1804 United States presidential election in Rhode Island

The 1804 United States presidential election in Rhode Island took place as part of the 1804 United States presidential election. Voters chose 4 representatives, or electors to the Electoral College who voted for president and vice president.

1804 United States presidential election in Rhode Island

← 1800 November 2 - December 5, 1804 1808 →
 
Nominee Thomas Jefferson Charles C. Pinckney
(not on ballot)
Party Democratic-Republican Federalist
Home state Virginia South Carolina
Running mate George Clinton Rufus King
Electoral vote 4 0
Popular vote 1,312
Percentage 100.00%

President before election

Thomas Jefferson
Democratic-Republican

Elected President

Thomas Jefferson
Democratic-Republican

Rhode Island voted for the Democratic-Republican candidate, Thomas Jefferson, over the Federalist candidate, Charles C. Pinckney. Jefferson won Rhode Island by a margin of 100.00%, the largest margin and landslide victory ever since in the state. Because Pinckney was not on the ballot.

Results

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1804 United States presidential election in Rhode Island[1]
Party Candidate Votes Percentage Electoral votes
Democratic-Republican Thomas Jefferson (incumbent) 1,312 100.00% 4
Federalist Charles C. Pinckney
(not on ballot)
0
Totals 1,312 100.0% 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.