The 1840 United States presidential election in Delaware was held on November 10, 1840 as part of the 1840 United States presidential election.[1] Voters chose three representatives, or electors to the Electoral College, who voted for President and Vice President.

Delaware voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Delaware by a margin of 10.1%.

Results

General Election Results[2][3]
Party Pledged to Elector Votes
Whig Party William Henry Harrison Peter F. Causey 5,967
Whig Party William Henry Harrison Benjamin Caulk 5,962
Whig Party William Henry Harrison Henry F. Hall 5,958
Democratic Party Martin Van Buren Thomas Jacobs 4,872
Democratic Party Martin Van Buren Christopher Vandergrift 4,871
Democratic Party Martin Van Buren Nehemiah Clarke 4,870
Write-in Scattering 13
Votes cast[a] 10,852

Results by county

County[2][3] William Henry Harrison
Whig		 Martin Van Buren
Democratic		 Margin Total votes cast[a]
# % # % # %
Kent 1,593 59.20% 1,095 40.69% 498 18.51% 2,691[b]
New Castle 2,321 51.28% 2,195 48.50% 126 2.78% 4,526[c]
Sussex 2,053 56.48% 1,582 43.52% 471 12.96% 3,635
Totals5,96754.99%4,87244.89%1,09510.09%10,852

Counties that flipped from Democratic to Whig

See also

Notes

  1. 1 2 Based on highest elector on each ticket
  2. ↑ Includes 3 Scattering votes
  3. ↑ Includes 10 Scattering votes

References

  1. ↑ Dubin, Michael J. (2002). United States Presidential Elections, 1788–1860: The Official Results by County and State. Jefferson, NC: McFarland & Co. p. xvi.
  2. 1 2 Governor's Register State of Delaware Volume One. Wilmington: Public Archives Commission of Delaware. 1926. pp. 379–380. Retrieved December 17, 2025.
  3. 1 2 The Whig Almanac and United States Register for 1845. New York: Greeley & McElrath. 1845. p. 55. Retrieved December 17, 2025.


source: https://en.wikipedia.org/wiki/1840_United_States_presidential_election_in_Delaware