1840 United States presidential election in Delaware
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| Elections in Delaware |
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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 3 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
| United States presidential election in Delaware, 1840[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% | ||||
References
- ^ "1840 Presidential General Election Results - Delaware". U.S. Election Atlas. Retrieved 23 December 2013.
State and district results of the 1840 United States presidential election | ||
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