1840 United States presidential election in New Jersey
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| Elections in New Jersey |
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A presidential election was held in New Jersey from November 3–4, 1840 as part of the 1840 United States presidential election.[1] Voters chose eight representatives, or electors to the Electoral College, who voted for President and Vice President.
New Jersey voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won New Jersey by a narrow margin of 3.59%.
Results
[edit]| 1840 United States presidential election in New Jersey[2] | ||||||||
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| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Whig | William Henry Harrison of Ohio | John Tyler of Virginia | 33,351 | 51.74% | 8 | 100.00% | ||
| Democratic | Martin Van Buren of New York | Richard M. Johnson of Kentucky | 31,034 | 48.15% | 0 | 0.00% | ||
| Liberty | James G. Birney of New York | Thomas Earle of Pennsylvania | 69 | 0.11% | 0 | 0.00% | ||
| Total | 64,454 | 100.00% | 8 | 100.00% | ||||
See also
[edit]References
[edit]- ^ Dubin, Michael J. (2002). United States Presidential Elections, 1788–1860: The Official Results by County and State. Jefferson, NC: McFarland & Co. p. xvi.
- ^ "1840 Presidential General Election Results - New Jersey". U.S. Election Atlas. Retrieved December 23, 2013.
State and district results of the 1840 United States presidential election | ||
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