1848 United States presidential election in Rhode Island
The 1848 United States presidential election in Rhode Island took place on November 7, 1844, as part of the 1848 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, Zachary Taylor, over Democratic candidate Lewis Cass and Free Soil candidate former president Martin Van Buren. Taylor won the state by a margin of 28.09%.
With 60.77% of the popular vote, Rhode Island would prove to be Taylor's strongest state in the election in terms of percentage in the popular vote.[1]
Results
1848 United States presidential election in Rhode Island[2] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Whig | Zachary Taylor of Louisiana | Millard Fillmore of New York | 6,779 | 60.77% | 4 | 100.00% | ||
Democratic | Lewis Cass of Michigan | William O. Butler of Kentucky | 3,646 | 32.68% | 0 | 0.00% | ||
Free Soil | Martin Van Buren of New York | Charles Francis Adams, Sr. of Massachusetts | 730 | 6.54% | 0 | 0.00% | ||
Total | 11,155 | 100.00% | 4 | 100.00% |
References
- "1848 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
- "1848 Presidential General Election Results - Rhode Island".
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