1828 United States presidential election in New York

← 1824 October 31 – December 2, 1828 1832 →
Turnout	80.2%[1]
Nominee Andrew Jackson John Quincy Adams Party Democratic National Republican Alliance - Anti-Masonic 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%
County results Jackson   50–60%   60–70%   70–80% Adams   50–60%   60–70%   70–80%

The1828 United States presidential election in New York took place between October 31 and December 2, 1828, as part of the1828 United States presidential election. Voters chose 36 representatives, or electors to theElectoral College, who voted forPresident andVice 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 choose its electoral votes through its State Legislature.^[2]

Adams support in New York aligned with theAnti-Masonic Party andThurlow Weed, his campaign manager in the state, was sympathetic to the anti-masonic movement.^[3]

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

• United States presidential elections in New York

• Howe, Daniel (2007).What Hath God Wrought: The Transformation of America, 1815–1848.Oxford University Press.ISBN 978-0-19-507894-7.

• 1828 United States presidential election by state
• United States presidential elections in New York (state)
• 1828 New York (state) elections

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