1840 United States presidential election in Delaware

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United States presidential election in Delaware, 1840

← 1836 October 30 - December 2, 1840 1844 →
  Harrison crop.jpg Van Buren.jpg
Nominee William Henry Harrison Martin Van Buren
Party Whig Democratic
Home state Ohio New York
Running mate John Tyler none
Electoral vote 3 0
Popular vote 5,967 4,872
Percentage 54.99% 44.89%

President before election

Martin Van Buren

Elected President

William Henry Harrison

The 1840 United States presidential election in Delaware took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. 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%.


United States presidential election in Delaware, 1840[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Whig William Henry Harrison of Ohio John Tyler of Virginia 5,967 54.99% 3 100.00%
Democratic Martin Van Buren of New York Richard M. Johnson of Kentucky 4,872 44.89% 0 0.00%
N/A Others Others 13 0.12% 0 0.00%
Total 10,852 100.00% 3 100.00%


  1. ^ "1840 Presidential General Election Results - Delaware". U.S. Election Atlas. Retrieved 23 December 2013.

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