1832 United States presidential election in New Jersey
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Elections in New Jersey |
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The 1832 United States presidential election in New Jersey took place between November 2 and December 5, 1832, as part of the 1832 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 Party candidate, Andrew Jackson, over the National Republican candidate, Henry Clay, and the Anti-Masonic Party candidate, William Wirt. Jackson won New Jersey by a margin of 0.76%.
Results
1832 United States presidential election in New Jersey[1] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Democratic | Andrew Jackson (incumbent) | 23,826 | 49.89% | 8 | |
National Republican | Henry Clay | 23,466 | 49.13% | 0 | |
Anti-Masonic | William Wirt | 468 | 0.98% | 0 | |
Totals | 47,760 | 100.0% | 8 |
See also
References
- ^ "1832 Presidential General Election Results - New Jersey". U.S. Election Atlas. Retrieved 12 April 2013.