1836 United States presidential election in Ohio
The 1836 United States presidential election in Ohio took place between November 3 and December 7, 1836, as part of the 1836 United States presidential election. Voters chose twenty-one representatives, or electors to the Electoral College, who voted for President and Vice President.
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Elections in Ohio |
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Ohio voted for Whig candidate William Henry Harrison over Democratic candidate Martin Van Buren. Harrison won Ohio by a narrow margin of 4.31%.
Results
1836 United States presidential election in Ohio[1] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Whig | William Henry Harrison of Ohio | Francis Granger of New York | 104,958 | 51.87% | 21 | 100.00% | ||
Democratic | Martin Van Buren of New York | Richard M. Johnson of Kentucky | 96,238 | 47.56% | 0 | 0.00% | ||
N/A | Others | Others | 1,137 | 0.56% | 0 | 0.00% | ||
Total | 202,333 | 100.00% | 21 | 100.00% |
References
- "1836 Presidential General Election Results - Ohio". U.S. Election Atlas. Retrieved 23 December 2013.
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