United States presidential election in Delaware, 1824

United States presidential election in Delaware, 1824
Delaware
October 26 – December 2, 1824
 
Nominee William H. Crawford John Quincy Adams
Party Democratic-Republican Democratic-Republican
Home state Georgia Massachusetts
Running mate Nathaniel Macon John C. Calhoun
Electoral vote 2 1
Elections in Delaware

The 1824 United States presidential election in Delaware took place between October 26 and December 2, 1824, as part of the 1824 United States presidential election. Voters chose three 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. Delaware cast two electoral votes for William H. Crawford and one for John Quincy Adams.

Results

United States presidential election in Delaware, 1824[1]
Party Candidate Votes Percentage Electoral votes
Democratic-Republican William H. Crawford 2
Democratic-Republican John Quincy Adams 1
Democratic-Republican Henry Clay 0
Democratic-Republican Andrew Jackson 0
Totals 3

References

  1. ↑ "Electoral Votes for President and Vice President 1821-1837". National Archives and Records Administration. Retrieved 28 February 2013.
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