1804 United States presidential election in Delaware

A presidential election was held in Delaware on November 12, 1804, as part of the 1804 United States presidential election.^[1] The Federalist Party's ticket defeated the Democratic-Republican Party's ticket in the Delaware General Assembly.^[2] The Federalist electors voted for former U.S. minister to France Charles Cotesworth Pinckney and former U.S. minister to Great Britain Rufus King over the incumbent president Thomas Jefferson and the former New York governor George Clinton.^[3]

Quick facts Nominee, Party ...

1804 United States presidential election in Delaware

← 1800 November 12, 1804 1808 â†’

← NY MD â†’

Nominee Charles Cotesworth Pinckney Thomas Jefferson Party Federalist Democratic-Republican Home state South Carolina Virginia Running mate Rufus King George Clinton Electoral vote 3 0 Legislative vote 18 5 Percentage 78.2% 21.7%

President before election Thomas Jefferson Democratic-Republican Elected President Thomas Jefferson Democratic-Republican

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Jefferson won the national election in a landslide. Delaware was one of two states to back the losing Federalist candidates, in addition to Connecticut and two single-member districts in Maryland.^[4]