1828 United States presidential election in New York

Main article: 1828 United States presidential election

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 choose its electoral votes through its State Legislature.^[2]

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%
Jackson   50–60%   60–70%   70–80% Adams   50–60%   60–70%   70–80%

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

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%.

Results

1828 United States presidential election in New York[4]
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

• United States presidential elections in New York