United States presidential election in Ohio, 1840
Main article: United States presidential election, 1840
Elections in Ohio | ||||||||||||||||
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Ballot measures
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Cincinnati
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The 1840 United States presidential election in Ohio took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose twenty-one 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%.
Results
United States presidential election in Ohio, 1840[1] | ||||||||
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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% | ||||
References
- ↑ "1840 Presidential General Election Results - Ohio". U.S. Election Atlas. Retrieved 23 December 2013.
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