United States presidential election in New Jersey, 1824

United States presidential election in New Jersey, 1824
New Jersey
October 26 – December 2, 1824

 
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%

President before election

James Monroe
Democratic-Republican

Elected President

John Quincy Adams
Democratic-Republican

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.

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 a margin of 10.19%.

Results

United States presidential election in New Jersey, 1824[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

References

  1. "1824 Presidential General Election Results - New Jersey". U.S. Election Atlas. Retrieved 27 February 2013.
This article is issued from Wikipedia - version of the Friday, March 04, 2016. The text is available under the Creative Commons Attribution/Share Alike but additional terms may apply for the media files.