1824 United States presidential election 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.

1824 United States presidential election in 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

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

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

1824 United States presidential election in Delaware[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

See also

• United States presidential elections in Delaware

References

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