SME
Grade 9 Science · Problem type
Dehydration synthesis: water and mass
Use this for “How many water molecules leave as the chain forms, or are needed to split it, and what is the polymer’s mass?”
Diagram
3 glucose molecules join by 2 bonds, and 2 water molecules given off.
3 − 1 = 2
This is an example. Type your own values to start a new problem.
M (g/mol)
=
n
×
m (g/mol)
−
18
×
w
b
Bonds formed
worked out (not typed)
Formulas
b
=
n
−
1
2 = 3 − 1
w
=
b
2 = 2
M
=
n
×
m
−
18
×
w
504 = 3 × 180 − 18 × 2
Assumptions
•
Each bond joining two monomers gives off one water molecule, 18 g/mol.
•
The monomers form one chain, not a ring, so a chain of n units has n − 1 bonds.
•
A fat follows the same count: one glycerol and three fatty acids are 4 units, 3 bonds, 3 water.
•
Glucose is 180 g/mol, so two glucose make maltose, 2 × 180 − 18 = 342 g/mol.
Step-by-step
WE KNOW
n = 3, m = 180 g/mol
FIND
bonds formed (b), water molecules given off (w), polymer molar mass (M)
Step 1 · Find bonds formed (b)
Use b = n − 1
A chain has one bond fewer than its units.
b
=
3
−
1
b
=
2
Step 2 · Find water molecules given off (w)
Use w = b
Every bond gives off one water molecule.
w
=
2
Step 3 · Find polymer molar mass (M)
Use M = n × m − 18 × w
Add the monomers’ molar masses, then take away 18 g/mol for each water molecule given off.
M
=
3
×
180
−
18
×
2
M
=
540
−
36
M
=
504
g/mol
Check
Put the numbers back in. Both sides should match.
2 = 3 − 1 ✓
2 = 2 ✓
504 = 3 × 180 − 18 × 2 ✓
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