1844 United States presidential election in New Jersey
| Elections in New Jersey |
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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%.
Results
| United States presidential election in New Jersey, 1844[1] | ||||||||
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| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Whig | Henry Clay of Kentucky | Theodore Frelinghuysen of New York | 38,318 | 50.46% | 7 | 100.00% | ||
| Democratic | James K. Polk of Tennessee | George M. Dallas of Pennsylvania | 37,495 | 49.37% | 0 | 0.00% | ||
| Liberty Party (1840s) | James G. Birney of Michigan | Thomas Morris of Ohio | 131 | 0.17% | 0 | 0.00% | ||
| Total | 75,944 | 100.00% | 7 | 100.00% | ||||
References
- ^ "1844 Presidential General Election Results - New Jersey". U.S. Election Atlas. Retrieved 23 December 2013.
State and district results of the 1844 United States presidential election | ||
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