United States presidential election in New Jersey, 1836
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Elections in New Jersey | ||||||||||
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The 1836 United States presidential election in New Jersey took place between November 3 and December 7, 1836, as part of the 1836 United States presidential election. Voters chose eight representatives, or electors to the Electoral College, who voted for President and Vice President.
New Jersey voted for the Democratic candidate, Martin Van Buren, over Whig candidate William Henry Harrison. Van Buren won New Jersey by a margin of 1.06%.
Results
United States presidential election in New Jersey, 1836[1] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Democratic | Martin Van Buren of New York | Richard M. Johnson of Kentucky | 26,137 | 50.53% | 8 | 100.00% | ||
Whig | William Henry Harrison of Ohio | Francis Granger of New York | 25,592 | 49.47% | 0 | 0.00% | ||
Total | 51,729 | 100.00% | 8 | 100.00% | ||||
References
- ↑ "1836 Presidential General Election Results - New Jersey". U.S. Election Atlas. Retrieved 23 December 2013.
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