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

← 1864 November 3, 1868 1872 →
 
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%

County Results
Grant
  60-70%
  70-80%


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

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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

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References

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  1. ^ "1868 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
  2. ^ "1868 Presidential General Election Results - Rhode Island".