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 8 representatives, or electors to the Electoral College, who voted for President and Vice President.

1832 United States presidential election in New Jersey

November 2 – December 5, 1832
 
Nominee Andrew Jackson Henry Clay
Party Democratic National Republican
Home state Tennessee Kentucky
Running mate Martin Van Buren John Sergeant
Electoral vote 8 0
Popular vote 23,826 23,466
Percentage 49.89% 49.13%

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]
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

  1. "1832 Presidential General Election Results - New Jersey". U.S. Election Atlas. Retrieved 12 April 2013.
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