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1840 United States presidential election in New Jersey

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Main article: 1840 United States presidential election
1840 United States presidential election in New Jersey

← 1836 October 30 - December 2, 1840 1844 →
 
Nominee William Henry Harrison Martin Van Buren
Party Whig Democratic
Home state Ohio New York
Running mate John Tyler none
Electoral vote 8 0
Popular vote 33,351 31,034
Percentage 51.74% 48.15%

County Results

Harrison

  50-60%   60-70%   80–90%

Van Buren

  50-60%   60-70%   70-80%


President before election

Martin Van Buren
Democratic

Elected President

William Henry Harrison
Whig

Elections in New Jersey
Federal government
U.S. President
U.S. Senate
U.S. House of Representatives
State government
General elections
Gubernatorial elections
Senate elections
General Assembly elections
Newark
Mayoral elections
Jersey City
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Mayoral elections

The 1840 United States presidential election in New Jersey took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose eight representatives, or electors to the Electoral College, who voted for President and Vice President.

New Jersey voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won New Jersey by a narrow margin of 3.59%.

Results

1840 United States presidential election in New Jersey
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Whig William Henry Harrison of Ohio John Tyler of Virginia 33,351 51.74% 8 100.00%
Democratic Martin Van Buren of New York Richard M. Johnson of Kentucky 31,034 48.15% 0 0.00%
Liberty James G. Birney of New York Thomas Earle of Pennsylvania 69 0.11% 0 0.00%
Total 64,454 100.00% 8 100.00%

See also

References

  1. "1840 Presidential General Election Results - New Jersey". U.S. Election Atlas. Retrieved December 23, 2013.
State and district results of the 1840 United States presidential election
Electoral map, 1840 election


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