1840 United States presidential election in Louisiana

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.

1840 United States presidential election in Louisiana

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 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

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

United States presidential election in Louisiana, 1840[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%

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.