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 3 representatives, or electors to the Electoral College, who voted for President and Vice President.
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Elections in Delaware | ||||||||||||
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During this election, the Democratic-Republican Party was the only major national party, and 4 different candidates from this party sought the Presidency. Delaware cast 2 electoral votes for William H. Crawford and 1 for John Quincy Adams.
Results
United States presidential election in Delaware, 1824[1] | |||||
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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
- "Electoral Votes for President and Vice President 1821-1837". National Archives and Records Administration. Retrieved 28 February 2013.
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