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Polymers

Materials Science · FE Reference Handbook section

Materials Science
0 formulas
10 exam-style examples
~45 min
All Materials Science lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

  • Polymers are classified as thermoplastics that can be melted and reformed. Thermosets cannot be melted and reformed.

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1
Polymers — degree of polymerization — solve for degree of polymerization — Polymers

A materials engineer computes the degree of polymerization for a batch of polymers. Given number-average molecular weight (Mn) = 113,000 g/mol; repeat unit molecular weight (M0) = 140.0 g/mol, determine the degree of polymerization (DP).

Given

  • number−averagemolecularweight(Mn)=113,000g/molnumber-average molecular weight (Mn) = 113,000 g/mol
  • repeatunitmolecularweight(M0)=140.0g/molrepeat unit molecular weight (M_{0}) = 140.0 g/mol

Find

degree of polymerization (DP)

Start with the thinking

  • The governing relation printed in this handbook section is Polymers — degree of polymerization.
  • Everything except DP is given, so isolate DP symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Polymers are characterized by their degree of polymerization, the ratio of chain molecular weight to repeat unit weight.

Step-by-step solution

  1. Step 1 — State the governing relation:

    DP=MnM0DP = \dfrac{M_n}{M_0}
  2. Step 2 — Rearrange symbolically for DP:

    DP=MnM0DP = \dfrac{M_n}{M_0}
  3. Step 3 — List the givens: number-average molecular weight (Mn) = 113,000 g/mol, repeat unit molecular weight (M0) = 140.0 g/mol.

  4. Step 4 — Substitute the given values:

    DP=MnM0DP = \dfrac{M_n}{M_0}
  5. Step 5 — Evaluate:

    DP=807.1DP = 807.1
  6. Step 6 — Check: returning DP = 807.1 to

    DP=MnM0DP = \dfrac{M_n}{M_0}

    reproduces the given quantities, and both sides carry the same units.

Answer:
DP=807.1DP = 807.1

Why the other options are there

  • 1,614 — kept a factor of two that cancels in the correct rearrangement.
  • 403.6 — dropped that same factor in the other direction.
  • 887.9 — rounded an intermediate value before the final step.

Reference: FE Handbook — Polymers

Example 2
Polymers — degree of polymerization — solve for number-average molecular weight — Polymers (2)

A student relates molecular weight to the degree of polymerization for a polymer chain. Given repeat unit molecular weight (M0) = 147.0 g/mol; degree of polymerization (DP) = 2,286, determine the number-average molecular weight (Mn) in g/mol.

Given

  • repeatunitmolecularweight(M0)=147.0g/molrepeat unit molecular weight (M_{0}) = 147.0 g/mol
  • degreeofpolymerization(DP)=2,286degree of polymerization (DP) = 2,286

Find

number-average molecular weight (Mn), in g/mol

Start with the thinking

  • The governing relation printed in this handbook section is Polymers — degree of polymerization.
  • Everything except Mn is given, so isolate Mn symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Polymers are characterized by their degree of polymerization, the ratio of chain molecular weight to repeat unit weight.

Step-by-step solution

  1. Step 1 — State the governing relation:

    DP=MnM0DP = \dfrac{M_n}{M_0}
  2. Step 2 — Rearrange symbolically for Mn:

    Mn=DP×M0Mn = DP \times M_0
  3. Step 3 — List the givens: repeat unit molecular weight (M0) = 147.0 g/mol, degree of polymerization (DP) = 2,286.

  4. Step 4 — Substitute the given values:

    Mn=2286×M0Mn = 2286 \times M_0
  5. Step 5 — Evaluate:

    Mn=336042 g/molMn = 336042\ \text{g/mol}
  6. Step 6 — Check: returning Mn = 336,042 g/mol to

    DP=MnM0DP = \dfrac{M_n}{M_0}

    reproduces the given quantities, and both sides carry the same units.

Answer:
Mn=336042 g/molMn = 336042\ \text{g/mol}

Why the other options are there

  • 672,084 — kept a factor of two that cancels in the correct rearrangement.
  • 168,021 — dropped that same factor in the other direction.
  • 369,646 — rounded an intermediate value before the final step.

