1824 United States presidential election in Vermont
(Redirected from United States presidential election in Vermont, 1824)
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| Elections in Vermont |
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The 1824 United States presidential election in Vermont took place between October 26 and December 2, 1824, as part of the 1824 United States presidential election. The state legislature chose seven representatives, or electors (the last time they would do this in Vermont) to the Electoral College, who voted for President and Vice President.[1]
During this election, the Democratic-Republican Party was the only major national party, and four different candidates from this party sought the Presidency. Vermont cast seven electoral votes for New England native John Quincy Adams.
Results
[edit]| 1824 United States presidential election in Vermont[2] | |||||
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| Party | Candidate | Votes | Percentage | Electoral votes | |
| Democratic-Republican | John Quincy Adams | 7 | |||
| Democratic-Republican | Henry Clay | 0 | |||
| Democratic-Republican | William H. Crawford | 0 | |||
| Democratic-Republican | Andrew Jackson | 0 | |||
| Totals | 7 | ||||
See also
[edit]References
[edit]- ^ Moore, John L., ed. (1985). Congressional Quarterly's Guide to U.S. Elections (2nd ed.). Washington, D.C.: Congressional Quarterly, Inc. pp. 254–56.
- ^ "Electoral Votes for President and Vice President 1821-1837". National Archives and Records Administration. Retrieved February 28, 2013.
State and district results of the 1824 United States presidential election | ||
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