1836 United States presidential election in Ohio

The 1836 United States presidential election in Ohio took place between November 3 and December 7, 1836, as part of the 1836 United States presidential election. Voters chose twenty-one representatives, or electors to the Electoral College, who voted for President and Vice President.

1836 United States presidential election in Ohio

November 3 – December 7, 1836
 
Nominee William Henry Harrison Martin Van Buren
Party Whig Democratic
Home state Ohio New York
Running mate Francis Granger Richard Johnson
Electoral vote 21 0
Popular vote 104,958 96,238
Percentage 51.87% 47.56%

President before election

Andrew Jackson
Democratic

Elected President

Martin Van Buren
Democratic

Ohio voted for Whig candidate William Henry Harrison over Democratic candidate Martin Van Buren. Harrison won Ohio by a narrow margin of 4.31%.

Results

1836 United States presidential election in Ohio[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Whig William Henry Harrison of Ohio Francis Granger of New York 104,958 51.87% 21 100.00%
Democratic Martin Van Buren of New York Richard M. Johnson of Kentucky 96,238 47.56% 0 0.00%
N/A Others Others 1,137 0.56% 0 0.00%
Total 202,333 100.00% 21 100.00%

See also

References

  1. "1836 Presidential General Election Results - Ohio". U.S. Election Atlas. Retrieved 23 December 2013.


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