1836 United States presidential election in Virginia explained

See main article: 1836 United States presidential election.

Election Name:1836 United States presidential election in Virginia
Country:Virginia
Flag Year:1788
Type:presidential
Ongoing:no
Previous Election:1832 United States presidential election in Virginia
Previous Year:1832
Next Election:1840 United States presidential election in Virginia
Next Year:1840
Election Date:November 3 – December 7, 1836
Image1:Martin Van Buren circa 1837 crop.jpg
Nominee1:Martin Van Buren
Party1:Democratic Party (United States)
Home State1:New York
Running Mate1:William Smith
Electoral Vote1:23
Popular Vote1:30,556
Percentage1:56.64%
Nominee2:Hugh L. White
Party2:Whig Party (United States)
Home State2:Tennessee
Running Mate2:John Tyler
Electoral Vote2:0
Popular Vote2:23,384
Percentage2:43.35%
Map Size:350px
President
Before Election:Andrew Jackson
Before Party:Democratic Party (United States)
After Election:Martin Van Buren
After Party:Democratic Party (United States)
Colour2:800080

The 1836 United States presidential election in Virginia took place between November 3 and December 7, 1836, as part of the 1836 United States presidential election. Voters chose 23 representatives, or electors to the Electoral College, who voted for President and Vice President.

Virginia voted for the Democratic candidate, Martin Van Buren, over Whig candidate Hugh White. Van Buren won Virginia by a margin of 13.29%. While Van Buren's national running mate was Richard Mentor Johnson, the Virginia Democratic electors refused to support his candidacy and voted for William Smith of South Carolina.

Results

United States presidential election in Virginia, 1836[1]
PartyCandidateVotesPercentageElectoral votes
DemocraticMartin Van Buren30,55656.64%23
WhigHugh L. White23,38443.35%0
N/AOthers50.01%0
Totals53,945100.00%23

Notes and References

  1. Web site: 1836 Presidential General Election Results - Virginia. U.S. Election Atlas. 12 April 2013.