1836 United States presidential election in Illinois
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.
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Elections in Illinois |
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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%.
Results
1836 United States presidential election in Illinois[1] | |||||
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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
- "1836 Presidential General Election Results - Illinois". U.S. Election Atlas. Retrieved 4 August 2012.
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