What is the dipole moment?
Enter the size of the partial charges and the distance between them, such as a bond length. The calculator gives the dipole moment in debyes and coulomb meters, or works back from a measured dipole moment to the charge, the percent ionic character, or the distance.
- Answer
- μ = 4.8032 D (1.60218 × 10⁻²⁹ C·m)
μ = 4.8032 D (1.60218 × 10⁻²⁹ C·m).
- Dipole moment (D)
- 4.8032
- Dipole moment (C·m)
- 1.60218 × 10⁻²⁹
- Charge (e)
- 1
- Charge (C)
- 1.60218 × 10⁻¹⁹
- Distance (pm)
- 100
- Distance (Å)
- 1
- Percent ionic character (%)
- 100
Answer: μ = 4.8032 D (1.60218 × 10⁻²⁹ C·m). μ = 4.8032 D (1.60218 × 10⁻²⁹ C·m).
How to calculate
Finds a bond’s dipole moment μ = Q × r in debyes and coulomb meters from the partial charge and the distance between the charges, or the charge (with the percent ionic character) or the distance from the other two.
Example with the default inputs (Find Dipole moment, Charge Q 1, Charge unit e, Distance r 100, Distance unit pm): μ = 4.8032 D (1.60218 × 10⁻²⁹ C·m).
Method: μ = Q × r; 1 D = 3.335641 × 10⁻³⁰ C·m; e = 1.602176634 × 10⁻¹⁹ C; percent ionic character = Q ÷ e × 100.
- A bond dipole: two equal and opposite point charges a distance r apart. A molecule’s dipole moment is the vector sum of its bond dipoles.
- Percent ionic character compares a measured dipole moment with the one a full electron charge at the bond length would give.
Worked examples
Each example is checked against the calculator on every build.
- Find Dipole moment, Charge Q 1, Charge unit e, Distance r 100, Distance unit pm gives Dipole moment (C·m) 1.60218 × 10⁻²⁹, Dipole moment (D) 4.8032, Answer μ = 4.8032 D (1.60218 × 10⁻²⁹ C·m).Source: OpenStax, Chemistry 2e, §7.6 Molecular Structure and Polarity (μ = Qr), https://openstax.org/books/chemistry-2e/pages/7-6-molecular-structure-and-polarity; NIST SP 811, Appendix B.8 (debye = 3.335 641 × 10⁻³⁰ C·m), https://www.nist.gov/pml/special-publication-811/nist-guide-si-appendix-b-conversion-factors/nist-guide-si-appendix-b8
- Find Charge, Dipole moment μ 1.109, Dipole moment unit D, Distance r 127.5, Distance unit pm gives Charge (e) 0.181088, Percent ionic character (%) 18.1088.Source: Averill and Eldredge, General Chemistry: Principles, Patterns, and Applications, §8.9 Polar Covalent Bonds (percent ionic character), https://saylordotorg.github.io/text_general-chemistry-principles-patterns-and-applications-v1.0/s12-09-polar-covalent-bonds.html: Q = 0.1811 e, about 18% ionic
- Find Charge, Dipole moment μ 9.001, Dipole moment unit D, Distance r 236.1, Distance unit pm gives Percent ionic character (%) 79.3713.Source: Averill and Eldredge, General Chemistry: Principles, Patterns, and Applications, §8.9 Polar Covalent Bonds (percent ionic character), https://saylordotorg.github.io/text_general-chemistry-principles-patterns-and-applications-v1.0/s12-09-polar-covalent-bonds.html, Example 11: 79.39% after rounding Q to 1.272 × 10⁻¹⁹ C
- Find Charge, Dipole moment μ 6.08, Dipole moment unit D, Distance r 228.1, Distance unit pm gives Percent ionic character (%) 55.4941.Source: Averill and Eldredge, General Chemistry: Principles, Patterns, and Applications, §8.9 Polar Covalent Bonds (percent ionic character), https://saylordotorg.github.io/text_general-chemistry-principles-patterns-and-applications-v1.0/s12-09-polar-covalent-bonds.html, Exercise: 55.5%
- Find Distance, Dipole moment μ 0, Dipole moment unit C·m, Charge Q 0, Charge unit C gives Distance (pm) 100, Distance (Å) 1, Answer r = 100 pm (1 Å).Source: OpenStax, Chemistry 2e, §7.6 Molecular Structure and Polarity (μ = Qr), https://openstax.org/books/chemistry-2e/pages/7-6-molecular-structure-and-polarity: r = μ ÷ Q
How it works
Dipole moment (OpenStax §7.6).
μ = Q × r
with Q the size of the partial charge at each end and r the distance between them. Pick what to find:
- Dipole moment: μ = Q × r.
- Charge: Q = μ ÷ r.
- Distance: r = μ ÷ Q.
