1828 United States presidential election in Ohio

The 1828 United States presidential election in Ohio took place between October 31 and December 2, 1828, as part of the 1828 United States presidential election. Voters chose 16 representatives, or electors to the Electoral College, who voted for President and Vice President.

1828 United States presidential election in Ohio

October 31 – December 2, 1828

Nominee Andrew Jackson John Quincy Adams Party Democratic National Republican Home state Tennessee Massachusetts Running mate John C. Calhoun Richard Rush Electoral vote 16 0 Popular vote 67,597 63,396 Percentage 51.60% 48.40%

President before election John Quincy Adams Democratic-Republican Elected President Andrew Jackson Democratic

Ohio voted for the Democratic candidate, Andrew Jackson, over the National Republican candidate, John Quincy Adams. Jackson won Ohio by a narrow margin of 3.2%.

Results

1828 United States presidential election in Ohio[1]
Party		Candidate	Votes	Percentage	Electoral votes
	Democratic	Andrew Jackson	67,597	51.60%	16
	National Republican	John Quincy Adams	63,396	48.40%	0
Totals			130,993	100.0%	16

See also

• United States presidential elections in Ohio

References

1. "1828 Presidential General Election Results - Ohio". U.S. Election Atlas. Retrieved 28 February 2013.