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]

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1884 United States presidential election in Missouri

← 1880 November 4, 1884 1888 â†’

Nominee Grover Cleveland James G. Blaine Party Democratic Republican Alliance Greenback 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

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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]