United States presidential election in Illinois, 1824

United States presidential election in Illinois, 1824

October 26 – December 2, 1824

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%   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 Democratic-Republican Elected President John Quincy Adams Democratic-Republican

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%.

Results

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

References

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