Reference: FE Handbook — Polymers

Example 3
Polymers — degree of polymerization — solve for repeat unit molecular weight — Polymers (3)

The polymer's degree of polymerization influences its mechanical strength. Given number-average molecular weight (Mn) = 163,000 g/mol; degree of polymerization (DP) = 3,984, determine the repeat unit molecular weight (M0) in g/mol.

Given

  • number−averagemolecularweight(Mn)=163,000g/molnumber-average molecular weight (Mn) = 163,000 g/mol
  • degreeofpolymerization(DP)=3,984degree of polymerization (DP) = 3,984

Find

repeat unit molecular weight (M0), in g/mol

Start with the thinking

  • The governing relation printed in this handbook section is Polymers — degree of polymerization.
  • Everything except M0 is given, so isolate M0 symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Polymers are characterized by their degree of polymerization, the ratio of chain molecular weight to repeat unit weight.

Step-by-step solution

  1. Step 1 — State the governing relation:

    DP=MnM0DP = \dfrac{M_n}{M_0}
  2. Step 2 — Rearrange symbolically for M0:

    M0=MnDPM_{0} = \dfrac{M_n}{DP}
  3. Step 3

    Listthegivens:number−averagemolecularweight(Mn)=163,000g/mol,degreeofpolymerization(DP)=3,984List the givens: number-average molecular weight (Mn) = 163,000 g/mol, degree of polymerization (DP) = 3,984
  4. Step 4 — Substitute the given values:

    M0=Mn3984M_{0} = \dfrac{M_n}{3984}
  5. Step 5 — Evaluate:

    M0=40.9137 g/molM_{0} = 40.9137\ \text{g/mol}
  6. Step 6 — Check: returning M0 = 40.9137 g/mol to

    DP=MnM0DP = \dfrac{M_n}{M_0}

    reproduces the given quantities, and both sides carry the same units.

Answer:
M0=40.9137 g/molM_{0} = 40.9137\ \text{g/mol}

Why the other options are there

  • 81.8273 — kept a factor of two that cancels in the correct rearrangement.
  • 20.4568 — dropped that same factor in the other direction.
  • 45.0050 — rounded an intermediate value before the final step.

Reference: FE Handbook — Polymers

Example 4
Polymers — degree of polymerization — solve for degree of polymerization (case 2) — Polymers (4)

A materials engineer computes the degree of polymerization for a batch of polymers. Given number-average molecular weight (Mn) = 23,500 g/mol; repeat unit molecular weight (M0) = 60.0000 g/mol, determine the degree of polymerization (DP).

Given

  • number−averagemolecularweight(Mn)=23,500g/molnumber-average molecular weight (Mn) = 23,500 g/mol
  • repeatunitmolecularweight(M0)=60.0000g/molrepeat unit molecular weight (M_{0}) = 60.0000 g/mol

Find

degree of polymerization (DP)

Start with the thinking

  • The governing relation printed in this handbook section is Polymers — degree of polymerization.
  • Everything except DP is given, so isolate DP symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Polymers are characterized by their degree of polymerization, the ratio of chain molecular weight to repeat unit weight.

Step-by-step solution

  1. Step 1 — State the governing relation:

    DP=MnM0DP = \dfrac{M_n}{M_0}
  2. Step 2 — Rearrange symbolically for DP:

    DP=MnM0DP = \dfrac{M_n}{M_0}
  3. Step 3 — List the givens: number-average molecular weight (Mn) = 23,500 g/mol, repeat unit molecular weight (M0) = 60.0000 g/mol.

  4. Step 4 — Substitute the given values:

    DP=MnM0DP = \dfrac{M_n}{M_0}
  5. Step 5 — Evaluate:

    DP=391.7DP = 391.7
  6. Step 6 — Check: returning DP = 391.7 to

    DP=MnM0DP = \dfrac{M_n}{M_0}

    reproduces the given quantities, and both sides carry the same units.

Answer:
DP=391.7DP = 391.7

Why the other options are there

  • 783.3 — kept a factor of two that cancels in the correct rearrangement.
  • 195.8 — dropped that same factor in the other direction.
  • 430.8 — rounded an intermediate value before the final step.

