1840 United States presidential election in Vermont

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.

1840 United States presidential election in Vermont

← 1836 October 30 - December 2, 1840 1844 →
  William Henry Harrison crop.jpg Martin Van Buren circa 1837 crop.jpg
Nominee William Henry Harrison Martin Van Buren
Party Whig Democratic
Home state Ohio New York
Running mate John Tyler none
Electoral vote 7 0
Popular vote 32,445 18,009
Percentage 63.90% 35.47%

Vermont Presidential Election Results 1840.svg
County Results
Harrison
  50–60%
  60–70%
  70–80%


President before election

Martin Van Buren
Democratic

Elected President

William Henry Harrison
Whig

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 the Green Mountain State against Van Buren four years earlier.

ResultsEdit

1840 United States presidential election in Vermont[2]
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 alsoEdit

ReferencesEdit

  1. ^ "1840 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
  2. ^ "1840 Presidential General Election Results - Vermont". U.S. Election Atlas. Retrieved 23 December 2013.