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1844 United States presidential election in Rhode Island

From Wikipedia, the free encyclopedia

1844 United States presidential election in Rhode Island

← 1840 November 1 - December 4, 1844 1848 →
 
Clay portrait.jpg
JamesKnoxPolk.jpg
Nominee Henry Clay James K. Polk
Party Whig Democratic
Home state Kentucky Tennessee
Running mate Theodore Frelinghuysen George M. Dallas
Electoral vote 4 0
Popular vote 7,322 4,867
Percentage 59.55% 39.58%

President before election

John Tyler
Independent

Elected President

James K. Polk
Democratic

The 1844 United States presidential election in Rhode Island took place between November 1 and December 4, 1844, as part of the 1844 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for President and Vice President.

Rhode Island voted for the Whig candidate, Henry Clay, over Democratic candidate James K. Polk. Clay won Rhode Island by a margin of 19.97%.

With 59.55% of the popular vote, Rhode Island would prove to be Henry Clay's strongest state in the nation.[1]

Results

1844 United States presidential election in Rhode Island[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Whig Henry Clay of Kentucky Theodore Frelinghuysen of New York 7,322 59.55% 4 100.00%
Democratic James K. Polk of Tennessee George M. Dallas of Pennsylvania 4,867 39.58% 0 0.00%
Liberty James G. Birney of Michigan Thomas Morris of Ohio 107 0.87% 0 0.00%
Total 12,296 100.00% 4 100.00%

References

  1. ^ "1844 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
  2. ^ "1844 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.


This page was last edited on 11 July 2019, at 18:35
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