United States presidential election in Missouri, 1860
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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
Party | Candidate | Votes | % | |
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Democratic | Stephen A. Douglas | 58,801 | 35.5% | |
Constitutional Union | John Bell | 58,372 | 35.3% | |
Southern Democratic | John C. Breckinridge | 31,362 | 18.9% | |
Republican | Abraham Lincoln | 17,028 | 10.3% | |
Total votes | 165,563 | 100% |
References
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