1888 United States presidential election in Missouri

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1888 United States presidential election in Missouri

←  1884 November 6, 1888 1892 →
  StephenGroverCleveland.png Benjamin Harrison 1896.jpg
Nominee Grover Cleveland Benjamin Harrison
Party Democratic Republican
Home state New York Indiana
Running mate Allen G. Thurman Levi P. Morton
Electoral vote 16 0
Popular vote 261,943 236,252
Percentage 50.24% 45.31%

Missouri Presidential Election Results 1888.svg
County Results

President before election

Grover Cleveland

Elected President

Benjamin Harrison

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%.


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



  1. ^ "1888 Presidential General Election Results - Missouri". U.S. Election Atlas. Retrieved 23 December 2013.