A presidential election was held in Rhode Island on November 6, 1844, as part of the 1844 United States presidential election.[1] 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.[2]

Results

1844 United States presidential election in Rhode Island[3]
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%

See also

References

  1. ↑ Presidential Elections, 1844. [Boston]. 1844.
  2. ↑ "1844 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
  3. ↑ "1844 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved December 23, 2013.
source: https://en.wikipedia.org/wiki/1844_United_States_presidential_election_in_Rhode_Island