1824 United States presidential election in Maryland

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1824 United States presidential election in Maryland

← 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 7 3 1
Popular vote 14,523 14,632 3,364
Percentage 43.73% 44.05% 10.13%

President before election

James Monroe

Elected President

John Quincy Adams

The 1824 United States presidential election in Maryland took place between October 26 and December 2, 1824, as part of the 1824 United States presidential election. Voters chose 11 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 Maryland voted for John Quincy Adams over Andrew Jackson, William H. Crawford and Henry Clay, only three electoral votes were assigned to Adams, while Jackson received seven and Crawford received one. Adams won Maryland by a very narrow margin of 0.32%.

This was also the first time in which the winner of the election didn't carry the state of Maryland.


1824 United States presidential election in Maryland[1]
Party Candidate Votes Percentage Electoral votes
Democratic-Republican Andrew Jackson 14,523 43.73% 7
Democratic-Republican John Quincy Adams 14,632 44.05% 3
Democratic-Republican William H. Crawford 3,364 10.13% 1
Democratic-Republican Henry Clay 695 2.09% 0
Totals 33,214 100.0% 11

See also[edit]


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