1808 United States presidential election in New York explained

See main article: 1808 United States presidential election.

Election Name:1808 United States presidential election in New York
Country:New York
Flag Year:1778
Type:presidential
Ongoing:no
Previous Election:1804 United States presidential election in New York
Previous Year:1804
Next Election:1812 United States presidential election in New York
Next Year:1812
Election Date:November 4 – December 7, 1808
Image1:James Madison.jpg
Nominee1:James Madison
Party1:Democratic-Republican Party
Home State1:Virginia
Running Mate1:George Clinton
Electoral Vote1:13
Percentage1:68.38%
Nominee2:George Clinton
Party2:Democratic-Republican Party
Home State2:New York
Running Mate2:James Madison/ James Monroe
Electoral Vote2:6
Percentage2:31.62%
President
Before Election:Thomas Jefferson
Before Party:Democratic-Republican Party
After Election:James Madison
After Party:Democratic-Republican Party

The 1808 United States presidential election in New York took place between November 4 and December 7, 1808, as part of the 1808 United States presidential election. The state legislature chose 19 representatives, or electors to the Electoral College, who voted for President and Vice President.

During this election, New York split its 19 electoral votes among two Democratic-Republican candidates. 13 electoral votes went to Secretary of State James Madison while the remaining 6 went to incumbent Vice President George Clinton. Clinton was also supported by a group of New York Democratic-Republicans for president even though he had remained the party's official vice presidential candidate.[1]

This would be the final election until 1972 in which New York did not have the largest number of electors in the Electoral College.

See also

Notes and References

  1. Book: Kaminski. John P.. George Clinton: Yeoman Politician of the New Republic. 1993. Rowman & Littlefield. 281–288. 9780945612186. 12 October 2015.