1868 United States presidential election in Rhode Island

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1868 United States presidential election in Rhode Island
Flag of Rhode Island.svg
  1864 November 3, 1868 1872  
  Ulysses S Grant by Brady c1870-restored (3x4 crop).jpg Hon. Horatio Seymour, N.Y - NARA - 528568 (cropped).jpg
Nominee Ulysses S. Grant Horatio Seymour
Party Republican Democratic
Home state Illinois New York
Running mate Schuyler Colfax Francis Preston Blair Jr.
Electoral vote40
Popular vote12,9936,548
Percentage66.49%33.51%

Rhode Island Presidential Election Results 1868.svg
County Results
Grant
  60–70%
  70–80%

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.

Contents

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]
PartyCandidateRunning matePopular voteElectoral vote
Count%Count%
Republican Ulysses S. Grant of Illinois Schuyler Colfax of Indiana 12,99366.49%4100.00%
Democratic Horatio Seymour of New York Francis Preston Blair Jr. of Missouri 6,54833.51%00.00%
Total19,541100.00%4100.00%

See also

References

  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".