1844 United States presidential election in New Jersey explained

See main article: 1844 United States presidential election.

Election Name:1844 United States presidential election in New Jersey
Country:New Jersey
Type:presidential
Ongoing:no
Previous Election:1840 United States presidential election in New Jersey
Previous Year:1840
Next Election:1848 United States presidential election in New Jersey
Next Year:1848
Election Date:November 1 - December 4, 1844
Image1:Clay 1848.jpg
Nominee1:Henry Clay
Party1:Whig Party (United States)
Home State1:Kentucky
Running Mate1:Theodore Frelinghuysen
Electoral Vote1:7
Popular Vote1:38,318
Percentage1:50.46%
Nominee2:James K. Polk
Party2:Democratic Party (United States)
Home State2:Tennessee
Running Mate2:George M. Dallas
Electoral Vote2:0
Popular Vote2:37,495
Percentage2:49.37%
Map Size:350px
President
Before Election:John Tyler
Before Party:Independent (politician)
After Election:James K. Polk
After Party:Democratic Party (United States)

The 1844 United States presidential election in New Jersey took place between November 1 and December 4, 1844, as part of the 1844 United States presidential election. Voters chose seven representatives, or electors to the Electoral College, who voted for President and Vice President.

New Jersey voted for the Whig candidate, Henry Clay, over Democratic candidate James K. Polk. Clay won New Jersey by a margin of 1.09 percent; however, Polk would be the last Democrat to carry Cape May County until Woodrow Wilson in 1912.[1] This is the last time a Democrat won without carrying Hudson County.

See also

Notes and References

  1. Menendez, Albert J.; The Geography of Presidential Elections in the United States, 1868-2004, pp. 258-259