1860 United States presidential election in Missouri

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.

Main article: 1860 United States presidential election
1860 United States presidential election in Missouri
November 2, 1860
 
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%
 
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

Elections in Missouri

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 win the popular vote and three of the seven electoral votes from New Jersey under a fusion ticket.

As of the 2016 presidential election, this is the last occasion when Putnam County voted for a Democratic presidential candidate and the last occasion when Taney County did not vote for the Republican candidate.[1]

Results
1860 United States presidential election in Missouri[2]
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%
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References
  1. Menendez, Albert J.; The Geography of Presidential Elections in the United States, 1868-2004, pp. 239-246 ISBN 0786422173
  2. "1860 Presidential Election Results – Missouri".


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