1824 United States presidential election in New Jersey

Main article: 1824 United States presidential election

The 1824 United States presidential election in New Jersey took place between October 26 and December 2, 1824, as part of the 1824 United States presidential election. Voters chose eight representatives, or electors to the Electoral College, who voted for President and Vice President.

1824 United States presidential election in New Jersey

← 1820 October 26 – December 2, 1824 1828 →

Nominee Andrew Jackson John Quincy Adams William H. Crawford Party Democratic-Republican Democratic-Republican Democratic-Republican Home state Tennessee Massachusetts Georgia Running mate John C. Calhoun John C. Calhoun Nathaniel Macon Electoral vote 8 0 0 Popular vote 10,332 8,309 1,196 Percentage 52.08% 41.89% 6.03%

County Results Jackson   40-50%   50-60%   60-70% Adams   40-50%   50-60%   70-80%

President before election James Monroe Democratic-Republican Elected President John Quincy Adams Democratic-Republican

During this election, the Democratic-Republican Party was the only major national party, and four different candidates from this party sought the Presidency. New Jersey voted for Andrew Jackson over John Quincy Adams, William H. Crawford, and Henry Clay. Jackson won New Jersey by over half of the vote.

Results

1824 United States presidential election in New Jersey[1]
Party		Candidate	Votes	Percentage	Electoral votes
	Democratic-Republican	Andrew Jackson	10,332	52.08%	8
	Democratic-Republican	John Quincy Adams	8,309	41.89%	0
	Democratic-Republican	William H. Crawford	1,196	6.03%	0
Totals			19,837	100.0%	8

See also

• United States presidential elections in New Jersey

References

1. "1824 Presidential General Election Results - New Jersey". U.S. Election Atlas. Retrieved February 27, 2013.