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1836 United States presidential election in Vermont

From Wikipedia, the free encyclopedia

1836 United States presidential election in Vermont

← 1832 November 3 - December 7, 1836 1840 →
William Henry Harrison by James Reid Lambdin, 1835 crop.jpg
Van Buren.jpg
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 35,031 14,037
Percentage 59.93% 40.07%

President before election

Andrew Jackson

Elected President

Martin Van Buren

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.

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  • ✪ The American Presidential Election of 1836
  • ✪ The American Presidential Election of 1928
  • ✪ The American Presidential Election of 1936
  • ✪ The American Presidential Election of 1860
  • ✪ The American Presidential Election of 1840



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%


  1. ^ "1836 Presidential General Election Results - Vermont". U.S. Election Atlas. Retrieved 23 December 2013.
This page was last edited on 23 December 2019, at 21:10
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