United States presidential election in Rhode Island, 1824

United States presidential election in Rhode Island, 1824

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 four 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 four 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

United States presidential election in Rhode Island, 1824[1]
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

References

1. "1824 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 27 February 2013.