1844 United States presidential election in Rhode Island
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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.
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November 6, 1844
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 County Results
Clay 50â60% 60â70% 70â80% 80â90%
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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.[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% | ||||
