1816 United States presidential election in Ohio explained

See main article: 1816 United States presidential election.

Election Name:1816 United States presidential election in Ohio
Country:Ohio
Type:presidential
Ongoing:no
Previous Election:1812 United States presidential election in Ohio
Previous Year:1812
Next Election:1820 United States presidential election in Ohio
Next Year:1820
Election Date:November 1 – December 4, 1816
Image1:John Vanderlyn - James Monroe - Google Art Project.jpg
Nominee1:James Monroe
Party1:Democratic-Republican Party
Home State1:Virginia
Running Mate1:Daniel D. Tompkins
Electoral Vote1:8
Popular Vote1:3,326
Percentage1:84.87%
Nominee2:Rufus King
Party2:Federalist Party (United States)
Running Mate2:John E. Howard
Home State2:New York
Electoral Vote2:0
Popular Vote2:593
Percentage2:15.13%
President
Before Election:James Madison
Before Party:Democratic-Republican Party
After Election:James Monroe
After Party:Democratic-Republican Party

The 1816 United States presidential election in Ohio took place between November 1 and December 4, 1816, as part of the 1816 United States presidential election. Voters chose 8 representatives, or electors to the Electoral College, who voted for President and Vice President.

Ohio elected Democratic-Republican candidate James Monroe over Federalist candidate Rufus King. Monroe won Ohio by a margin of 69.74%.

Results

1816 United States presidential election in Ohio[1]
PartyCandidateVotesPercentageElectoral votes
Democratic-RepublicanJames Monroe3,32684.87%8
FederalistRufus King59315.13%0
Totals3,919100.0%8

See also

Notes and References

  1. Book: Ohio. Secretary of State. . 1909 . Ohio election statistics: the general election. . 3 . Springfield, Ohio . The Springfield Publishing Company, State Printers . 5470967 . 2019-12-08.