1888 United States presidential election in Missouri

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1888 United States presidential election in Missouri
Flag of Missouri.svg
  1884 November 6, 1888 1892  
  StephenGroverCleveland.jpg 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 vote160
Popular vote261,943236,252
Percentage50.24%45.31%

Missouri Presidential Election Results 1888.svg
County results

President before election

Grover Cleveland
Democratic

Elected President

Benjamin Harrison
Republican

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.

Contents

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%.

Results

1888 United States presidential election in Missouri [1]
PartyCandidateRunning matePopular voteElectoral vote
Count%Count%
Democratic Grover Cleveland of New York (incumbent) Allen Granberry Thurman of Ohio 261,94350.24%16100.00%
Republican Benjamin Harrison of Indiana Levi Parsons Morton of New York 236,25245.31%00.00%
Labor Alson Streeter of Illinois Charles E. Cunningham of Arkansas 18,6263.57%00.00%
Prohibition Clinton Fisk of New Jersey John A. Brooks of Missouri 4,5390.87%00.00%
Total521,360100.00%16100.00%

See also

Notes

    References

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