1860 United States presidential election in Missouri
Election in Missouri
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 County Results[a]
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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, by a very narrow margin of 0.26%. The state was the only one to fully give its votes to Douglas, though he won the popular vote and three of the seven electoral votes from New Jersey under a fusion ticket.
As of the 2020 presidential election[update], this is the last occasion when Putnam County, Ozark County, and Taney County voted for a candidate running as a Democrat, with the former voting for Douglas and the latter two voting for Breckinridge. This was the first time in American history that four candidates each won at least one county in the same state, something that has only been repeated in 1912.[1]
Results
| Party | Candidate | Votes | % | |
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| Democratic | Stephen A. Douglas | 58,801 | 35.52% | |
| 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% | ||
See also
Notes
- ^ Current county borders
References
State results of the 1860 U.S. presidential election | ||
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