Jump to content

1800 United States presidential election in Rhode Island

From Wikipedia, the free encyclopedia

1800 United States presidential election in Rhode Island

← 1796 October 31 - December 3, 1800 1804 →
 
Nominee John Adams Thomas Jefferson
Party Federalist Democratic-Republican
Home state Massachusetts Virginia
Electoral vote 4 0
Popular vote 2,353 2,159
Percentage 52.15% 47.85%

 
Nominee Charles C. Pinckney John Jay
Party Federalist Federalist
Home state South Carolina New York
Electoral vote 3 1

County Results

President before election

John Adams
Federalist

Elected President

Thomas Jefferson
Democratic-Republican

The 1800 United States presidential election in Rhode Island took place as part of the 1800 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 Federalist candidate, John Adams, over the Democratic-Republican candidate, Thomas Jefferson. Adams won Rhode Island by a margin of 4.3%. All 4 Adams electors received more votes than the 4 Jefferson electors and the electoral vote was all for Adams in Rhode Island.

Results

[edit]
1800 United States presidential election in Rhode Island [1]
Party Candidate Votes Percentage Electoral votes
Federalist John Adams (incumbent) 2,353 52.15% 4
Democratic-Republican Thomas Jefferson 2,159 47.85% 0
Totals 4,512 100.0% 4

See also

[edit]

References

[edit]
  1. ^ "A New Nation Votes". elections.lib.tufts.edu. Retrieved April 25, 2022.