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1828 United States presidential election in New York

From Wikipedia, the free encyclopedia

1828 United States presidential election in New York

← 1824 October 31 – December 2, 1828 1832 →
Andrew Jackson.jpg
John Quincy Adams 1858 crop.jpg
Nominee Andrew Jackson John Quincy Adams
Party Democratic National Republican
Home state Tennessee Massachusetts
Running mate John C. Calhoun Richard Rush
Electoral vote 20 16
Popular vote 139,412 131,563
Percentage 51.45% 48.55%

The 1828 United States presidential election in New York took place between October 31 and December 2, 1828, as part of the 1828 United States presidential election. Voters chose 36 representatives, or electors to the Electoral College, who voted for President and Vice President. For this election, New York used the Congressional District Method of choosing electors, with 34 of its electors being chosen by the winner in each of the state's congressional districts, and the remaining two being chosen by the first 34 electors. Jackson won 18 congressional districts while Adams won 16. This election marks the first time New York did not chose its electoral votes through its State Legislature.[2]

New York voted for the Democratic candidate, Andrew Jackson, over the National Republican candidate, John Quincy Adams. Jackson won New York by a margin of 2.9%.

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1828 United States presidential election in New York[3]
Party Candidate Votes Percentage Electoral votes
Democratic Andrew Jackson 139,412 51.45% 20
National Republican John Quincy Adams (incumbent) 131,563 48.55% 16
Totals 270,975 100.0% 36

See also


  1. ^ Bicentennial Edition: Historical Statistics of the United States, Colonial Times to 1970, part 2, p. 1072.
  2. ^ Moore, John L., ed. (1985). Congressional Quarterly's Guide to U.S. Elections (2nd ed.). Washington, D.C.: Congressional Quarterly, Inc. pp. 254–56.
  3. ^ "1828 Presidential General Election Results - New York". U.S. Election Atlas. Retrieved 28 February 2013.
This page was last edited on 6 June 2022, at 06:48
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