1844 United States presidential election in Rhode Island

1844 United States presidential election in Rhode Island

← 1840 November 1 - December 4, 1844 1848 →
 
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%

County Results
Clay
  50-60%
  60-70%
  70-80%
  80-90%


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]

  1. ^ "1844 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.