1836 United States presidential election in Vermont
The 1836 United States presidential election in Vermont took place between November 3 and December 7, 1836, as part of the 1836 United States presidential election. Voters chose seven representatives, or electors to the Electoral College, who voted for President and Vice President.
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Vermont voted for Whig candidate William Henry Harrison over Democratic candidate Martin Van Buren. Harrison won Vermont by a margin of 19.86%.
This would be the final time a Democratic candidate would carry Essex County until Franklin D. Roosevelt won it 104 years later in 1940.
Results
| 1836 United States presidential election in Vermont[1] | ||||||||
|---|---|---|---|---|---|---|---|---|
| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Whig | William Henry Harrison of Ohio | Francis Granger of New York | 20,994 | 59.93% | 7 | 100.00% | ||
| Democratic | Martin Van Buren of New York | Richard M. Johnson of Kentucky | 14,037 | 40.07% | 0 | 0.00% | ||
| Total | 35,031 | 100.00% | 7 | 100.00% | ||||
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References
- "1836 Presidential General Election Results - Vermont". U.S. Election Atlas. Retrieved 23 December 2013.
State and district results of the 1836 United States presidential election | ||
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