1840 United States presidential election in Rhode Island
(Redirected from United States presidential election in Rhode Island, 1840)
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 County Results
Harrison 50-60% 60-70% 70-80%
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| 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%.
With 61.22% of the popular vote, Rhode Island would be Harrison's third strongest state in the 1840 election after Kentucky and Vermont.[1]
Results
[edit]| 1840 United States presidential election in Rhode Island[2] | ||||||||
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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% | ||||
See also
[edit]References
[edit]- ^ "1840 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
- ^ "1840 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved December 23, 2013.
State and district results of the 1840 United States presidential election | ||
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