1844 United States presidential election in Rhode Island
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.
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Elections in Rhode Island |
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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] | ||||||||
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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
- "1844 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
- "1844 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.
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