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United States presidential election in Missouri, 1896

From Wikipedia, the free encyclopedia

United States presidential election in Missouri, 1896

← 1892 November 3, 1896 1900 →

William McKinley by Courtney Art Studio, 1896.jpg
Nominee William J. Bryan William McKinley
Party Democratic Republican
Home state Nebraska Ohio
Running mate Arthur Sewall Garret Hobart
Electoral vote 17 0
Popular vote 363,667 304,940
Percentage 53.96% 45.25%

President before election

Grover Cleveland

Elected President

William McKinley

The 1896 United States presidential election in Missouri took place on November 3, 1896. All contemporary 45 states were part of the 1896 United States presidential election. Missouri voters chose seventeen electors to the Electoral College, which selected the president and vice president.

Missouri was won by the Democratic nominees, former U.S. Representative William Jennings Bryan of Nebraska and his running mate Arthur Sewall of Maine. Four electors cast their Vice Presidential ballots for Thomas E. Watson. This is the only election since the Civil War where Douglas County voted for a Democratic presidential candidate.[1]


United States presidential election in Missouri, 1896[2]
Party Candidate Votes Percentage Electoral votes
Democratic William Jennings Bryan 363,667 53.96% 17
Republican William McKinley 304,940 45.25% 0
National Democratic John M. Palmer 2,365 0.35% 0
Prohibition Joshua Levering 2,043 0.30% 0
Socialist Labor Charles Matchett 599 0.09% 0
National Prohibition Charles Bentley 292 0.04% 0
Totals 673,906 100.00% 17
Voter turnout



  1. ^ Menendez, Albert J.; The Geography of Presidential Elections in the United States, 1868-2004, pp. 31, 239-246 ISBN 0786422173
  2. ^ Dave Leip’s U.S. Election Atlas; Presidential General Election Results – Missouri
This page was last edited on 15 April 2018, at 23:50
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