1824 United States presidential election in Alabama

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United States presidential election in Alabama, 1824

← 1820 October 26 – December 2, 1824 1828 →
  Andrew Jackson.jpg JohnQAdams.jpg WilliamHCrawford.jpg
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[edit]

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 13,423 100.0% 5

References[edit]

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