1819 Delaware gubernatorial election

The 1819 Delaware gubernatorial election was held on October 5, 1819. Incumbent Federalist Governor John Clark was unable to seek re-election due to term limits. State Senator Henry Molleston ran as Clark's successor, winning the Federalist nomination. He faced Manaen Bull, Clark's 1816 opponent, and the Democratic-Republican nominee. Molleston won by a fairly wide margin, but died on November 11, 1819, prior to assuming office. State Senate President Jacob Stout became Governor and a special election was held in 1820.[1]

1819 Delaware gubernatorial election

← 1816 October 5, 1819 1820 (special) →
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Nominee Henry Molleston Manaen Bull
Party Federalist Democratic-Republican
Popular vote 3,823 3,185
Percentage 54.55% 45.45%

Governor before election

John Clark
Federalist

Elected Governor

Henry Molleston
Federalist

General electionEdit

ResultsEdit

1819 Delaware gubernatorial election[2]
Party Candidate Votes % ±%
Federalist Henry Molleston 3,823 54.55% +1.29%
Democratic-Republican Manaen Bull 3,185 45.45% -1.29%
Majority 638 9.10% +2.58%
Turnout 7,008 100.00%
Federalist hold

ReferencesEdit

  1. ^ Conrad, Henry C. (1908). History of the State of Delaware: From the Earliest Settlements to the Year 1907. Vol. 1. Lancaster, Pa.: Wickersham Company, Printers and Binders. p. 176.
  2. ^ "Delaware 1819 Governor". A New Nation Votes: American Election Returns 1787-1825. Tufts University. Retrieved June 17, 2021.

BibliographyEdit

  • Gubernatorial Elections, 1787-1997. Washington, D.C.: Congressional Quarterly Inc. 1998. ISBN 1-56802-396-0.
  • Glashan, Roy R. (1979). American Governors and Gubernatorial Elections, 1775-1978. Meckler Books. ISBN 0-930466-17-9.
  • Dubin, Michael J. (2003). United States Gubernatorial Elections, 1776-1860: The Official Results by State and County. Jefferson, North Carolina: McFarland. ISBN 978-0-7864-1439-0.