1824 United States presidential election in Illinois
The 1824 United States presidential election in Illinois took place between October 26 and December 2, 1824, as part of the 1824 United States presidential election. Voters chose three 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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During this election, the Democratic-Republican Party was the only major national party, and four different candidates from this party sought the Presidency. Although Illinois voted for John Quincy Adams over Andrew Jackson, Henry Clay, and William H. Crawford, only one of the state's electoral votes were assigned to Adams, while the remaining two were assigned to Jackson. Adams won Illinois by a margin of 5.23%.
Results
1824 United States presidential election in Illinois[1] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Democratic-Republican | Andrew Jackson | 1,272 | 27.23% | 2 | |
Democratic-Republican | John Quincy Adams | 1,516 | 32.46% | 1 | |
Democratic-Republican | Henry Clay | 1,036 | 22.18% | 0 | |
Democratic-Republican | William H. Crawford | 847 | 18.13% | 0 | |
Totals | 4,671 | 100.0% | 3 | ||
References
- "1824 Presidential General Election Results - Illinois". U.S. Election Atlas. Retrieved 27 February 2013.
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