United States presidential election in Illinois, 1824

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United States presidential election in Illinois, 1824

← 1820 October 26 – December 2, 1824 1828 →
  Andrew Jackson.jpg JohnQAdams.jpg
Nominee Andrew Jackson John Quincy Adams
Party Democratic-Republican Democratic-Republican
Home state Tennessee Massachusetts
Running mate John C. Calhoun John C. Calhoun
Electoral vote 2 1
Popular vote 1,272 1,516
Percentage 27.23% 32.46%

  Henry Clay.JPG WilliamHCrawford.jpg
Nominee Henry Clay William H. Crawford
Party Democratic-Republican Democratic-Republican
Home state Kentucky Georgia
Running mate Nathan Sanford Nathaniel Macon
Electoral vote 0 0
Popular vote 1,036 847
Percentage 22.18% 18.13%

President before election

James Monroe

Elected President

John Quincy Adams

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.

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%.


United States presidential election in Illinois, 1824[1]
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


  1. ^ "1824 Presidential General Election Results - Illinois". U.S. Election Atlas. Retrieved 27 February 2013.