1836 United States presidential election in Vermont

Election in Vermont

1836 United States presidential election in Vermont

← 1832 November 3 - December 7, 1836 1840 →
 
Nominee William Henry Harrison Martin Van Buren
Party Whig Democratic
Home state Ohio New York
Running mate Francis Granger Richard Mentor Johnson
Electoral vote 7 0
Popular vote 20,994 14,037
Percentage 59.93% 40.07%

County results

Harrison

  50-60%
  60-70%
  70-80%

Van Buren

  50-60%


President before election

Andrew Jackson
Democratic

Elected President

Martin Van Buren
Democratic

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The 1836 United States presidential election in Vermont took place between November 3 and December 7, 1836, as part of the 1836 United States presidential election. Voters chose seven representatives, or electors to the Electoral College, who voted for President and Vice President.

Vermont voted for Whig candidate William Henry Harrison over Democratic candidate Martin Van Buren. Harrison won Vermont by a margin of 19.86%.

This would be the final time a Democratic candidate would carry Essex County until Franklin D. Roosevelt won it 104 years later in 1940.

1836 would stand as the strongest performance for a Democratic candidate in Vermont until 96 years later in 1932, when Franklin D. Roosevelt performed slightly better with 41.08%.

Harrison would later win Vermont again four years later when he successfully defeated Van Buren.

Results

1836 United States presidential election in Vermont[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Whig William Henry Harrison of Ohio Francis Granger of New York 20,994 59.93% 7 100.00%
Democratic Martin Van Buren of New York Richard M. Johnson of Kentucky 14,037 40.07% 0 0.00%
Total 35,031 100.00% 7 100.00%

See also

References

  1. ^ "1836 Presidential General Election Results - Vermont". U.S. Election Atlas. Retrieved December 23, 2013.