1824 United States presidential election in North Carolina explained

See main article: 1824 United States presidential election.

Election Name:1824 United States presidential election in North Carolina
Country:North Carolina
Type:presidential
Ongoing:no
Previous Election:1820 United States presidential election in North Carolina
Previous Year:1820
Next Election:1828 United States presidential election in North Carolina
Next Year:1828
Election Date:October 26 – December 2, 1824
Image1:Andrew Jackson.jpg
Nominee1:Andrew Jackson
Party1:Democratic-Republican Party
Home State1:Tennessee
Running Mate1:John C. Calhoun
Electoral Vote1:15
Popular Vote1:20,231
Percentage1:56.03%
Nominee2:William H. Crawford
Party2:Democratic-Republican Party
Home State2:Georgia
Running Mate2:Nathaniel Macon
Electoral Vote2:0
Popular Vote2:15,622
Percentage2:43.26%
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 North Carolina took place between October 26 and December 2, 1824, as part of the 1824 United States presidential election. Voters chose 15 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. North Carolina voted for Andrew Jackson over William H. Crawford, Henry Clay, and John Quincy Adams. Jackson won North Carolina by a margin of 12.77%.

Results

1824 United States presidential election in North Carolina[1]
PartyCandidateVotesPercentageElectoral votes
Democratic-RepublicanAndrew Jackson20,23156.03%15
Democratic-RepublicanWilliam H. Crawford15,62243.26%0
N/AOthers2560.71%0
Totals36,109100.0%15

Notes and References

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