1888 United States presidential election in Missouri

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 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 Cleveland   40–50%   50–60%   60–70%   70–80% Harrison   40–50%   50–60%   60–70%   70–80%

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

1888 United States presidential election in Missouri[1]
Party		Candidate	Running mate	Popular vote		Electoral vote
				Count	%	Count	%
	Democratic	Grover Cleveland of New York (incumbent)	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

• United States presidential elections in Missouri

Notes