Reference: FE Handbook — Polymers

Example 5
Polymers — degree of polymerization — solve for number-average molecular weight (case 2) — Polymers (5)

A student relates molecular weight to the degree of polymerization for a polymer chain. Given repeat unit molecular weight (M0) = 145.0 g/mol; degree of polymerization (DP) = 4,965, determine the number-average molecular weight (Mn) in g/mol.

Given

  • repeatunitmolecularweight(M0)=145.0g/molrepeat unit molecular weight (M_{0}) = 145.0 g/mol
  • degreeofpolymerization(DP)=4,965degree of polymerization (DP) = 4,965

Find

number-average molecular weight (Mn), in g/mol

Start with the thinking

  • The governing relation printed in this handbook section is Polymers — degree of polymerization.
  • Everything except Mn is given, so isolate Mn symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Polymers are characterized by their degree of polymerization, the ratio of chain molecular weight to repeat unit weight.

Step-by-step solution

  1. Step 1 — State the governing relation:

    DP=MnM0DP = \dfrac{M_n}{M_0}
  2. Step 2 — Rearrange symbolically for Mn:

    Mn=DP×M0Mn = DP \times M_0
  3. Step 3 — List the givens: repeat unit molecular weight (M0) = 145.0 g/mol, degree of polymerization (DP) = 4,965.

  4. Step 4 — Substitute the given values:

    Mn=4965×M0Mn = 4965 \times M_0
  5. Step 5 — Evaluate:

    Mn=719925 g/molMn = 719925\ \text{g/mol}
  6. Step 6 — Check: returning Mn = 719,925 g/mol to

    DP=MnM0DP = \dfrac{M_n}{M_0}

    reproduces the given quantities, and both sides carry the same units.

Answer:
Mn=719925 g/molMn = 719925\ \text{g/mol}

Why the other options are there

  • 1,439,850 — kept a factor of two that cancels in the correct rearrangement.
  • 359,963 — dropped that same factor in the other direction.
  • 791,918 — rounded an intermediate value before the final step.

Reference: FE Handbook — Polymers

Example 6
Polymers — degree of polymerization — solve for repeat unit molecular weight (case 2) — Polymers (6)

The polymer's degree of polymerization influences its mechanical strength. Given number-average molecular weight (Mn) = 21,500 g/mol; degree of polymerization (DP) = 1,171, determine the repeat unit molecular weight (M0) in g/mol.

Given

  • number−averagemolecularweight(Mn)=21,500g/molnumber-average molecular weight (Mn) = 21,500 g/mol
  • degreeofpolymerization(DP)=1,171degree of polymerization (DP) = 1,171

Find

repeat unit molecular weight (M0), in g/mol

Start with the thinking

  • The governing relation printed in this handbook section is Polymers — degree of polymerization.
  • Everything except M0 is given, so isolate M0 symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Polymers are characterized by their degree of polymerization, the ratio of chain molecular weight to repeat unit weight.

Step-by-step solution

  1. Step 1 — State the governing relation:

    DP=MnM0DP = \dfrac{M_n}{M_0}
  2. Step 2 — Rearrange symbolically for M0:

    M0=MnDPM_{0} = \dfrac{M_n}{DP}
  3. Step 3

    Listthegivens:number−averagemolecularweight(Mn)=21,500g/mol,degreeofpolymerization(DP)=1,171List the givens: number-average molecular weight (Mn) = 21,500 g/mol, degree of polymerization (DP) = 1,171
  4. Step 4 — Substitute the given values:

    M0=Mn1171M_{0} = \dfrac{M_n}{1171}
  5. Step 5 — Evaluate:

    M0=18.3604 g/molM_{0} = 18.3604\ \text{g/mol}
  6. Step 6 — Check: returning M0 = 18.3604 g/mol to

    DP=MnM0DP = \dfrac{M_n}{M_0}

    reproduces the given quantities, and both sides carry the same units.

Answer:
M0=18.3604 g/molM_{0} = 18.3604\ \text{g/mol}

Why the other options are there

  • 36.7208 — kept a factor of two that cancels in the correct rearrangement.
  • 9.1802 — dropped that same factor in the other direction.
  • 20.1964 — rounded an intermediate value before the final step.

