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1836 United States presidential election in New York

From Wikipedia, the free encyclopedia

United States presidential election in New York, 1836

← 1832 November 3 - December 7, 1836 1840 →
Turnout70.5%[1] Decrease 13.7 pp
 
Van Buren.jpg
William Henry Harrison by James Reid Lambdin, 1835 crop.jpg
Nominee Martin Van Buren William Henry Harrison
Party Democratic Whig
Home state New York Ohio
Running mate Richard Mentor Johnson Francis Granger
Electoral vote 42 0
Popular vote 166,795 138,548
Percentage 54.63% 45.47%

President before election

Andrew Jackson
Democratic

Elected President

Martin Van Buren
Democratic

The 1836 United States presidential election in New York took place between November 3 and December 7, 1836, as part of the 1836 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 Democratic candidate, Martin Van Buren, over Whig candidate William Henry Harrison. Van Buren won New York by a margin of 9.26%.

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  • ✪ The American Presidential Election of 1948
  • ✪ The American Presidential Election of 1860
  • ✪ The American Presidential Election of 1932
  • ✪ The American Presidential Election of 1888
  • ✪ The American Presidential Election of 1964

Transcription

Results

United States presidential election in New York, 1836[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Democratic Martin Van Buren of New York Richard M. Johnson of Kentucky 166,795 54.63% 42 100.00%
Whig William Henry Harrison of Ohio Francis Granger of New York 138,548 45.37% 0 0.00%
Total 305,343 100.00% 42 100.00%

References

  1. ^ Bicentennial Edition: Historical Statistics of the United States, Colonial Times to 1970, part 2, p. 1072.
  2. ^ "1836 Presidential General Election Results - New York". U.S. Election Atlas. Retrieved 23 December 2013.
This page was last edited on 26 April 2019, at 13:02
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