11

is:

M

sucrose

= 12 ร— 12.01 (C) + 22 ร— 1.008 (H) + 11 ร— 16.00 (O) = 342.296 g mol

-

As

o o o

rxn f f

โˆ† H = m โˆ† H (products) โˆ’ n โˆ† H (reac tan ts) โˆ†โˆ† == โˆ†โˆ† โˆ’โˆ’ โˆ†โˆ†

, the โˆ† โˆ†โˆ†

comb

H

o

(sucrose)

is given by:

o o o

comb f f

o o

f 2 f 2

o o

f 12 22 11 f 2

1

H m H (products) n H (reac tan ts)

(12 H (CO (g)) 11 H (H O(l)))

( H (C H O (g)) 12 H (O (g)))

(12 393.5 11 285.8) ( 2221.7 12 0) 5644.1kJ mol

โˆ’โˆ’โˆ’โˆ’

==== ร— โˆ’ร— โˆ’ร— โˆ’ร— โˆ’ ++++ ร— โˆ’ร— โˆ’ร— โˆ’ร— โˆ’ โˆ’ โˆ’โˆ’ โˆ’โˆ’ โˆ’โˆ’ โˆ’ ++++ ร—ร—ร—ร— = โˆ’= โˆ’= โˆ’= โˆ’

As a standard can generates 795 kJ of energy, it must contain

1

795kJ

5644.1kJ mol

โˆ’ โˆ’โˆ’

โˆ’

0.141 mol of sucrose. Using the molar mass from above, this corresponds to a

mass of:

mass = number of moles ร— molar mass = 0.141 ร— 342.296 = 48.2 g

As tagatose only has 92% of the sweetness of sucrose, more is required to match

the sweetness provided by sucrose:

mass of tagatose required =

52.4g

Answer: 52.4 g

ANSWER CONTINUES ON THE NEXT PAGE

CHEM1101 2007-N-8 November 2007

How much energy will a person obtain from this reduced-calorie can of soft drink?

As tagatose has a calorific value of 6.2 kJ g

-

, 52.4 g will provide:

energy obtained = 52.4 g ร— 6.2 kJ g

-

= 330 kJ

Answer: 330 kJ