1808 United States presidential election in Rhode Island

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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.

Quick facts Nominee, Party ...
1808 United States presidential election in Rhode Island

← 1804
November 4 – December 7, 1808
1812 â†’
 
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

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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

More information Party, Candidate ...
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
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See also

References

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