1824 United States presidential election in Rhode Island

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1824 United States presidential election in Rhode Island
Flag of Rhode Island.svg
  1820 October 26 – December 2, 1824 1828  
  John Quincy Adams 1858 crop.jpg WilliamHCrawford.jpg
Nominee John Quincy Adams William H. Crawford
Party Democratic-Republican Democratic-Republican
Home state Massachusetts Georgia
Running mate John C. Calhoun Nathaniel Macon
Electoral vote40
Popular vote2,145200
Percentage91.47%8.53%

Rhode Island Presidential Election Results 1824.svg
County Results
Adams
  80–90%
  90–100%

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.

Contents

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-Republican John Quincy Adams 2,14591.47%4
Democratic-Republican William H. Crawford 2008.53%0
Totals2,345100.0%4

See also

References

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