1848 United States presidential election in Rhode Island

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.

1848 United States presidential election in Rhode Island

November 7, 1848
 
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 vote 4 0 0
Popular vote 6,779 3,646 730
Percentage 60.8% 32.7% 6.5%

President before election

James K. Polk
Democratic

Elected President

Zachary Taylor
Whig

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]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Whig Zachary Taylor of Louisiana Millard Fillmore of New York 6,779 60.77% 4 100.00%
Democratic Lewis Cass of Michigan William O. Butler of Kentucky 3,646 32.68% 0 0.00%
Free Soil Martin Van Buren of New York Charles Francis Adams, Sr. of Massachusetts 730 6.54% 0 0.00%
Total 11,155 100.00% 4 100.00%

See also

References

  1. "1848 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
  2. "1848 Presidential General Election Results - Rhode Island".


This article is issued from Wikipedia. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.