1840 United States presidential election in Louisiana

← 1836 October 30 – December 2, 1840 1844 →
 
William Henry Harrison crop.jpg
Martin Van Buren circa 1837 crop.jpg
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

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

1840 United States presidential election in Louisiana[2]
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

References

  1. ^ "1840 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
  2. ^ "1840 Presidential General Election Results - Louisiana". U.S. Election Atlas. Retrieved 23 December 2013.