1840 United States presidential election in Vermont
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A presidential election was held in Vermont on November 10, 1840 as part of the 1840 United States presidential election.[1] Voters chose seven representatives, or electors to the Electoral College, who voted for President and Vice President.
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November 10, 1840
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 County Results
Harrison 50â60% 60â70% 70â80%
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Vermont voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Vermont by a margin of 28.43%.
Harrison's 28.43% margin of victory made it his strongest victory in the election while he carried 63.90% of the popular vote made Vermont his second strongest state after Kentucky.[2]
Harrison had previously won Vermont against Van Buren four years earlier.
Results
| 1840 United States presidential election in Vermont[3] | ||||||||
|---|---|---|---|---|---|---|---|---|
| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Whig | William Henry Harrison of Ohio | John Tyler of Virginia | 32,445 | 63.90% | 7 | 100.00% | ||
| Democratic | Martin Van Buren of New York | Richard Mentor Johnson of Kentucky | 18,009 | 35.47% | 0 | 0.00% | ||
| Liberty | James G. Birney of New York | Thomas Earle of Pennsylvania | 319 | 0.63% | 0 | 0.00% | ||
| Total | 50,773 | 100.00% | 7 | 100.00% | ||||
