United States presidential election in Louisiana, 1840

October 30 - December 2, 1840


Nominee	William Henry Harrison	Martin Van Buren
Party	Whig	Democratic
Home state	Ohio	New York
Running mate	John Tyler	none
Electoral vote	4	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

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%.

Results

United States presidential election in Louisiana, 1840^[1]
Party		Candidate	Running mate	Popular vote		Electoral vote
Party		Candidate	Running mate	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%

References

↑ "1840 Presidential General Election Results - Louisiana". U.S. Election Atlas. Retrieved 23 December 2013.

State Results of the 1840 U.S. presidential election
Candidates	William Henry Harrison Martin Van Buren James G. Birney
General articles	Election timeline
Local results	Alabama Arkansas Connecticut Delaware Georgia Illinois Indiana Kentucky Louisiana Maine Maryland Massachusetts Michigan Mississippi Missouri New Hampshire New Jersey New York North Carolina Ohio Pennsylvania Rhode Island South Carolina Tennessee Vermont Virginia
Other 1840 elections	House Senate Gubernatorial

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