Units. Charge in elementary charges (e = 1.602176634 × 10⁻¹⁹ C, exact) or coulombs; distance in picometers (10⁻¹² m), ångströms (10⁻¹⁰ m) or nanometers (10⁻⁹ m); dipole moment in debyes (1 D = 3.335641 × 10⁻³⁰ C·m, NIST SP 811) or coulomb meters. The page works in C and m and shows μ in D and C·m, Q in e and C, and r in pm and Å.
Percent ionic character = Q ÷ e × 100, the dipole moment as a percent of e × r. It is shown in every mode; it means most in the charge mode, where μ is a measured dipole moment and r the bond length.
Rules the page checks. μ from 10⁻⁴⁰ to 1,000,000 in its unit; Q from 10⁻³⁰ to 1,000,000 in its unit; r from 0.000001 to 1,000,000,000 in its unit. All three are above 0.
How results are written. Every result is worked exactly from the typed decimals and the constants above and rounded once: 6 significant figures, trailing zeros dropped, thousands commas, and m × 10ⁿ below 0.000001 or from 10²¹ (in the rows, from 10¹⁵). The percent ionic character row is in percent (%): 1 × 10⁻²⁸ for 10⁻³⁰ e. The sentence reads “μ = 4.8032 D (1.60218 × 10⁻²⁹ C·m)”, “Q = 0.181088 e (2.90135 × 10⁻²⁰ C): 18.1088% ionic” or “r = 100 pm (1 Å)”.
Assumptions. Two equal and opposite point charges; a molecule’s dipole moment is the vector sum of its bond dipoles, which this page does not add up.
Worked examples by hand
One electron charge at 100 pm. μ = 1.602176634 × 10⁻¹⁹ C × 1 × 10⁻¹⁰ m = 1.60218 × 10⁻²⁹ C·m; ÷ 3.335641 × 10⁻³⁰ = 4.8032 D.
HCl (Averill and Eldredge). μ = 1.109 D = 1.109 × 3.335641 × 10⁻³⁰ = 3.69923 × 10⁻³⁰ C·m; r = 127.5 pm = 1.275 × 10⁻¹⁰ m. Q = 3.69923 × 10⁻³⁰ ÷ 1.275 × 10⁻¹⁰ = 2.90135 × 10⁻²⁰ C = 0.181088 e: 18.1088% ionic (book: 0.1811 e, about 18%).
NaCl (Example 11). Q = 9.001 × 3.335641 × 10⁻³⁰ ÷ 236.1 × 10⁻¹² = 1.27167 × 10⁻¹⁹ C = 0.793713 e: 79.3713% ionic. The book rounds Q to 1.272 × 10⁻¹⁹ C and gets 79.39%.
AgCl (exercise). Q = 6.08 × 3.335641 × 10⁻³⁰ ÷ 228.1 × 10⁻¹² = 0.554941 e: 55.4941% ionic (book: 55.5%).
Distance. μ = 1.60218 × 10⁻²⁹ C·m with Q = 1.60218 × 10⁻¹⁹ C: r = 10⁻¹⁰ m = 100 pm (1 Å).
Other questions people ask
How do I calculate a dipole moment?
Multiply the charge by the distance between the charges: μ = Q × r. A full electron charge (1.602 × 10⁻¹⁹ C) 100 pm from its opposite gives 1.602 × 10⁻¹⁹ × 1 × 10⁻¹⁰ = 1.602 × 10⁻²⁹ C·m, which is 4.80 D.
What is a debye?
The usual unit of dipole moment for molecules: 1 D = 3.335 641 × 10⁻³⁰ C·m (NIST). Molecular dipole moments are a few debyes; HCl’s is 1.109 D.
How do I calculate percent ionic character?
Divide the measured dipole moment by the one a full electron charge at the bond length would give, and multiply by 100. That is the same as the charge Q = μ ÷ r in units of e, times 100. HCl: 1.109 D at 127.5 pm gives Q = 0.181 e, so the bond is about 18% ionic.
Why does my percent ionic character differ from the textbook?
Textbooks round along the way. For NaCl (9.001 D, 236.1 pm), Averill and Eldredge round Q to 1.272 × 10⁻¹⁹ C and get 79.39%; without rounding, the page gives 79.3713%. They also use 1 D = 3.3356 × 10⁻³⁰ C·m and e = 1.6022 × 10⁻¹⁹ C.
Is a molecule’s dipole moment the same as its bond dipole?
Only for a two-atom molecule. A bigger molecule’s dipole moment is the vector sum of its bond dipoles, so symmetric molecules such as CO₂ have polar bonds but no overall dipole (OpenStax §7.6). This page works with one pair of charges.
What sets the size of the partial charges?
The electronegativity difference between the two atoms: the larger it is, the larger the partial charges and the dipole moment. HF has a larger dipole moment than CO, because the difference is larger (OpenStax §7.6).