1832 United States presidential election in Louisiana explained

See main article: 1832 United States presidential election.

Election Name:1832 United States presidential election in Louisiana
Country:Louisiana
Type:presidential
Ongoing:no
Previous Election:1828 United States presidential election in Louisiana
Previous Year:1828
Next Election:1836 United States presidential election in Louisiana
Next Year:1836
Election Date:November 2 – December 5, 1832
Image1:Andrew jackson head.jpg
Nominee1:Andrew Jackson
Party1:Democratic Party (United States)
Home State1:Tennessee
Running Mate1:Martin Van Buren
Electoral Vote1:5
Popular Vote1:3,908
Percentage1:61.67%
Nominee2:Henry Clay
Party2:National Republican Party (United States)
Home State2:Kentucky
Running Mate2:John Sergeant
Electoral Vote2:0
Popular Vote2:2,429
Percentage2:38.33%
President
Before Election:Andrew Jackson
Before Party:Democratic Party (United States)
Before Color:FFF402
After Election:Andrew Jackson
After Party:Democratic Party (United States)
After Color:FFF402

The 1832 United States presidential election in Louisiana took place between November 2 and December 5, 1832, as part of the 1832 United States presidential election. Voters chose five representatives, or electors to the Electoral College, who voted for President and Vice President.

Louisiana voted for the Democratic Party candidate, Andrew Jackson, over the National Republican candidate, Henry Clay. Jackson won Louisiana by a margin of 23.34%.

Results

1832 United States presidential election in Louisiana[1]
PartyCandidateVotesPercentageElectoral votes
DemocraticAndrew Jackson (incumbent)3,90861.67%5
width: 3px" National RepublicanHenry Clay2,42938.33%0
Totals6,337100.0%5

See also

Notes and References

  1. Web site: 1832 Presidential General Election Results - Louisiana. U.S. Election Atlas. 12 April 2013.