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1836 United States presidential election in Delaware

From Wikipedia, the free encyclopedia

United States presidential election in Delaware, 1836

← 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 Johnson
Electoral vote 3 0
Popular vote 4,736 4,154
Percentage 53.24% 46.70%

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

Delaware voted for Whig candidate William Henry Harrison over the Democratic candidate, Martin Van Buren. Harrison won Delaware by a margin of 6.54%.

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  • ✪ The American Presidential Election of 2008
  • ✪ The American Presidential Election of 1880
  • ✪ The American Presidential Election of 1792
  • ✪ The American Presidential Election of 1988
  • ✪ The American Presidential Election of 1840



United States presidential election in Delaware, 1836[1]
Party Candidate Votes Percentage Electoral votes
Whig William Henry Harrison 4,736 53.24% 3
Democratic Martin Van Buren 4,154 46.70% 0
N/A Other 5 0.06% 0
Totals 8,895 100.00% 3


  1. ^ "1836 Presidential General Election Results - Delaware". U.S. Election Atlas. Retrieved 4 August 2012.

This page was last edited on 2 January 2020, at 01:45
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