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The1868 United States presidential election in Rhode Island took place on November 3, 1868, as part of the1868 United States presidential election. Voters chose four representatives, or electors to theElectoral College, who voted forpresident andvice president.
Rhode Island voted for theRepublican nominee,Ulysses S. Grant, over theDemocratic 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 afterVermont,Massachusetts,Kansas andTennessee.[1]
| 1868 United States presidential election in Rhode Island[2] | ||||||||
|---|---|---|---|---|---|---|---|---|
| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Republican | Ulysses S. Grant ofIllinois | Schuyler Colfax ofIndiana | 12,993 | 66.49% | 4 | 100.00% | ||
| Democratic | Horatio Seymour ofNew York | Francis Preston Blair Jr. ofMissouri | 6,548 | 33.51% | 0 | 0.00% | ||
| Total | 19,541 | 100.00% | 4 | 100.00% | ||||
State results of the1868 U.S. presidential election | ||
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