1844 United States presidential election in Rhode Island

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1844 United States presidential election in Rhode Island
Flag of Rhode Island.svg
  1840 November 1 - December 4, 1844 1848  
  Clay 1848.jpg Polk 1849.jpg
Nominee Henry Clay James K. Polk
Party Whig Democratic
Home state Kentucky Tennessee
Running mate Theodore Frelinghuysen George M. Dallas
Electoral vote40
Popular vote7,3224,867
Percentage59.55%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.

Contents

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]
PartyCandidateRunning matePopular voteElectoral vote
Count%Count%
Whig Henry Clay of Kentucky Theodore Frelinghuysen of New York 7,32259.55%4100.00%
Democratic James K. Polk of Tennessee George M. Dallas of Pennsylvania 4,86739.58%00.00%
Liberty James G. Birney of Michigan Thomas Morris of Ohio 1070.87%00.00%
Total12,296100.00%4100.00%

See also

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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.