1800 United States presidential election in Rhode Island

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1800 United States presidential election in Rhode Island
Flag of Rhode Island.svg
  1796 October 31 – December 3, 1800 1804  
  Gilbert Stuart, John Adams, c. 1800-1815, NGA 42933.jpg Thomas Jefferson by Rembrandt Peale, 1800.jpg
Nominee John Adams Thomas Jefferson
Party Federalist Democratic-Republican
Home state Massachusetts Virginia
Electoral vote40
Popular vote2,3532,159
Percentage52.15%47.85%

  CharlesCPinckney crop.jpg Gilbert Stuart, John Jay, 1794, NGA 75023.jpg
Nominee Charles Cotesworth Pinckney John Jay
Party Federalist Federalist
Home state South Carolina New York
Electoral vote31

Rhode Island Presidential Election Results 1800.svg
County results

President before election

John Adams
Federalist

Elected President

Thomas Jefferson
Democratic-Republican

The 1800 United States presidential election in Rhode Island took place 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.

Contents

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 four Adams electors received more votes than the four 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

1800 United States presidential election in Rhode Island [1]
PartyCandidateVotesPercentageElectoral votes
Federalist John Adams (incumbent)2,35352.15%4
Democratic-Republican Thomas Jefferson 2,15947.85%0
Totals4,512100.0%4

See also

    References

    1. "A New Nation Votes". elections.lib.tufts.edu. Retrieved April 25, 2022.