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

← 1832 November 3 – December 7, 1836 1840 →
 
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%

County Results

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%. Ohio was the home state of William Henry Harrison.

Results

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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

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References

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