1868 United States presidential election in Missouri explained

See main article: 1868 United States presidential election.

Election Name:1868 United States presidential election in Missouri
Country:Missouri
Type:presidential
Ongoing:no
Previous Election:1864 United States presidential election in Missouri
Previous Year:1864
Next Election:1872 United States presidential election in Missouri
Next Year:1872
Election Date:November 3, 1868
Image1:Ulysses S Grant by Brady c1870-restored (3x4 crop).jpg
Nominee1:Ulysses S. Grant
Party1:Republican Party (United States)
Home State1:Ohio
Running Mate1:Schuyler Colfax
Electoral Vote1:11
Popular Vote1:86,860
Percentage1:56.96%
Nominee2:Horatio Seymour
Party2:Democratic Party (United States)
Home State2:New York
Running Mate2:Francis Preston Blair Jr.
Electoral Vote2:0
Popular Vote2:65,628
Percentage2:43.04%
Map Size:340px
President
Before Election:Andrew Johnson
Before Party:Democratic Party (United States)
After Election:Ulysses S. Grant
After Party:Republican Party (United States)

The 1868 United States presidential election in Missouri took place on November 3, 1868, as part of the 1868 United States presidential election. Voters chose 11 representatives, or electors, to the Electoral College, who voted for president and vice president.[1]

Missouri was won by Ulysses S. Grant, formerly the 6th Commanding General of the United States Army (R-Ohio), running with Speaker of the House Schuyler Colfax, with 56.96% of the popular vote, against the 18th governor of New York, Horatio Seymour (DNew York), running with former Senator Francis Preston Blair Jr., with 43.04% of the vote.[1]

Grant's victory in Missouri made him the first Republican presidential candidate to win the state, although President Abraham Lincoln had won the state on the National Union ticket in 1864. Grant would also be the only Republican to carry the state until Theodore Roosevelt won it in 1904.

See also

Notes and References

  1. Web site: 1868 Presidential Election Results Missouri.