1824 United States presidential election in New Jersey

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 →
  Andrew Jackson.jpg John Quincy Adams 1858 crop.jpg WilliamHCrawford.jpg
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

Elected President

John Quincy Adams

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.


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 alsoEdit


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