1808 United States presidential election in Rhode Island

Main article: 1808 United States presidential election
1808 United States presidential election in Rhode Island
November 4 – December 7, 1808
 
Nominee Charles Cotesworth Pinckney James Madison
Party Federalist Democratic-Republican
Home state South Carolina Virginia
Running mate Rufus King George Clinton
Electoral vote 4 0
Popular vote 3,072 2,692
Percentage 53.30% 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.

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]
Party Candidate Votes Percentage Electoral votes
Federalist Charles Cotesworth Pinckney 3,072 53.30% 4
Democratic-Republican James Madison 2,692 46.70% –
Totals 5,764 100.00% 4

See also

References

  1. ^ "A New Nation Votes". elections.lib.tufts.edu. Retrieved August 31, 2024.
This article is issued from Wikipedia. The text is available under Creative Commons Attribution-Share Alike 4.0 unless otherwise noted. Additional terms may apply for the media files.