United States presidential election in Maryland, 1824

United States presidential election in Maryland, 1824
Maryland
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 7 3
Popular vote 14,523 14,632
Percentage 43.73% 44.05%
 
Nominee William H. Crawford Henry Clay
Party Democratic-Republican Democratic-Republican
Home state Georgia Kentucky
Running mate Nathaniel Macon Nathan Sanford
Electoral vote 1 0
Popular vote 3,364 695
Percentage 10.13% 2.09%

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 eleven 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 margin of 0.32%.

Results

United States presidential election in Maryland, 1824[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% 0
Democratic-Republican Henry Clay 695 2.09% 0
Totals 33,214 100.0% 11

References

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