1908 United States presidential election in Delaware

Summary

The 1908 United States presidential election in Delaware took place on November 3, 1908. All contemporary 46 states were part of the 1908 United States presidential election. State voters chose three electors to the Electoral College, which selected the president and vice president.

1908 United States presidential election in Delaware

← 1904 November 3, 1908 1912 →
 
Nominee William Howard Taft William Jennings Bryan
Party Republican Democratic
Home state Ohio Nebraska
Running mate James S. Sherman John W. Kern
Electoral vote 3 0
Popular vote 25,014 22,055
Percentage 52.10% 45.94%

County Results
Taft
  40-50%
  50-60%


President before election

Theodore Roosevelt
Republican

Elected President

William Howard Taft
Republican

Delaware was won by the Republican nominees, former Secretary of War William Howard Taft of Ohio and his running mate James S. Sherman of New York. They defeated the Democratic nominees, former U.S. Representative William Jennings Bryan of Nebraska and his running mate John W. Kern. Taft won the state by a margin of 6.16%.

Bryan had previously lost Delaware to William McKinley in both 1896 and 1900.

Results edit

1908 United States presidential election in Delaware[1]
Party Candidate Votes Percentage Electoral votes
Republican William Howard Taft 25,014 52.10% 3
Democratic William Jennings Bryan 22,055 45.94% 0
Prohibition Eugene W. Chafin 670 1.40% 0
Social Democratic Eugene V. Debs 239 0.50% 0
Independence Thomas L. Hisgen 29 0.06% 0
Totals 48,007 100.00% 3
Voter turnout

Results by county edit

County William Howard Taft
Republican
William Jennings Bryan
Democrat
Various candidates
Others
Total
votes
% # % # % #
Kent 49.75% 4,158 49.00% 4,095 1.25% 104 8,357
New Castle 52.29% 14,979 45.26% 12,964 2.45% 701 28,644
Sussex 53.25% 5,870 45.48% 5,013 1.27% 140 11,023

See also edit

Notes edit

References edit

  1. ^ Dave Leip's U.S. Election Atlas; Presidential General Election Results – Delaware