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1888 United States presidential election in Missouri

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.

1888 United States presidential election in Missouri

← 1884 November 6, 1888 1892 →
 
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%

County Results

President before election

Grover Cleveland
Democratic

Elected President

Benjamin Harrison
Republican

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 edit

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 edit

Notes edit

References edit

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


1888, united, states, presidential, election, missouri, main, article, 1888, united, states, presidential, election, took, place, november, 1888, part, 1888, united, states, presidential, election, voters, chose, representatives, electors, electoral, college, . Main article 1888 United States presidential election 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 1888 United States presidential election in Missouri 1884 November 6 1888 1892 Nominee Grover Cleveland Benjamin HarrisonParty Democratic RepublicanHome state New York IndianaRunning mate Allen G Thurman Levi P MortonElectoral vote 16 0Popular vote 261 943 236 252Percentage 50 24 45 31 County Results Cleveland 40 50 50 60 60 70 70 80 Harrison 40 50 50 60 60 70 70 80 President before electionGrover ClevelandDemocratic Elected President Benjamin HarrisonRepublicanMissouri 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 Contents 1 Results 2 See also 3 Notes 4 ReferencesResults edit1888 United States presidential election in Missouri 1 Party Candidate Running mate Popular vote Electoral voteCount 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 editUnited States presidential elections in MissouriNotes editReferences edit 1888 Presidential General Election Results Missouri U S Election Atlas Retrieved December 23 2013 nbsp This Missouri elections related article is a stub You can help Wikipedia by expanding it vte Retrieved from https en wikipedia org w index php title 1888 United States presidential election in Missouri amp oldid 1215753356, wikipedia, wiki, book, books, library,

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