United States presidential election in Rhode Island, 1840
Main article: United States presidential election, 1840
| Elections in Rhode Island | |||||||||
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The 1840 United States presidential election in Rhode Island took place between October 30 and December 2, 1840, as part of the 1840 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, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Rhode Island by a margin of 22.93%.
Results
| United States presidential election in Rhode Island, 1840[1] | ||||||||
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| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Whig | William Henry Harrison of Ohio | John Tyler of Virginia | 5,278 | 61.22% | 4 | 100.00% | ||
| Democratic | Martin Van Buren of New York | Richard M. Johnson of Kentucky | 3,301 | 38.29% | 0 | 0.00% | ||
| Liberty | James G. Birney of New York | Thomas Earle of Pennsylvania | 42 | 0.49% | 0 | 0.00% | ||
| Total | 8,621 | 100.00% | 4 | 100.00% | ||||
References
- ↑ "1840 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.
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