United States presidential election in Vermont, 1836
Main article: United States presidential election, 1836
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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.
Vermont voted for Whig candidate William Henry Harrison over Democratic candidate Martin Van Buren. Harrison won Vermont by a margin of 19.86%.
Results
| United States presidential election in Vermont, 1836[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% | ||||
References
- ↑ "1836 Presidential General Election Results - Vermont". U.S. Election Atlas. Retrieved 23 December 2013.
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