1884 United States presidential election in Missouri

The 1884 United States presidential election in Missouri took place on November 4, 1884. All contemporary 38 states were part of the 1884 United States presidential election. Voters chose 16 electors to the Electoral College, which selected the president and vice president.^[1]

Main article: 1884 United States presidential election

1884 United States presidential election in Missouri

← 1880 November 4, 1884 1888 →

Nominee Grover Cleveland James G. Blaine Party Democratic Republican Home state New York Maine Running mate Thomas A. Hendricks John A. Logan Electoral vote 16 0 Popular vote 236,023 203,081 Percentage 53.49% 46.02%

County Results Cleveland   40-50%   50-60%   60-70%   70-80%   80-90% Blaine   50-60%   60-70%   70-80%

President before election Chester A. Arthur Republican Elected President Grover Cleveland Democratic

Missouri was won by Governor Grover Cleveland of New York, and Governor Thomas A. Hendricks of Indiana, with 53.49% of the vote, against former Secretary of State and Senator James G. Blaine of Maine and his running mate Senator John A. Logan of Illinois, with 46.02% of the vote.^[1]

Results

1884 United States presidential election in Missouri
Party		Candidate	Votes	Percentage	Electoral votes
	Democratic	Grover Cleveland	236,023	53.49%	16
	Republican	James G. Blaine	203,081	46.02%	0
	Prohibition	John St. John	2,164	0.49%	0

See also

• United States presidential elections in Missouri

References

1. "1884 Presidential Election Results Missouri".