1844 United States presidential election in Vermont
Appearance
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Elections in Vermont |
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The 1844 United States presidential election in Vermont took place between November 1 and December 4, 1844, as part of the 1844 United States presidential election. Voters chose six representatives, or electors to the Electoral College, who voted for President and Vice President.
Vermont voted for the Whig candidate, Henry Clay, over Democratic candidate James K. Polk and Liberty candidate James G. Birney. Clay won Vermont by a margin of 17.88%.
Results
United States presidential election in Vermont, 1844[1] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Whig | Henry Clay of Kentucky | Theodore Frelinghuysen of New York | 26,780 | 54.84% | 6 | 100.00% | ||
Democratic | James K. Polk of Tennessee | George M. Dallas of Pennsylvania | 18,049 | 36.96% | 0 | 0.00% | ||
Liberty Party (1840s) | James G. Birney of Michigan | Thomas Morris of Ohio | 3,970 | 8.13% | 0 | 0.00% | ||
N/A | Others | Others | 30 | 0.06% | 0 | 0.00% | ||
Total | 48,829 | 100.00% | 6 | 100.00% |
References
- ^ "1844 Presidential General Election Results - Vermont". U.S. Election Atlas. Retrieved 23 December 2013.