United States presidential election in Missouri, 1996

United States presidential election in Missouri, 1996
Missouri
November 5, 1996

 
Nominee Bill Clinton Bob Dole Ross Perot
Party Democratic Republican Reform
Home state Arkansas Kansas Texas
Running mate Al Gore Jack Kemp Pat Choate
Electoral vote 11 0 0
Popular vote 1,025,935 890,016 217,188
Percentage 47.5% 41.2% 10.1%

County Results
  Clinton—80-90%
  Clinton—70-80%
  Clinton—60-70%
  Clinton—50-60%
  Dole—40-50%
  Dole—50-60%
President before election

Bill Clinton
Democratic

 Elected President

Bill Clinton
Democratic

The 1996 United States presidential election in Missouri took place on November 5, 1996 as part of the 1996 United States presidential election. Voters chose 11 representatives, or electors to the Electoral College, who voted for President and Vice President.

Missouri was won by President Bill Clinton (D) over Senator Bob Dole (R-KS), with Clinton winning 47.54% to 41.24% by a margin of 6.3%. Billionaire businessman Ross Perot (Reform Party of the United States of America-TX) finished in third with 10.06% of the popular vote.[1] Since 1904, this state has been carried by the winner of the presidential election, with the exceptions of the elections of 1956, 2008, and 2012. This election is the most recent in which Missouri voted for the Democrat.

Results

United States presidential election in Missouri, 1996
Party Candidate Running mate Votes Percentage Electoral votes
Democratic Bill Clinton (incumbent) Al Gore 1,025,935 47.54% 11
Republican Robert Dole Jack Kemp 890,016 41.24% 0
Reform Ross Perot Patrick Choate 217,188 10.06% 0
U.S. Taxpayer Howard Phillips Herbert Titus 11,521 0.53% 0
Libertarian Harry Browne Jo Jorgensen 10,522 0.49% 0
Natural Law Dr. John Hagelin Dr. V. Tompkins 2,287 0.11% 0
Write-in Ralph Nader Winona LaDuke 534 0.02% 0
Write-in Charles Collins Rosemary Giumarra 62 0.00% 0

References

  1. ↑ Dave Leip's Atlas of United States Presidential Election Results - 1996 Missouri Results

See also

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