1808 United States presidential election in Rhode Island

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.

1808 United States presidential election in Rhode Island

← 1804 November 4 – December 7, 1808 1812 →
 
Nominee Charles C. 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

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

Results

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1808 United States presidential election in New Hampshire[1]
Party Candidate Votes Percentage Electoral votes
Federalist Charles C. Pinckney 3,072 53.30% 4
Democratic-Republican James Madison 2,692 46.70%
Totals 5,764 100.00% 4

See also

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References

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  1. ^ "A New Nation Votes". elections.lib.tufts.edu. Retrieved 2024-08-31.