1840 United States presidential election in New Jersey
The 1840 United States presidential election in New Jersey took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose eight representatives, or electors to the Electoral College, who voted for President and Vice President.
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Elections in New Jersey |
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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
1840 United States presidential election in New Jersey[1] | ||||||||
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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% |
References
- "1840 Presidential General Election Results - New Jersey". U.S. Election Atlas. Retrieved 23 December 2013.
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