Reference: FE Handbook — Polymers

Example 7
Polymers — degree of polymerization — solve for degree of polymerization (case 3) — Polymers (7)

A materials engineer computes the degree of polymerization for a batch of polymers. Given number-average molecular weight (Mn) = 146,500 g/mol; repeat unit molecular weight (M0) = 70.0000 g/mol, determine the degree of polymerization (DP).

Given

  • number−averagemolecularweight(Mn)=146,500g/molnumber-average molecular weight (Mn) = 146,500 g/mol
  • repeatunitmolecularweight(M0)=70.0000g/molrepeat unit molecular weight (M_{0}) = 70.0000 g/mol

Find

degree of polymerization (DP)

Start with the thinking

  • The governing relation printed in this handbook section is Polymers — degree of polymerization.
  • Everything except DP is given, so isolate DP symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Polymers are characterized by their degree of polymerization, the ratio of chain molecular weight to repeat unit weight.

Step-by-step solution

  1. Step 1 — State the governing relation:

    DP=MnM0DP = \dfrac{M_n}{M_0}
  2. Step 2 — Rearrange symbolically for DP:

    DP=MnM0DP = \dfrac{M_n}{M_0}
  3. Step 3 — List the givens: number-average molecular weight (Mn) = 146,500 g/mol, repeat unit molecular weight (M0) = 70.0000 g/mol.

  4. Step 4 — Substitute the given values:

    DP=MnM0DP = \dfrac{M_n}{M_0}
  5. Step 5 — Evaluate:

    DP=2093DP = 2093
  6. Step 6 — Check: returning DP = 2,093 to

    DP=MnM0DP = \dfrac{M_n}{M_0}

    reproduces the given quantities, and both sides carry the same units.

Answer:
DP=2093DP = 2093

Why the other options are there

  • 4,186 — kept a factor of two that cancels in the correct rearrangement.
  • 1,046 — dropped that same factor in the other direction.
  • 2,302 — rounded an intermediate value before the final step.

Reference: FE Handbook — Polymers

Example 8
Polymers — degree of polymerization — solve for number-average molecular weight (case 3) — Polymers (8)

A student relates molecular weight to the degree of polymerization for a polymer chain. Given repeat unit molecular weight (M0) = 42.0000 g/mol; degree of polymerization (DP) = 5,711, determine the number-average molecular weight (Mn) in g/mol.

Given

  • repeatunitmolecularweight(M0)=42.0000g/molrepeat unit molecular weight (M_{0}) = 42.0000 g/mol
  • degreeofpolymerization(DP)=5,711degree of polymerization (DP) = 5,711

Find

number-average molecular weight (Mn), in g/mol

Start with the thinking

  • The governing relation printed in this handbook section is Polymers — degree of polymerization.
  • Everything except Mn is given, so isolate Mn symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Polymers are characterized by their degree of polymerization, the ratio of chain molecular weight to repeat unit weight.

Step-by-step solution

  1. Step 1 — State the governing relation:

    DP=MnM0DP = \dfrac{M_n}{M_0}
  2. Step 2 — Rearrange symbolically for Mn:

    Mn=DP×M0Mn = DP \times M_0
  3. Step 3 — List the givens: repeat unit molecular weight (M0) = 42.0000 g/mol, degree of polymerization (DP) = 5,711.

  4. Step 4 — Substitute the given values:

    Mn=5711×M0Mn = 5711 \times M_0
  5. Step 5 — Evaluate:

    Mn=239862 g/molMn = 239862\ \text{g/mol}
  6. Step 6 — Check: returning Mn = 239,862 g/mol to

    DP=MnM0DP = \dfrac{M_n}{M_0}

    reproduces the given quantities, and both sides carry the same units.

Answer:
Mn=239862 g/molMn = 239862\ \text{g/mol}

Why the other options are there

  • 479,724 — kept a factor of two that cancels in the correct rearrangement.
  • 119,931 — dropped that same factor in the other direction.
  • 263,848 — rounded an intermediate value before the final step.

