1824 United States presidential election in Mississippi explained

See main article: 1824 United States presidential election.

Election Name:1824 United States presidential election in Mississippi
Country:Mississippi
Flag Year:1822
Type:presidential
Ongoing:no
Previous Election:1820 United States presidential election in Mississippi
Previous Year:1820
Next Election:1828 United States presidential election in Mississippi
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 Vote1:3
Popular Vote1:3,121
Percentage1:63.77%
Nominee2:John Quincy Adams
Party2:Democratic-Republican Party
Home State2:Massachusetts
Running Mate2:John C. Calhoun
Electoral Vote2:0
Popular Vote2:1,654
Percentage2:33.80%
Map Size:250
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 Mississippi took place between October 26 and December 2, 1824, as part of the 1824 United States presidential election. Voters chose three 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. Mississippi voted for Andrew Jackson over John Quincy Adams, William H. Crawford and Henry Clay. Jackson won Mississippi by a margin of 29.97%.

Results

United States presidential election in Mississippi, 1824[1]
PartyCandidateVotesPercentageElectoral votes
Democratic-RepublicanAndrew Jackson3,12163.77%3
Democratic-RepublicanJohn Quincy Adams1,65433.80%0
Democratic-RepublicanWilliam H. Crawford1192.43%0
Totals4,894100.0%3

Notes and References

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