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| Elections in Missouri |
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The 1888 United States presidential election in Missouri took place on November 6, 1888, as part of the 1888 United States presidential election. Voters chose 16 representatives, or electors to the Electoral College, who voted for president and vice president.
Missouri voted for the Democratic nominee, incumbent President Grover Cleveland, over the Republican nominee, Benjamin Harrison. Cleveland won the state by a margin of 4.93%.
Results
| 1888 United States presidential election in Missouri[1] | ||||||||
|---|---|---|---|---|---|---|---|---|
| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Democratic | Grover Cleveland of New York | Allen Granberry Thurman of Ohio | 261,943 | 50.24% | 16 | 100.00% | ||
| Republican | Benjamin Harrison of Indiana | Levi Parsons Morton of New York | 236,252 | 45.31% | 0 | 0.00% | ||
| Labor | Alson Streeter of Illinois | Charles E. Cunningham of Arkansas | 18,626 | 3.57% | 0 | 0.00% | ||
| Prohibition | Clinton Fisk of New Jersey | John A. Brooks of Missouri | 4,539 | 0.87% | 0 | 0.00% | ||
| Total | 521,360 | 100.00% | 16 | 100.00% | ||||
See also
Notes
References
- ^ "1888 Presidential General Election Results - Missouri". U.S. Election Atlas. Retrieved December 23, 2013.
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This page was last edited on 26 March 2024, at 22:33