1852 United States presidential election in Delaware

The 1852 United States presidential election in Delaware took place on November 2, 1852, as part of the 1852 United States presidential election. Voters chose 3 representatives, or electors to the Electoral College, who voted for President and Vice President.

November 2, 1852


Nominee	Franklin Pierce	Winfield Scott
Party	Democratic	Whig
Home state	New Hampshire	New Jersey
Running mate	William R. King	William A. Graham
Electoral vote	3	0
Popular vote	6,318	6,293
Percentage	49.85%	49.66%

President before election

Millard Fillmore
Whig

Elected President

Franklin Pierce
Democratic

Elections in Delaware

Federal government

U.S. President
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U.S. Senate
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U.S. House
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Special elections
Senate, 2010

State government

General elections
2010 2012 2014 2016 2018
Gubernatorial elections
1964 1968 1972 1976 1980 1984 1988 1992 1996 2000 2004 2008 2012 2016 2020
Lieutenant gubernatorial elections
2004 2008 2012 2016 2020
Senate elections
2018
House of Representatives elections
2008 2010 2012 2014 2016 2018
Special elections
10th Senate district, 2017

Wilmington

Mayoral elections
2012 2016 2020

Delaware voted for the Democratic candidate, Franklin Pierce, over Whig candidate Winfield Scott. Pierce won Delaware by a margin of 0.19%.

Results

United States presidential election in Delaware, 1852[1]
Party		Candidate	Running mate	Popular vote		Electoral vote
Party		Candidate	Running mate	Count	%	Count	%
	Democratic	Franklin Pierce of New Hampshire	William R. King of Alabama	6,318	49.85%	3	100.00%
	Whig	Winfield Scott of New Jersey	William Alexander Graham of North Carolina	6,293	49.66%	0	0.00%
	Free Soil	John P. Hale of New Hampshire	George Washington Julian of Indiana	62	0.49%	0	0.00%
Total				12,673	100.00%	3	100.00%

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References

"1852 Presidential General Election Results - Delaware". U.S. Election Atlas. Retrieved 17 March 2015.

State results of the 1852 U.S. presidential election
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