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.
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 County Results[a]
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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 2020 presidential election[update], 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]
ResultsEdit
| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| 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 alsoEdit
NotesEdit
- ^ Current county borders
ReferencesEdit
- ^ Menendez, Albert J.; The Geography of Presidential Elections in the United States, 1868-2004, pp. 239-246 ISBN 0786422173
- ^ "1860 Presidential Election Results – Missouri".
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