Election in Missouri
1884 United States presidential election in Missouri
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| | | Nominee | Grover Cleveland | James G. Blaine | | Party | Democratic | Republican | Home state | New York | Maine | Running mate | Thomas A. Hendricks | John A. Logan | Electoral vote | 16 | 0 | Popular vote | 236,023 | 203,081 | Percentage | 53.49% | 46.02% | |
County Results Cleveland 40-50% 50-60% 60-70% 70-80% 80-90% | Blaine 50-60% 60-70% 70-80% | |
President before election Chester A. Arthur Republican | Elected President Grover Cleveland Democratic | |
Elections in Missouri |
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The 1884 United States presidential election in Missouri took place on November 4, 1884. All contemporary 38 states were part of the 1884 United States presidential election. Voters chose 16 electors to the Electoral College, which selected the president and vice president.[1]
Missouri was won by Governor Grover Cleveland of New York, and Governor Thomas A. Hendricks of Indiana, with 53.49% of the vote, against former Secretary of State and Senator James G. Blaine of Maine and his running mate Senator John A. Logan of Illinois, with 46.02% of the vote.[1]
Results
1884 United States presidential election in Missouri |
Party | Candidate | Votes | Percentage | Electoral votes |
| Democratic | Grover Cleveland | 236,023 | 53.49% | 16 |
| Republican | James G. Blaine | 203,081 | 46.02% | 0 |
| Prohibition | John St. John | 2,164 | 0.49% | 0 |
See also
References
- ^ a b "1884 Presidential Election Results Missouri".
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