United States presidential election in Alabama, 1824

United States presidential election in Alabama, 1824

October 26 – December 2, 1824

Nominee Andrew Jackson John Quincy Adams William H. Crawford Party Democratic-Republican Democratic-Republican Democratic-Republican Home state Tennessee Massachusetts Georgia Running mate John C. Calhoun John C. Calhoun Nathaniel Macon Electoral vote 5 0 0 Popular vote 9,429 2,422 1,656 Percentage 69.32% 17.80% 12.17%

President before election James Monroe Democratic-Republican Elected President John Quincy Adams Democratic-Republican

The 1824 United States presidential election in Alabama took place between October 26 and December 2, 1824, as part of the 1824 United States presidential election. Voters chose five 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. Alabama voted for Andrew Jackson over John Quincy Adams, William H. Crawford and Henry Clay. Jackson won Alabama by a margin of 51.52%.

Results

United States presidential election in Alabama, 1824[1]
Party		Candidate	Votes	Percentage	Electoral votes
	Democratic-Republican	Andrew Jackson	9,429	69.32%	5
	Democratic-Republican	John Quincy Adams	2,422	17.80%	0
	Democratic-Republican	William H. Crawford	1,656	12.17%	0
	Democratic-Republican	Henry Clay	96	0.71%	0
Totals			13,603	100.0%	5

References

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