1840 United States presidential election in Louisiana information
Election in Louisiana
Main article: 1840 United States presidential election
1840 United States presidential election in Louisiana
← 1836
October 30 – December 2, 1840
1844 →
Nominee
William Henry Harrison
Martin Van Buren
Party
Whig
Democratic
Home state
Ohio
New York
Running mate
John Tyler
none
Electoral vote
5
0
Popular vote
11,296
7,616
Percentage
59.73%
40.27%
President before election
Martin Van Buren
Democratic
Elected President
William Henry Harrison
Whig
Preview warning: Page using Template:Infobox election with unknown parameter "after_color"
Preview warning: Page using Template:Infobox election with unknown parameter "before_color"
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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]
^"1840 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
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