A presidential election was held in Rhode Island between October 31 and December 3, 1800, as part of the 1800 United States presidential election. Voters chose four 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 four Adams electors received more votes than the four Jefferson electors and the electoral vote was all for Adams in Rhode Island. Adams’s running mate Charles Cotesworth Pinckney received three electoral votes, and John Jay received one electoral vote. Rhode Island was the only state in the election of 1800 in which an elector “threw away” a vote by not voting for both candidates on a party’s ticket.
Results 1800 United States presidential election in Rhode Island PartyCandidateVotesPercentageElectoral votesFederalistJohn Adams (incumbent)2,35352.15%4Democratic-RepublicanThomas Jefferson2,15947.85%0Totals4,512100.0%4
Results by county +1800 United States presidential election in Rhode IslandCountyJohn AdamsFederalistThomas JeffersonDemocratic-RepublicanMarginTotalvotes#%#%#% Bristol21375.27%7024.73%14350.54%283 Kent35963.20%20936.80%15026.40%568 Newport45445.81%53754.19%-83-8.38%991 Providence1,00856.53%77543.47%23313.06%1,783 Washington31135.79%55864.21%-247-28.42%869Total2,34552.18%2,14947.82%1964.36%4,494
See also United States presidential elections in Rhode Island
References