1836 United States presidential election in Illinois

Main article: 1836 United States presidential election

1836 United States presidential election in Illinois

←  1832 November 3 – December 7, 1836 1840  →

Nominee Martin Van Buren William Henry Harrison Party Democratic Whig Home state New York Ohio Running mate Richard Johnson Francis Granger Electoral vote 5 0 Popular vote 18,369 15,220 Percentage 54.69% 45.31%

County Results Van Buren   50-60%   60-70%   70-80%   80-90%   90-100% Harrison   50-60%   60-70%   70-80% No Data/Vote

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

Illinois voted for the Democratic candidate, Martin Van Buren, over Whig candidate William Henry Harrison. Van Buren won Illinois by a margin of 9.38%.

Harrison won Chicago by six votes. ^[1]

Results

1836 United States presidential election in Illinois [2]
Party		Candidate	Votes	Percentage	Electoral votes
	Democratic	Martin Van Buren	18,369	54.69%	5
	Whig	William Henry Harrison	15,220	45.31%	0
Totals			33,589	100.0%	5

See also

• United States presidential elections in Illinois

References

1. Merriner 2004, p. 21.
2. "1836 Presidential General Election Results - Illinois". U.S. Election Atlas. Retrieved August 4, 2012.

Works cited

• Merriner, James (2004). Grafters and Goo Goos: Corruption and Reform in Chicago, 1833-2003. Southern Illinois University Press. ISBN   0809325713.