Reference: FE Handbook — Polymers

Example 9
Polymers — degree of polymerization — solve for repeat unit molecular weight (case 3) — Polymers (9)

The polymer's degree of polymerization influences its mechanical strength. Given number-average molecular weight (Mn) = 129,000 g/mol; degree of polymerization (DP) = 4,522, determine the repeat unit molecular weight (M0) in g/mol.

Given

  • number−averagemolecularweight(Mn)=129,000g/molnumber-average molecular weight (Mn) = 129,000 g/mol
  • degreeofpolymerization(DP)=4,522degree of polymerization (DP) = 4,522

Find

repeat unit molecular weight (M0), in g/mol

Start with the thinking

  • The governing relation printed in this handbook section is Polymers — degree of polymerization.
  • Everything except M0 is given, so isolate M0 symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Polymers are characterized by their degree of polymerization, the ratio of chain molecular weight to repeat unit weight.

Step-by-step solution

  1. Step 1 — State the governing relation:

    DP=MnM0DP = \dfrac{M_n}{M_0}
  2. Step 2 — Rearrange symbolically for M0:

    M0=MnDPM_{0} = \dfrac{M_n}{DP}
  3. Step 3

    Listthegivens:number−averagemolecularweight(Mn)=129,000g/mol,degreeofpolymerization(DP)=4,522List the givens: number-average molecular weight (Mn) = 129,000 g/mol, degree of polymerization (DP) = 4,522
  4. Step 4 — Substitute the given values:

    M0=Mn4522M_{0} = \dfrac{M_n}{4522}
  5. Step 5 — Evaluate:

    M0=28.5272 g/molM_{0} = 28.5272\ \text{g/mol}
  6. Step 6 — Check: returning M0 = 28.5272 g/mol to

    DP=MnM0DP = \dfrac{M_n}{M_0}

    reproduces the given quantities, and both sides carry the same units.

Answer:
M0=28.5272 g/molM_{0} = 28.5272\ \text{g/mol}

Why the other options are there

  • 57.0544 — kept a factor of two that cancels in the correct rearrangement.
  • 14.2636 — dropped that same factor in the other direction.
  • 31.3799 — rounded an intermediate value before the final step.

Reference: FE Handbook — Polymers

Example 10
Polymers — degree of polymerization — solve for degree of polymerization (case 4) — Polymers (10)

A materials engineer computes the degree of polymerization for a batch of polymers. Given number-average molecular weight (Mn) = 136,000 g/mol; repeat unit molecular weight (M0) = 78.0000 g/mol, determine the degree of polymerization (DP).

Given

  • number−averagemolecularweight(Mn)=136,000g/molnumber-average molecular weight (Mn) = 136,000 g/mol
  • repeatunitmolecularweight(M0)=78.0000g/molrepeat unit molecular weight (M_{0}) = 78.0000 g/mol

Find

degree of polymerization (DP)

Start with the thinking

  • The governing relation printed in this handbook section is Polymers — degree of polymerization.
  • Everything except DP is given, so isolate DP symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Polymers are characterized by their degree of polymerization, the ratio of chain molecular weight to repeat unit weight.

Step-by-step solution

  1. Step 1 — State the governing relation:

    DP=MnM0DP = \dfrac{M_n}{M_0}
  2. Step 2 — Rearrange symbolically for DP:

    DP=MnM0DP = \dfrac{M_n}{M_0}
  3. Step 3 — List the givens: number-average molecular weight (Mn) = 136,000 g/mol, repeat unit molecular weight (M0) = 78.0000 g/mol.

  4. Step 4 — Substitute the given values:

    DP=MnM0DP = \dfrac{M_n}{M_0}
  5. Step 5 — Evaluate:

    DP=1744DP = 1744
  6. Step 6 — Check: returning DP = 1,744 to

    DP=MnM0DP = \dfrac{M_n}{M_0}

    reproduces the given quantities, and both sides carry the same units.

Answer:
DP=1744DP = 1744

Why the other options are there

  • 3,487 — kept a factor of two that cancels in the correct rearrangement.
  • 871.8 — dropped that same factor in the other direction.
  • 1,918 — rounded an intermediate value before the final step.

Reference: FE Handbook — Polymers

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