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

← 1820 October 26 – December 2, 1824 1828 →
 
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

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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

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References

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