1816 United States presidential election in Rhode Island

The 1816 United States presidential election in Rhode Island took place as part of the 1816 United States presidential election. Voters chose 4 representatives, or electors to the Electoral College who voted for president and vice president.

1816 United States presidential election in Rhode Island

← 1812 November 1 – December 4, 1816 1820 →
 
Nominee James Monroe
Party Democratic-Republican
Home state Virginia
Running mate Daniel D. Tompkins
Electoral vote 4
Popular vote 1,236
Percentage 100.00%

President before election

James Madison
Democratic-Republican

Elected President

James Monroe
Democratic-Republican

Rhode Island voted for the Democratic-Republican candidate, James Monroe. Monroe won Rhode Island by a margin of 100.00%. Because the Federalist challenger in the nation, Rufus King, was not on the ballot.

Results

edit
1816 United States presidential election in Rhode Island[1]
Party Candidate Votes Percentage Electoral votes
Democratic-Republican James Monroe 1,236 100.00% 4
Federalist Rufus King (not on ballot) 0
Totals 1,236 100.0% 4

See also

edit

References

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