Main article: 1824 United States presidential election
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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 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
| 1824 United States presidential election in Delaware | |||||
|---|---|---|---|---|---|
| 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
References
- "Electoral Votes for President and Vice President 1821-1837". National Archives and Records Administration. Retrieved February 28, 2013.
| Elections in Delaware | |
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| General | |
| Delaware Senate | |
| Delaware House | |
| Governor |
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| U.S. President |
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| U.S. Senate Class 1 |
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| U.S. Senate Class 2 |
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| U.S. House |
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| Wilmington mayor | |
| New Castle County Executive | |
| See also: Political party strength in Delaware | |
| Elections in the United States | |
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| State and district results of the 1824 United States presidential election | ||
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