1840 United States presidential election in New York
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| ← 1836 | October 30 - December 2, 1840 | 1844 → |
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| Turnout | 91.9%[1]  21.4 pp |
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| |  |  | | Nominee | William Henry Harrison | Martin Van Buren | | | Party | Whig | Democratic | | Home state | Ohio | New York | | Running mate | John Tyler | none | | Electoral vote | 42 | 0 | | Popular vote | 226,001 | 212,733 | | Percentage | 51.18% | 48.18% | |
 County Results | Harrison 40-50% 50-60% 60-70% | Van Buren 40-50% 50-60% 60-70% 70-80% | |
President before election
Martin Van Buren
Democratic | Elected President
William Henry Harrison
Whig | |
| Elections in New York State |
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The 1840 United States presidential election in New York took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose 42 representatives, or electors to the Electoral College, who voted for President and Vice President.
New York voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won New York by a narrow margin of 3.00%.
Results
See also
References
- ^ Bicentennial Edition: Historical Statistics of the United States, Colonial Times to 1970, part 2, p. 1072.
- ^ "1840 Presidential General Election Results - New York". U.S. Election Atlas. Retrieved December 23, 2013.
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- See also
- Presidential elections
- Senate elections
- House elections
- Gubernatorial elections
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21.4 pp
