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The1888 United States presidential election in Missouri took place on November 6, 1888, as part of the1888 United States presidential election. Voters chose 16 representatives, or electors, to theElectoral College, who voted forpresident andvice president.
Missouri voted for theDemocratic nominee, incumbentPresidentGrover Cleveland, over theRepublican nominee,Benjamin Harrison. Cleveland won the state by a margin of 4.93%.
| 1888 United States presidential election in Missouri[1] | ||||||||
|---|---|---|---|---|---|---|---|---|
| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Democratic | Grover Cleveland ofNew York(incumbent) | Allen Granberry Thurman ofOhio | 261,943 | 50.24% | 16 | 100.00% | ||
| Republican | Benjamin Harrison ofIndiana | Levi Parsons Morton ofNew York | 236,252 | 45.31% | 0 | 0.00% | ||
| Labor | Alson Streeter ofIllinois | Charles E. Cunningham ofArkansas | 18,626 | 3.57% | 0 | 0.00% | ||
| Prohibition | Clinton Fisk ofNew Jersey | John A. Brooks ofMissouri | 4,539 | 0.87% | 0 | 0.00% | ||
| Total | 521,360 | 100.00% | 16 | 100.00% | ||||
State and district results of the1888 United States presidential election | ||
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