1840 United States presidential election in Louisiana
(Redirected from United States presidential election in Louisiana, 1840)
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| Elections in Louisiana |
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 Government |
The 1840 United States presidential election in Louisiana took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose five representatives, or electors to the Electoral College, who voted for President and Vice President.
Louisiana voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Louisiana by a margin of 19.46%.
With 59.73% of the popular vote, Louisiana would prove to be Harrison's fourth strongest state after Kentucky, Vermont and Rhode Island.[1]
Results
[edit]| 1840 United States presidential election in Louisiana[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 | 11,296 | 59.73% | 5 | 100.00% | ||
| Democratic | Martin Van Buren of New York | Richard M. Johnson of Kentucky | 7,616 | 40.27% | 0 | 0.00% | ||
| Total | 18,912 | 100.00% | 5 | 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 - Louisiana". U.S. Election Atlas. Retrieved December 23, 2013.
State and district results of the 1840 United States presidential election | ||
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