1840 United States presidential election in Delaware
(Redirected from United States presidential election in Delaware, 1840)
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.
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County Results
Harrison 50-60%
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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
edit1840 United States presidential election in Delaware[1] | ||||||||
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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
editCounty | 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
editReferences
edit- ^ "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.