1800 United States presidential election in Kentucky
October 31 – December 11, 1800
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The 1800 United States presidential election in Kentucky took place between October 31 and December 11, 1800, as part of the 1800 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for President and Vice President.[1]
Kentucky cast four electoral votes for the Democratic-Republican candidate and incumbent Vice President Thomas Jefferson over the Federalist candidate and incumbent President John Adams. The electoral votes for Vice president were cast for Jefferson's running mate Aaron Burr from New York. The state was divided into two electoral districts with two electors each, whereupon each district's voters chose the electors.[2] The only true popular vote figures still extant today are those of Franklin County.[3]
Results
| 1800 United States presidential election in Kentucky[4][5][6][7] | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| Democratic-Republican | Thomas Jefferson | 75 | 63.00% | 4 | |
| Federalist | John Adams (incumbent) | 44 | 37.00% | 0 | |
| Totals | 119 | 100.00% | 4 | ||
See also
References
- ^ "A New Nation Votes". elections.lib.tufts.edu. Retrieved 2023-07-14.
- ^ "1800 Presidential General Election Results". U.S. Election Atlas. Retrieved 2023-07-14.
- ^ "A New Nation Votes". elections.lib.tufts.edu. Retrieved 2025-10-12.
- ^ "A New Nation Votes". elections.lib.tufts.edu. Retrieved 2025-10-12.
- ^ "A New Nation Votes". elections.lib.tufts.edu. Retrieved 2025-10-12.
- ^ "A New Nation Votes". elections.lib.tufts.edu. Retrieved 2025-10-12.
- ^ "A New Nation Votes". elections.lib.tufts.edu. Retrieved 2025-10-12.