1824 United States presidential election in Maryland
|Elections in Maryland|
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|
|Democratic-Republican||John Quincy Adams||14,632||44.05%||3|
|Democratic-Republican||William H. Crawford||3,364||10.13%||1|
- "1824 Presidential General Election Results - Maryland". U.S. Election Atlas. Retrieved 27 February 2013.