1840 United States presidential election in New York

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United States presidential election in New York, 1840

← 1836 October 30 - December 2, 1840 1844 →
  Harrison crop.jpg Van Buren.jpg
Nominee William Henry Harrison Martin Van Buren
Party Whig Democratic
Home state Ohio New York
Running mate John Tyler none
Electoral vote 42 0
Popular vote 226,001 212,733
Percentage 51.18% 48.18%

President before election

Martin Van Buren
Democratic

Elected President

William Henry Harrison
Whig

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

New York voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won New York by a margin of 3.00%.

Results[edit]

United States presidential election in New York, 1840[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Whig William Henry Harrison of Ohio John Tyler of Virginia 226,001 51.18% 42 100.00%
Democratic Martin Van Buren of New York Richard M. Johnson of Kentucky 212,733 48.18% 0 0.00%
Liberty James G. Birney of New York Thomas Earle of Pennsylvania 2,809 0.64% 0 0.00%
Total 441,543 100.00% 42 100.00%

References[edit]

  1. ^ "1840 Presidential General Election Results - New York". U.S. Election Atlas. Retrieved 23 December 2013.