1840 United States presidential election in Pennsylvania

Election in Pennsylvania

1840 United States presidential election in Pennsylvania

← 1836 October 30 – December 2, 1840 1844 →
 
Nominee William Henry Harrison Martin Van Buren
Party Whig Democratic
Home state Ohio New York
Running mate John Tyler none
Electoral vote 30 0
Popular vote 144,010 143,676
Percentage 50.00% 49.88%

County Results

Harrison

  50-60%
  60-70%
  70–80%

Van Buren

  50-60%
  60-70%
  70-80%
  80-90%


President before election

Martin Van Buren
Democratic

Elected President

William Henry Harrison
Whig

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The 1840 United States presidential election in Pennsylvania took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose 30 representatives, or electors to the Electoral College, who voted for President and Vice President.

Pennsylvania voted for Whig challenger William Henry Harrison over Democratic incumbent Martin Van Buren by just 334 votes, a margin of 0.12%. It is the narrowest margin of victory in a presidential election in Pennsylvania history, with Donald Trump's 2016 win following close behind at 0.72%.

Results

1840 United States presidential election in Pennsylvania[1]
Party Candidate Votes Percentage Electoral votes
Whig William Henry Harrison 144,010 50.00% 30
Democratic Martin Van Buren (incumbent) 143,676 49.88% 0
Liberty James G. Birney 340 0.12% 0
Totals 288,026 100.0% 30

See also

References

  1. ^ "1840 Presidential General Election Results - Pennsylvania". U.S. Election Atlas. Retrieved August 4, 2012.


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