1884 United States presidential election in Missouri

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. Missouri voters chose sixteen electors to the Electoral College, which selected the president and vice president.[1]

Main article: 1884 United States presidential election
1884 United States presidential election in Missouri
November 4, 1884
 
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%
President before election

Chester A. Arthur
Republican

 Elected President

Grover Cleveland
Democratic

Elections in Missouri

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
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References
  1. "1884 Presidential Election Results Missouri".
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