1868 United States presidential election in Delaware

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1868 United States presidential election in Delaware
Flag of Delaware.svg
  1864 November 3, 1868 1872  
  Hon. Horatio Seymour, N.Y - NARA - 528568 (cropped).jpg Ulysses S Grant by Brady c1870-restored (3x4 crop).jpg
Nominee Horatio Seymour Ulysses S. Grant
Party Democratic Republican
Home state New York Illinois
Running mate Francis Preston Blair, Jr. Schuyler Colfax
Electoral vote30
Popular vote10,9577,614
Percentage59.00%41.00%

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

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.

Contents

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]

Results

1868 United States presidential election in Delaware [2]
PartyCandidateRunning matePopular voteElectoral vote
Count%Count%
Democratic Horatio Seymour of New York Francis Preston Blair, Jr. of Missouri 10,95759.00%3100.00%
Republican Ulysses S. Grant of Illinois Schuyler Colfax of Indiana 7,61441.00%00.00%
Total18,571100.00%3100.00%

See also

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References

  1. "1868 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
  2. "1868 Presidential General Election Results - Delaware".