1800 United States presidential election in Rhode Island
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| Elections in Rhode Island |
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The 1800 United States presidential election in Rhode Island took place as part of the 1800 United States presidential election. Voters chose 4 representatives, or electors to the Electoral College who voted for president and vice president.
Rhode Island voted for the Federalist candidate, John Adams, over the Democratic-Republican candidate, Thomas Jefferson. Adams won Rhode Island by a margin of 4.3%. All 4 Adams electors received more votes than the 4 Jefferson electors and the electoral vote was all for Adams in Rhode Island. Adams’s running mate Charles Cotesworth Pinckney received three electoral votes and John Jay received one electoral vote. Rhode Island was the only state in the election of 1800 in which an elector “threw away” a vote by not voting for both candidates on a party’s ticket.
Results
[edit]| 1800 United States presidential election in Rhode Island [1] | |||||
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| Party | Candidate | Votes | Percentage | Electoral votes | |
| Federalist | John Adams (incumbent) | 2,353 | 52.15% | 4 | |
| Democratic-Republican | Thomas Jefferson | 2,159 | 47.85% | 0 | |
| Totals | 4,512 | 100.0% | 4 | ||
See also
[edit]References
[edit]- ^ "A New Nation Votes". elections.lib.tufts.edu. Retrieved April 25, 2022.
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| Other 1800 elections | ||
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