1844 United States presidential election in Vermont

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.

1844 United States presidential election in Vermont

← 1840 November 1 - December 4, 1844 1848 →
  Clay 1848.jpg Polk 1849.jpg James Birney(Cropped).jpg
Nominee Henry Clay James K. Polk James G. 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
Independent

Elected President

James K. Polk
Democratic

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]

ResultsEdit

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%

See alsoEdit

ReferencesEdit

  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.