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1868 United States presidential election in Delaware

From Wikipedia, the free encyclopedia

1868 United States presidential election in Delaware

← 1864 November 3, 1868 1872 →
 
Horatio Seymour - Brady-Handysmall.jpg
Ulysses S Grant by Brady c1870-restored (cropped).jpg
Nominee Horatio Seymour Ulysses S. Grant
Party Democratic Republican
Home state New York Illinois
Running mate Francis Preston Blair, Jr. Schuyler Colfax
Electoral vote 3 0
Popular vote 10,957 7,614
Percentage 59.00% 41.00%

Delaware Presidential Election Results 1868.svg
County Results
Seymour
  50-60%
  60-70%


President before election

Andrew Johnson
Democratic

Elected President

Ulysses S. Grant
Republican

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

Delaware voted for the Democratic nominee, Horatio Seymour, over the Republican nominee, Ulysses S. Grant. Seymour won the state by a margin of 18%.

With 59% of the popular vote, Delaware would prove to be Seymour's fifth strongest state in terms of popular vote percentage after Kentucky, Louisiana, Maryland and Georgia.[1]

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Transcription

Results

1868 United States presidential election in Delaware[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Democratic Horatio Seymour of New York Francis Preston Blair, Jr. of Missouri 10,957 59.00% 3 100.00%
Republican Ulysses S. Grant of Illinois Schuyler Colfax of Indiana 7,614 41.00% 0 0.00%
Total 18,571 100.00% 3 100.00%

See also

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 - Delaware".


This page was last edited on 27 February 2021, at 12:20
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