1840 United States presidential election in Louisiana

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

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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]

  1. ^ "1840 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.

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