1840 United States presidential election in Delaware

Election in Delaware

1840 United States presidential election in Delaware

← 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 3 0
Popular vote 5,967 4,872
Percentage 54.99% 44.89%

County Results
Harrison
  50-60%


President before election

Martin Van Buren
Democratic

Elected President

William Henry Harrison
Whig

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

Delaware voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Delaware by a margin of 10.1%.

Results

1840 United States presidential election in Delaware[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Whig William Henry Harrison of Ohio John Tyler of Virginia 5,967 54.99% 3 100.00%
Democratic Martin Van Buren of New York Richard M. Johnson of Kentucky 4,872 44.89% 0 0.00%
N/A Others Others 13 0.12% 0 0.00%
Total 10,852 100.00% 3 100.00%

Results by county

1840 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 1,593 59.20% 1,095 40.69% 3 0.11% 498 18.51% 2,691
New Castle 2,321 51.28% 2,195 48.50% 10 0.22% 126 2.78% 4,526
Sussex 2,053 56.48% 1,582 43.52% 0 0.00% 471 12.96% 3,635
Total: 5,967 54.99% 4,872 44.89% 13 0.12% 1,095 10.09% 10,852

See also

References

  1. ^ "1840 Presidential General Election Results - Delaware". U.S. Election Atlas. Retrieved December 23, 2013.
  2. ^ Burnham, Walter Dean (1955). Presidential ballots, 1836-1892. Internet Archive. Baltimore, Johns Hopkins Press. p. 320.
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