1828 United States presidential election in Missouri explained

See main article: 1828 United States presidential election.

Election Name:1828 United States presidential election in Missouri
Country:Missouri
Type:presidential
Ongoing:no
Previous Election:1824 United States presidential election in Missouri
Previous Year:1824
Next Election:1832 United States presidential election in Missouri
Next Year:1832
Election Date:October 31 – December 2, 1828
Image1:Andrew Jackson.jpg
Nominee1:Andrew Jackson
Party1:Democratic Party (United States)
Home State1:Tennessee
Running Mate1:John C. Calhoun
Electoral Vote1:3
Popular Vote1:8,232
Percentage1:70.64%
Nominee2:John Quincy Adams
Party2:National Republican Party (United States)
Home State2:Massachusetts
Running Mate2:Richard Rush
Electoral Vote2:0
Popular Vote2:3,422
Percentage2:29.36%

The 1828 United States presidential election in Missouri took place between October 31 and December 2, 1828, as part of the 1828 United States presidential election. Voters chose three representatives, or electors to the Electoral College, who voted for President and Vice President.

Missouri voted for the Democratic candidate, Andrew Jackson, over the National Republican candidate, John Quincy Adams. Jackson won Missouri by a margin of 41.28%.

Results

1828 United States presidential election in Missouri[1]
PartyCandidateVotesPercentageElectoral votes
DemocraticAndrew Jackson8,23270.64%3
National RepublicanJohn Quincy Adams (incumbent)3,42229.36%0
Totals11,654100.0%3

See also

Notes and References

  1. Web site: 1828 Presidential General Election Results - Missouri. U.S. Election Atlas. 28 February 2013.