United States presidential election in Maryland, 1824

United States presidential election in Maryland, 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 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%

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

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.