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1840 United States presidential election in Delaware

Main article: 1840 United States presidential election

A presidential election was held in Delaware on November 10, 1840 as part of the 1840 United States presidential election.[1] 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[2]
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[3]
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. ^ Dubin, Michael J. (2002). United States Presidential Elections, 1788–1860: The Official Results by County and State. Jefferson, NC: McFarland & Co. p. xvi.
  2. ^ "1840 Presidential General Election Results - Delaware". U.S. Election Atlas. Retrieved December 23, 2013.
  3. ^ Burnham, Walter Dean (1955). Presidential ballots, 1836-1892. Internet Archive. Baltimore, Johns Hopkins Press. p. 320.