United States presidential election in Missouri, 1996

United States presidential election in Missouri, 1996

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.

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