1844 United States presidential election in Vermont

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1844 United States presidential election in Vermont

← 1840 November 1 - December 4, 1844 1848 →
  Clay portrait.jpg JamesKnoxPolk.jpg James Birney by Asa Park.jpg
Nominee Henry Clay James K. Polk James Gillespie Birney
Party Whig Democratic Liberty
Home state Kentucky Tennessee Michigan
Running mate Theodore Frelinghuysen George M. Dallas Thomas Morris
Electoral vote 6 0 0
Popular vote 26,780 18,049 3,970
Percentage 54.84% 36.96% 8.13%

President before election

John Tyler

Elected President

James K. Polk

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%.

With 54.84% of the popular vote, Vermont would prove to be Henry Clay's second strongest state after Rhode Island. The Green Mountain State would also prove to be James G. Birney's third strongest state after New Hampshire and Massachusetts.[1]


1844 United States presidential election in Vermont[2]
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 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%


  1. ^ "1844 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
  2. ^ "1844 Presidential General Election Results - Vermont". U.S. Election Atlas. Retrieved 23 December 2013.