1816 United States presidential election in Rhode Island explained

See main article: 1816 United States presidential election.

Election Name:1816 United States presidential election in Rhode Island
Country:Rhode Island
Type:presidential
Ongoing:no
Previous Election:1812 United States presidential election in Rhode Island
Previous Year:1812
Next Election:1820 United States presidential election in Rhode Island
Next Year:1820
Election Date:November 1 – December 4, 1816
Image1:John Vanderlyn - James Monroe - Google Art Project.jpg
Nominee1:James Monroe
Party1:Democratic-Republican Party
Home State1:Virginia
Running Mate1:Daniel D. Tompkins
Electoral Vote1:4
Popular Vote1:1,236
Percentage1:100.00%
President
Before Election:James Madison
Before Party:Democratic-Republican Party
After Election:James Monroe
After Party:Democratic-Republican Party
Flag Year:1816

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.

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

1816 United States presidential election in Rhode Island[1]
PartyCandidateVotesPercentageElectoral votes
Democratic-RepublicanJames Monroe1,236100.00%4
FederalistRufus King (not on ballot)0
Totals1,236100.0%4

See also

References

  1. Web site: A New Nation Votes . 2024-08-31 . elections.lib.tufts.edu.