1832 United States presidential election in New Jersey
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.
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Elections in New Jersey |
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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 | 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 | ||
References
- "1832 Presidential General Election Results - New Jersey". U.S. Election Atlas. Retrieved 12 April 2013.
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