1868 United States presidential election in Rhode Island
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 County Results
Grant 60-70% 70-80%
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| Elections in Rhode Island |
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The 1868 United States presidential election in Rhode Island took place on November 3, 1868, as part of the 1868 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.
Rhode Island voted for the Republican nominee, Ulysses S. Grant, over the Democratic nominee, Horatio Seymour. Grant won the state by a margin of 32.98%.
With 66.49% of the popular vote, Rhode Island would be Grant's fifth strongest victory in terms of popular vote percentage after Vermont, Massachusetts, Kansas and Tennessee.[1]
Results
[edit]| 1868 United States presidential election in Rhode Island[2] | ||||||||
|---|---|---|---|---|---|---|---|---|
| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Republican | Ulysses S. Grant of Illinois | Schuyler Colfax of Indiana | 12,993 | 66.49% | 4 | 100.00% | ||
| Democratic | Horatio Seymour of New York | Francis Preston Blair Jr. of Missouri | 6,548 | 33.51% | 0 | 0.00% | ||
| Total | 19,541 | 100.00% | 4 | 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 - Rhode Island".
