1848 United States presidential election in Rhode Island

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1848 United States presidential election in Rhode Island
Flag of Rhode Island.svg
  1844 November 7, 1848 1852  
  Zachary Taylor cropped.jpg Lewis Cass circa 1855.jpg Portrait of Martin Van Buren (cropped).jpg
Nominee Zachary Taylor Lewis Cass Martin Van Buren
Party Whig Democratic Free Soil
Home state Louisiana Michigan New York
Running mate Millard Fillmore William O. Butler Charles Francis Adams, Sr.
Electoral vote400
Popular vote6,7793,646730
Percentage60.77%32.68%6.54%

Rhode Island Presidential Election Results 1848.svg
County Results
Taylor
  50-60%
  60-70%

The 1848 United States presidential election in Rhode Island took place on November 7, 1844, as part of the 1848 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 Whig candidate, Zachary Taylor, over Democratic candidate Lewis Cass and Free Soil candidate former president Martin Van Buren. Taylor won the state by a margin of 28.09%.

With 60.77% of the popular vote, Rhode Island would prove to be Taylor's strongest state in the election in terms of percentage in the popular vote. [1]

Results

1848 United States presidential election in Rhode Island [2]
PartyCandidateRunning matePopular voteElectoral vote
Count%Count%
Whig Zachary Taylor of Louisiana Millard Fillmore of New York 6,77960.77%4100.00%
Democratic Lewis Cass of Michigan William O. Butler of Kentucky 3,64632.68%00.00%
Free Soil Martin Van Buren of New York Charles Francis Adams, Sr. of Massachusetts 7306.54%00.00%
Total11,155100.00%4100.00%

See also

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References

  1. "1848 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
  2. "1848 Presidential General Election Results - Rhode Island".