1824 United States presidential election in Rhode Island

Main article: 1824 United States presidential election

1824 United States presidential election in Rhode Island

October 26 – December 2, 1824

Nominee John Quincy Adams William H. Crawford Party Democratic-Republican Democratic-Republican Home state Massachusetts Georgia Running mate John C. Calhoun Nathaniel Macon Electoral vote 4 0 Popular vote 2,145 200 Percentage 91.47% 8.53%

President before election James Monroe Democratic-Republican Elected President John Quincy Adams Democratic-Republican

The 1824 United States presidential election in Rhode Island took place between October 26 and December 2, 1824, as part of the 1824 United States presidential election. Voters chose 4 representatives, or electors to the Electoral College, who voted for President and Vice President.

During this election, the Democratic-Republican Party was the only major national party, and 4 different candidates from this party sought the Presidency. Rhode Island voted for John Quincy Adams over William H. Crawford, Henry Clay, and Andrew Jackson. Adams won Rhode Island by a margin of 82.94%.

Results

1824 United States presidential election in Rhode Island
Party		Candidate	Votes	Percentage	Electoral votes
	Democratic-Republican	John Quincy Adams	2,145	91.47%	4
	Democratic-Republican	William H. Crawford	200	8.53%	0
Totals			2,345	100.0%	4