1808 United States presidential election in Rhode Island

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1808 United States presidential election in Rhode Island
Flag of Rhode Island.svg
  1804 November 4 – December 7, 1808 1812  
  CharlesCPinckney (cropped).png James Madison.jpg
Nominee Charles Cotesworth Pinckney James Madison
Party Federalist Democratic-Republican
Home state South Carolina Virginia
Running mate Rufus King George Clinton
Electoral vote40
Popular vote3,0722,692
Percentage53.30%46.70%

President before election

Thomas Jefferson
Democratic-Republican

Elected President

James Madison
Democratic-Republican

The 1808 United States presidential election in Rhode Island took place as part of the 1808 United States presidential election. Voters chose 4 representatives, or electors to the Electoral College who voted for president and vice president.

Contents

Rhode Island voted for the Federalist candidate, Charles Cotesworth Pinckney, over the Democratic-Republican candidate, James Madison. Pinckney won Rhode Island by a margin of 53.30%.

Results

1808 United States presidential election in Rhode Island [1]
PartyCandidateVotesPercentageElectoral votes
Federalist Charles Cotesworth Pinckney 3,07253.30%4
Democratic-Republican James Madison 2,69246.70%
Totals5,764100.00%4

See also

References

  1. "A New Nation Votes". elections.lib.tufts.edu. Retrieved August 31, 2024.