SME
Organic Chemistry II
Synthetic polymers
Use this for “A polymer sample holds 2 mol of chains at 15,000 g/mol, 3 mol at 30,000 g/mol and 1 mol at 60,000 g/mol. Find Mn, Mw and the dispersity.”
Chart
This is an example. Type your own values to start a new problem.
n₁
Moles in fraction 1
example · mol
M₁
Molar mass of fraction 1
example · g/mol
n₂
Moles in fraction 2
example · mol
M₂
Molar mass of fraction 2
example · g/mol
n₃
Moles in fraction 3
example · mol
M₃
Molar mass of fraction 3
example · g/mol
Mn
Number-average molar mass
worked out (not typed) · g/mol
Mw
Weight-average molar mass
worked out (not typed) · g/mol
Đ
Dispersity
worked out (not typed)
Formulas
Mn
=
(
n₁
×
M₁
+
n₂
×
M₂
+
n₃
×
M₃
)
÷
(
n₁
+
n₂
+
n₃
)
22,500 = (1 × 10,000 + 2 × 20,000 + 1 × 40,000) ÷ (1 + 2 + 1)
Mw
=
(
n₁
×
M₁
2
+
n₂
×
M₂
2
+
n₃
×
M₃
2
)
÷
(
n₁
×
M₁
+
n₂
×
M₂
+
n₃
×
M₃
)
27,778
=
(1
×
10,
000
2
+
2
×
20,
000
2
+
1
×
40,
000
2
)
÷
(1
×
10,000
+
2
×
20,000
+
1
×
40,000)
Đ
=
Mw
Mn
1.235
=
27,778
22,500
Assumptions
•
The sample is split into three fractions, and every chain in one fraction has that fraction’s molar mass.
•
Mn weights each chain equally by count; Mw weights each chain by its mass, so long chains count for more.
•
The dispersity Đ = Mw ÷ Mn is 1 when every chain is the same length and grows as the lengths spread.
Step-by-step
WE KNOW
n₁ = 1 mol, M₁ = 10,000 g/mol, n₂ = 2 mol, M₂ = 20,000 g/mol, n₃ = 1 mol, M₃ = 40,000 g/mol
FIND
number-average molar mass (Mn), weight-average molar mass (Mw), dispersity (Đ)
Step 1 · Find number-average molar mass (Mn)
Use Mn = (n₁ × M₁ + n₂ × M₂ + n₃ × M₃) ÷ (n₁ + n₂ + n₃)
Add up the grams of every fraction, then divide by the total moles of chains.
Mn = (1 × 10,000 + 2 × 20,000 + 1 × 40,000) ÷ (1 + 2 + 1)
Mn
=
10,000 + 40,000 + 40,000
3 + 1
Mn
=
50,000 + 40,000
4
Mn
=
90,000
4
Mn = 22,500 g/mol
Step 2 · Find weight-average molar mass (Mw)
Use Mw = (n₁ × M₁² + n₂ × M₂² + n₃ × M₃²) ÷ (n₁ × M₁ + n₂ × M₂ + n₃ × M₃)
Weight each molar mass by the grams of its fraction: square it, times the moles, then divide by the total grams.
Mw
=
(1
×
10,
000
2
+
2
×
20,
000
2
+
1
×
40,
000
2
)
÷
(1
×
10,000
+
2
×
20,000
+
1
×
40,000)
Mw
=
(
10
8
+
2
×
(4
×
10
8
)
+
(1.6
×
10
9
))
÷
(10,000
+
40,000
+
40,000)
Mw
=
(
10
8
+
(8
×
10
8
)
+
(1.6
×
10
9
))
÷
(50,000
+
40,000)
Mw
=
(9 ×
10
8
) + (1.6 ×
10
9
)
90,000
Mw
=
2.5 ×
10
9
90,000
Mw = 27,778 g/mol
Step 3 · Find dispersity (Đ)
Use Đ = Mw ÷ Mn
Divide the weight average by the number average.
Đ
=
27,778
22,500
Đ
=
1.235
→
Đ
is
above
1:
the
long
chains
weigh
more
in
Mw
than
in
Mn,
so
the
spread
widens
it.
Check
Put the numbers back in. Both sides should match.
22,500 = (1 × 10,000 + 2 × 20,000 + 1 × 40,000) ÷ (1 + 2 + 1) ✓
27,778
=
(1
×
10,
000
2
+
2
×
20,
000
2
+
1
×
40,
000
2
)
÷
(1
×
10,000
+
2
×
20,000
+
1
×
40,000)
✓
1.235
=
27,778
22,500
✓
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