1868 United States presidential election in Delaware
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.
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County Results
Seymour 50-60% 60-70%
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Elections in Delaware |
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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] | ||||||||
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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% |
References
- "1868 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
- "1868 Presidential General Election Results - Delaware".
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