1824 United States presidential election in Rhode Island explained

See main article: 1824 United States presidential election.

Election Name:1824 United States presidential election in Rhode Island
Country:Rhode Island
Type:presidential
Ongoing:no
Previous Election:1820 United States presidential election in Rhode Island
Previous Year:1820
Next Election:1828 United States presidential election in Rhode Island
Next Year:1828
Election Date:October 26 – December 2, 1824
Image1:John Quincy Adams 1858 crop.jpg
Nominee1:John Quincy Adams
Party1:Democratic-Republican Party
Home State1:Massachusetts
Running Mate1:John C. Calhoun
Electoral Vote1:4
Popular Vote1:2,145
Percentage1:91.47%
Nominee2:William H. Crawford
Party2:Democratic-Republican Party
Home State2:Georgia
Running Mate2:Nathaniel Macon
Electoral Vote2:0
Popular Vote2:200
Percentage2:8.53%
Map Size:250px
President
Before Election:James Monroe
Before Party:Democratic-Republican Party
After Election:John Quincy Adams
After Party:Democratic-Republican Party

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 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[1]
PartyCandidateVotesPercentageElectoral votes
Democratic-RepublicanJohn Quincy Adams2,14591.47%4
Democratic-RepublicanWilliam H. Crawford2008.53%0
Totals2,345100.0%4

See also

Notes and References

  1. Web site: 1824 Presidential General Election Results - Rhode Island. U.S. Election Atlas. 27 February 2013.