1884 United States presidential election in Missouri

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1884 United States presidential election in Missouri

←  1880 November 4, 1884 1888 →
  StephenGroverCleveland.jpg Unsuccessful 1884.jpg
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%

Missouri Presidential Election Results 1884.svg
County Results

President before election

Chester A. Arthur
Republican

Elected President

Grover Cleveland
Democratic

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

  1. ^ a b "1884 Presidential Election Results Missouri".