1804 United States presidential election in Delaware
November 2 – December 5, 1804
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The 1804 United States presidential election in Delaware took place between November 2 and December 5, 1804, as part of the 1804 United States presidential election. Voters chose three representatives, or electors to the Electoral College, who voted for President and Vice President.
Delaware cast three electoral votes for the Federalist candidate Charles Cotesworth Pinckney over the Democratic-Republican candidate and incumbent President Thomas Jefferson. These electors were elected by the Delaware General Assembly, the state legislature, rather than by popular vote. The three electoral votes for Vice president were cast for Pinckney's running mate Rufus King from New York.[1]
Results
| 1804 United States presidential election in Delaware[2] | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| Federalist | Charles Cotesworth Pinckney | – | – | 3 | |
| Democratic-Republican | Thomas Jefferson (incumbent) | – | – | 0 | |
| Totals | – | – | 3 | ||
See also
References
- ^ "1804 Presidential General Election Results". U.S. Election Atlas. Retrieved 2023-07-09.
- ^ "1804 Presidential Election". 270towin.com. Retrieved 2023-07-09.