1840 United States presidential election in Delaware
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![]() County Results
Harrison 50-60%
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A presidential election was held in Delaware 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
[edit]1840 United States presidential election in Delaware[2] | ||||||||
---|---|---|---|---|---|---|---|---|
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% |
Results by county
[edit]County | William Henry Harrison
Whig |
Martin Van Buren
Democratic |
Various candidates
Other parties |
Margin | Total votes cast | ||||
# | % | # | % | # | % | # | % | ||
Kent | 1,593 | 59.20% | 1,095 | 40.69% | 3 | 0.11% | 498 | 18.51% | 2,691 |
New Castle | 2,321 | 51.28% | 2,195 | 48.50% | 10 | 0.22% | 126 | 2.78% | 4,526 |
Sussex | 2,053 | 56.48% | 1,582 | 43.52% | 0 | 0.00% | 471 | 12.96% | 3,635 |
Total: | 5,967 | 54.99% | 4,872 | 44.89% | 13 | 0.12% | 1,095 | 10.09% | 10,852 |
See also
[edit]References
[edit]- ^ Dubin, Michael J. (2002). United States Presidential Elections, 1788–1860: The Official Results by County and State. Jefferson, NC: McFarland & Co. p. xvi.
- ^ "1840 Presidential General Election Results - Delaware". U.S. Election Atlas. Retrieved December 23, 2013.
- ^ Burnham, Walter Dean (1955). Presidential ballots, 1836-1892. Internet Archive. Baltimore, Johns Hopkins Press. p. 320.