1832 United States presidential election in New Jersey

Main article: 1832 United States presidential election

1832 United States presidential election in New Jersey

←  1828 November 2 – December 5, 1832 1836  →

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%

County Results Jackson   50–60%   60–70%   70–80% Clay   50–60%   60–70%

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%. This was the closest election in the state's history.

Results

1832 United States presidential election in New Jersey [1]
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

• United States presidential elections in New Jersey

References

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