1840 United States presidential election in Ohio

Main article: 1840 United States presidential election

1840 United States presidential election in Ohio

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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 21 0 Popular vote 148,157 124,782 Percentage 54.10% 45.57%

County Results Congressional District Results Results Harrison   40–50%   50–60%   60–70%   70–80% Van Buren   50–60%   60–70%   70–80%

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

A presidential election was held in Ohio on October 30, 1840 as part of the 1840 United States presidential election. ^[1] Voters chose 21 representatives, or electors to the Electoral College, who voted for President and Vice President.

Ohio voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Ohio by a margin of 8.53%. Ohio was the home state of William Henry Harrison, Harrison improved his margin of victory from the last election over Van Buren by +4.22%

Results

1840 United States presidential election in Ohio [2]
Party		Candidate	Running mate	Popular vote		Electoral vote
				Count	%	Count	%
	Whig	William Henry Harrison of Ohio	John Tyler of Virginia	148,157	54.10%	21	100.00%
	Democratic	Martin Van Buren of New York	Richard M. Johnson of Kentucky	124,782	45.57%	0	0.00%
	Liberty	James G. Birney of New York	Thomas Earle of Pennsylvania	903	0.33%	0	0.00%
Total				273,842	100.00%	21	100.00%

See also

• United States presidential elections in Ohio

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 - Ohio". U.S. Election Atlas. Retrieved December 23, 2013.