1840 United States presidential election in New York

Main article: 1840 United States presidential election

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 42 representatives, or electors to the Electoral College, who voted for President and Vice President.

1840 United States presidential election in New York

← 1836 October 30 – December 2, 1840 1844 →
Turnout	91.9%[1] 21.4 pp
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%
County Results Harrison   40–50%   50–60%   60–70% Van Buren   40–50%   50–60%   60–70%   70–80%
President before election Martin Van Buren Democratic Elected President William Henry Harrison Whig

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

Results

1840 United States presidential election in New York[2]
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%

See also

• United States presidential elections in New York

References

1. Bicentennial Edition: Historical Statistics of the United States, Colonial Times to 1970, part 2, p. 1072.
2. "1840 Presidential General Election Results - New York". U.S. Election Atlas. Retrieved December 23, 2013.