1816 United States presidential election in Kentucky
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The 1816 United States presidential election in Kentucky took place between 1 November and 4 December 1816, as part of the 1816 United States presidential election. Voters chose twelve representatives, or electors to the Electoral College, who voted for President and Vice President.[1]
Kentucky cast twelve electoral votes for the Democratic-Republican candidate James Monroe over the Federalist candidate Rufus King. The electoral votes for Vice president were cast for Monroe's running mate Daniel D. Tompkins from New York. The state was divided into three electoral districts with four electors each, whereupon each district's voters chose the electors.[2]
Results
| 1816 United States presidential election in Kentucky[3] | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| Democratic-Republican | James Monroe | 1,864 | 100.0% | 12 | |
| Federalist | Rufus King | 0 | 0.00% | 0 | |
| Totals | 1,864 | 100.0% | 12 | ||
See also
References
- ^ "A New Nation Votes". elections.lib.tufts.edu. Retrieved July 14, 2023.
- ^ "1816 Presidential General Election Results". U.S. Election Atlas. Retrieved July 14, 2023.
- ^ "1816 Presidential Election". 270towin.com. Retrieved July 14, 2023.
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