1808 United States presidential election in Kentucky
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Elections in Kentucky |
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Government |
The 1808 United States presidential election in Kentucky took place between 4 November and 7 December 1808, as part of the 1808 United States presidential election. Voters chose eight representatives, or electors to the Electoral College, who voted for President and Vice President.[1]
Kentucky cast seven electoral votes, as one elector didn't vote, for the Democratic-Republican candidate James Madison over the Federalist candidate Charles C. Pinckney. The electoral votes for Vice president were cast for Madison's running mate George Clinton from New York. The state was divided into two electoral districts with four electors each, whereupon each district's voters chose the electors.[2]
Results
1808 United States presidential election in Kentucky[3] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Democratic-Republican | James Madison | 2,679 | 98.02% | 7 | |
Federalist | Charles C. Pinckney | 54 | 1.98% | 0 | |
None | Not Cast | – | – | 1 | |
Totals | 2,733 | 100.0% | 8 |
See also
References
- ^ "A New Nation Votes". elections.lib.tufts.edu. Retrieved July 14, 2023.
- ^ "1808 Presidential General Election Results". U.S. Election Atlas. Retrieved July 14, 2023.
- ^ "1808 Presidential Election". 270towin.com. Retrieved July 14, 2023.
Categories:
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- Elections using electoral votes
- 1808 United States presidential election by state
- United States presidential elections in Kentucky
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- 1808 elections in the United States
- November 1808 events
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- 1808 in Kentucky