1840 United States presidential election in Vermont
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 County Results Harrison 50–60% 60–70% 70–80%
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| Elections in Vermont |
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The 1840 United States presidential election in Vermont took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose seven representatives, or electors to the Electoral College, who voted for President and Vice President.
Vermont voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Vermont by a margin of 28.43%.
Harrison's 28.43% margin of victory made it his strongest victory in the election while he carried 63.90% of the popular vote made Vermont his second strongest state after Kentucky.[1]
Harrison had previously won Vermont against Van Buren four years earlier.
Results
[edit]| 1840 United States presidential election in Vermont[2] | ||||||||
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| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Whig | William Henry Harrison of Ohio | John Tyler of Virginia | 32,445 | 63.90% | 7 | 100.00% | ||
| Democratic | Martin Van Buren of New York | Richard M. Johnson of Kentucky | 18,009 | 35.47% | 0 | 0.00% | ||
| Liberty | James G. Birney of New York | Thomas Earle of Pennsylvania | 319 | 0.63% | 0 | 0.00% | ||
| Total | 50,773 | 100.00% | 7 | 100.00% | ||||
See also
[edit]References
[edit]- ^ "1840 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
- ^ "1840 Presidential General Election Results - Vermont". U.S. Election Atlas. Retrieved December 23, 2013.
State and district results of the 1840 United States presidential election | ||
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