1840 United States presidential election in Kentucky

Main article: 1840 United States presidential election

1840 United States presidential election in Kentucky

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Nominee William Henry Harrison Martin Van Buren Party Whig Democratic Home state Ohio New York Running mate John Tyler none Electoral vote 15 0 Popular vote 58,488 32,616 Percentage 64.20% 35.80%

County Results Harrison   50–60%   60–70%   70–80%   80–90%   90–100% Van Buren   50–60%   60–70%   70–80%   80–90%

President before election Martin Van Buren Democratic Elected President William Henry Harrison Whig

The United States presidential election in Kentucky was held on November 2, 1840, as part of the 1840 United States presidential election. ^[1] Voters chose 15 representatives, or electors to the Electoral College, who voted for President and Vice President.

Kentucky voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Kentucky by a margin of 28.4%.

With 64.20% of the popular vote, Kentucky would prove to be Harrison's strongest state in the 1840 election. ^[2]

Results

1840 United States presidential election in Kentucky [3]
Party		Candidate	Running mate	Popular vote		Electoral vote
				Count	%	Count	%
	Whig	William Henry Harrison of Ohio	John Tyler of Virginia	58,488	64.20%	15	100.00%
	Democratic	Martin Van Buren of New York	Richard M. Johnson of Kentucky	32,616	35.80%	0	0.00%
Total				91,104	100.00%	15	100.00%

References

1. Dubin, Michael J. (2002). United States Presidential Elections, 1788–1860: The Official Results by County and State. Jefferson, NC: McFarland & Co. p. xvi.
2. "1840 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
3. "1840 Presidential General Election Results - Kentucky". U.S. Election Atlas. Retrieved December 23, 2013.