1872 United States presidential election in Missouri explained

See main article: 1872 United States presidential election.

Election Name:1872 United States presidential election in Missouri
Country:Missouri
Type:presidential
Ongoing:no
Previous Election:1868 United States presidential election in Missouri
Previous Year:1868
Next Election:1876 United States presidential election in Missouri
Next Year:1876
Election Date:November 5, 1872
Image1:Horace Greeley restored (cropped).jpg
Nominee1:Horace Greeley
Party1:Liberal Republican Party (United States)
Home State1:New York
Running Mate1:Benjamin G. Brown
Electoral Vote1:0 (8 invalidated)
Popular Vote1:151,434
Percentage1:55.46%
Nominee2:Ulysses S. Grant
Party2:Republican Party (United States)
Home State2:Illinois
Running Mate2:Henry Wilson
Electoral Vote2:0
Popular Vote2:119,196
Percentage2:43.65%
Nominee3:Thomas A. Hendricks (Given electoral votes due to death of Greeley)
Party3:Democratic Party (United States)
Home State3:Indiana
Electoral Vote3: 6
President
Before Election:Ulysses S. Grant
Before Party:Republican Party (United States)
After Election:Ulysses S. Grant
After Party:Republican Party (United States)
Map Size:340px

The 1872 United States presidential election in Missouri took place on November 5, 1872, as part of the 1872 United States presidential election. Voters chose 15 representatives, or electors to the Electoral College, who voted for president and vice president.

Missouri voted for the Liberal Republican candidate, Horace Greeley, over Republican candidate, Ulysses S. Grant. Greeley won Missouri by a margin of 11.81%, but died prior to the Electoral College meeting, allowing Missouri's fifteen electors to vote for the candidate of their choice.[1]

See also

Notes and References

  1. Web site: American presidential election, 1872. Encyclopædia Britannica. 5 December 2017.