1824 United States presidential election in New York
(Redirected from United States presidential election in New York, 1824)
The 1824 United States presidential election in New York took place between October 26 and December 2, 1824, as part of the 1824 United States presidential election. The state legislature chose 36 representatives, or electors to the Electoral College, who voted for President and Vice President.
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During this election, the Democratic-Republican Party was the only major national party, and 4 different candidates from this party sought the Presidency. New York cast 26 electoral votes for John Quincy Adams, 5 for William H. Crawford, 4 for Henry Clay and 1 for Andrew Jackson. This election marks the last time the New York State Legislature chose the state's electors as opposed to using some form of popular vote method.[2]
Results
edit| 1824 United States presidential election in New York[3] | |||||
|---|---|---|---|---|---|
| Party | Candidate | Electoral votes | |||
| Democratic-Republican | John Quincy Adams | 26 | |||
| Democratic-Republican | William H. Crawford | 5 | |||
| Democratic-Republican | Henry Clay | 4 | |||
| Democratic-Republican | Andrew Jackson | 1 | |||
| Totals | 36 | ||||
See also
editReferences
edit- ^ While Crawford's official running mate was Nathaniel Macon, all of the New York presidential electors who voted for Crawford cast their vice-presidential votes for Van Buren
- ^ 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.