United States presidential election in Rhode Island, 1844
Main article: United States presidential election, 1844
| Elections in Rhode Island | |||||||||
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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%.
Results
| United States presidential election in Rhode Island, 1844[1] | ||||||||
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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 General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.
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