1840 United States presidential election in Pennsylvania

Main article: 1840 United States presidential election

1840 United States presidential election in Pennsylvania

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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 30 0 Popular vote 144,010 143,676 Percentage 49.99% 49.88%

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

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

A presidential election was held in Pennsylvania on October 30, 1840 as part of the 1840 United States presidential election. ^[1] Voters chose 30 representatives, or electors to the Electoral College, who voted for President and Vice President.

Pennsylvania voted for Whig challenger William Henry Harrison over Democratic incumbent Martin Van Buren by just 334 votes, a margin of 0.12%. It is the narrowest margin of victory in a presidential election in Pennsylvania history, with Donald Trump's 2016 win following close behind at 0.72%.

Results

1840 United States presidential election in Pennsylvania [2]
Party		Candidate	Votes	Percentage	Electoral votes
	Whig	William Henry Harrison	144,010	49.99%	30
	Democratic	Martin Van Buren (incumbent)	143,676	49.89%	0
	Liberty	James G. Birney	340	0.12%	0
Totals			288,026	100.0%	30

See also

• List of United States presidential elections in Pennsylvania

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 General Election Results - Pennsylvania". U.S. Election Atlas. Retrieved August 4, 2012.