1844 United States presidential election in Rhode Island explained

See main article: 1844 United States presidential election.

Election Name:1844 United States presidential election in Rhode Island
Country:Rhode Island
Type:presidential
Ongoing:no
Previous Election:1840 United States presidential election in Rhode Island
Previous Year:1840
Next Election:1848 United States presidential election in Rhode Island
Next Year:1848
Election Date:November 1 - December 4, 1844
Image1:Clay 1848.jpg
Nominee1:Henry Clay
Party1:Whig Party (United States)
Home State1:Kentucky
Running Mate1:Theodore Frelinghuysen
Electoral Vote1:4
Popular Vote1:7,322
Percentage1:59.55%
Nominee2:James K. Polk
Party2:Democratic Party (United States)
Home State2:Tennessee
Running Mate2:George M. Dallas
Electoral Vote2:0
Popular Vote2:4,867
Percentage2:39.58%
Map Size:250px
President
Before Election:John Tyler
Before Party:Independent (politician)
After Election:James K. Polk
After Party:Democratic Party (United States)

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]

See also

Notes and References

  1. Web site: 1844 Presidential Election Statistics. Dave Leip’s Atlas of U.S. Presidential Elections. 2018-03-05.