1868 United States presidential election in Rhode Island

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.

1868 United States presidential election in Rhode Island

November 3, 1868

Nominee Ulysses S. Grant Horatio Seymour Party Republican Democratic Home state Illinois New York Running mate Schuyler Colfax Francis Preston Blair, Jr. Electoral vote 4 0 Popular vote 12,993 6,548 Percentage 66.49% 33.51%

President before election Andrew Johnson Democratic Elected President Ulysses S. Grant Republican

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

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

• United States presidential elections in Rhode Island

References

1. "1868 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
2. "1868 Presidential General Election Results - Rhode Island".