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1836 United States presidential election in Illinois

From Wikipedia, the free encyclopedia

1836 United States presidential election in Illinois

← 1832 November 3 – December 7, 1836 1840 →
 
Van Buren.jpg
William Henry Harrison by James Reid Lambdin, 1835 crop.jpg
Nominee Martin Van Buren William Henry Harrison
Party Democratic Whig
Home state New York Ohio
Running mate Richard Johnson Francis Granger
Electoral vote 5 0
Popular vote 18,369 15,220
Percentage 54.69% 45.31%

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

Illinois voted for the Democratic candidate, Martin Van Buren, over Whig candidate William Henry Harrison. Van Buren won Illinois by a margin of 9.38%.

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Transcription

Results

1836 United States presidential election in Illinois[1]
Party Candidate Votes Percentage Electoral votes
Democratic Martin Van Buren 18,369 54.69% 5
Whig William Henry Harrison 15,220 45.31% 0
Totals 33,589 100.0% 5

References

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


This page was last edited on 21 January 2020, at 19:18
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