1860 United States presidential election in Missouri

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United States presidential election in Missouri, 1860

← 1856 November 2, 1860 1864 →
  BradyHandy-StephenADouglas restored.jpg John-bell-brady-handy-cropped restored.jpg
Nominee Stephen A. Douglas John Bell
Party Democratic Constitutional Union
Home state Illinois Tennessee
Running mate Herschel Vespasian Johnson Edward Everett
Electoral vote 9 0
Popular vote 58,801 58,372
Percentage 35.52% 35.26%

  John C Breckinridge-04775-restored.jpg Abraham Lincoln O-26 by Hesler, 1860 (cropped).jpg
Nominee John C. Breckinridge Abraham Lincoln
Party Southern Democratic Republican
Home state Kentucky Illinois
Running mate Joseph Lane Hannibal Hamlin
Electoral vote 0 0
Popular vote 31,362 17,028
Percentage 18.94% 10.28%

President before election

James Buchanan
Democratic

Elected President

Abraham Lincoln
Republican

The 1860 United States presidential election in Missouri took place on November 2, 1860, as part of the 1860 United States presidential election. Voters chose nine representatives, or electors to the Electoral College, who voted for president and vice president.

Missouri was won by Democratic candidate, Stephen A. Douglas. He won the state by a very narrow margin of 0.26%. The state was the only one to fully give its votes to Douglas, though he would the popular vote and three of the seven electoral votes from New Jersey under a fusion ticket.

Results[edit]

United States presidential election in Missouri, 1860[1]
Party Candidate Votes %
Democratic Stephen A. Douglas 58,801 35.73%
Constitutional Union John Bell 58,372 35.26%
Southern Democratic John C. Breckinridge 31,362 18.94%
Republican Abraham Lincoln 17,028 10.28%
Total votes 165,563 100%

References[edit]

  1. ^ "1860 Presidential Election Results Missouri".