1840 United States presidential election in Ohio
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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.
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October 30, 1840
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Results
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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 Mentor 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% | ||||
