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1868 United States presidential election in Vermont

From Wikipedia, the free encyclopedia

1868 United States presidential election in Vermont

← 1864 November 3, 1868 1872 →
 
Ulysses S Grant by Brady c1870-restored (cropped).jpg
Horatio Seymour - Brady-Handysmall.jpg
Nominee Ulysses S. Grant Horatio Seymour
Party Republican Democratic
Home state Illinois New York
Running mate Schuyler Colfax Francis Preston Blair, Jr.
Electoral vote 5 0
Popular vote 44,167 12,045
Percentage 78.57% 21.43%

President before election

Andrew Johnson
Democratic

Elected President

Ulysses S. Grant
Republican

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

Vermont voted for the Republican nominee, Ulysses S. Grant, over the Democratic nominee, Horatio Seymour. Grant won the state by a margin of 57.14%.

With 78.57% of the popular vote, Vermont would be Grant's strongest victory in terms of percentage in the popular vote.[1]

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  • ✪ The American Presidential Election of 1876
  • ✪ The American Presidential Election of 1820
  • ✪ The American Presidential Election of 1836
  • ✪ The American Presidential Election of 1860
  • ✪ The American Presidential Election of 1900

Transcription

Results

1868 United States presidential election in Vermont[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican Ulysses S. Grant of Illinois Schuyler Colfax of Indiana 44,167 78.57% 5 100.00%
Democratic Horatio Seymour of New York Francis Preston Blair, Jr. of Missouri 12,045 21.43% 0 0.00%
Total 56,212 100.00% 5 100.00%


References

  1. ^ "1868 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
  2. ^ "1868 Presidential General Election Results - Vermont".
This page was last edited on 1 March 2020, at 21:05
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