1868 United States presidential election in Delaware
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 County Results
Seymour 50-60% 60-70%
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| Elections in Delaware |
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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]
Results
[edit]| 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
[edit]References
[edit]- ^ "1868 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
- ^ "1868 Presidential General Election Results - Delaware".
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