1868 United States presidential election in Vermont

The 1868 United States presidential election in Vermont took place on November 3, 1868, as part of the 1868 United States presidential election. Voters chose five representatives, or electors to the Electoral College, who voted for president and vice president.

1868 United States presidential election in Vermont

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 5 0 Popular vote 44,167 12,045 Percentage 78.57% 21.43%

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

Vermont voted for the Republican nominee, Ulysses S. Grant, over the Democratic nominee, Horatio Seymour. Grant won the state by a margin of 57.14%.

With 78.57% of the popular vote, Vermont would be Grant's strongest victory in terms of percentage in the popular vote.[1] In addition, Grant's performance in Vermont in popular vote percentage was the second best for a Republican presidential candidate only after William McKinley's 80.08% in 1896.

Results

1868 United States presidential election in Vermont[2]
Party		Candidate	Running mate	Popular vote		Electoral vote
				Count	%	Count	%
	Republican	Ulysses S. Grant of Illinois	Schuyler Colfax of Indiana	44,167	78.57%	5	100.00%
	Democratic	Horatio Seymour of New York	Francis Preston Blair, Jr. of Missouri	12,045	21.43%	0	0.00%
Total				56,212	100.00%	5	100.00%

See also

• United States presidential elections in Vermont

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 - Vermont".