1836 United States presidential election in Delaware

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

1836 United States presidential election in Delaware

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 3 0 Popular vote 4,736 4,154 Percentage 53.24% 46.70%

Delaware voted for Whig candidate William Henry Harrison over the Democratic candidate, Martin Van Buren. Harrison won Delaware by a margin of 6.54%.

Results

1836 United States presidential election in Delaware[1]
Party		Candidate	Votes	Percentage	Electoral votes
	Whig	William Henry Harrison	4,736	53.24%	3
	Democratic	Martin Van Buren	4,154	46.70%	0
	N/A	Other	5	0.06%	0
Totals			8,895	100.00%	3

1836 United States presidential election in Delaware[2]

County	William Henry Harrison Whig		Martin Van Buren Democratic		Various candidates Other parties		Margin		Total votes cast
	#	%	#	%	#	%	#	%
Kent	1,205	53.70%	1,039	46.30%	0	0.00%	166	7.40%	2,244
New Castle	1,673	47.91%	1,814	51.95%	5	0.14%	-141	-4.04%	3,492
Sussex	1,858	58.82%	1,301	41.18%	0	0.00%	557	17.63%	3,159
Total:	4,736	53.24%	4,154	46.70%	5	0.06%	582	6.54%	8,895

See also

• United States presidential elections in Delaware

References

1. "1836 Presidential General Election Results - Delaware". U.S. Election Atlas. Retrieved 4 August 2012.
2. Burnham, Walter Dean (1955). Presidential ballots, 1836-1892. Internet Archive. Baltimore, Johns Hopkins Press. p. 320.