1824 United States presidential election in Delaware
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| Elections in Delaware |
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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.
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.
State and district results of the 1824 United States presidential election | ||
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