United States presidential election in Rhode Island, 1916

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United States presidential election in Rhode Island, 1916

← 1912 November 7, 1916 1920 →
  Governor Charles Evans Hughes.jpg Woodrow Wilson-H&E.jpg
Nominee Charles Evans Hughes Woodrow Wilson
Party Republican Democratic
Home state New York New Jersey
Running mate Charles W. Fairbanks Thomas R. Marshall
Electoral vote 5 0
Popular vote 44,858 40,394
Percentage 51.08% 46.00%

President before election

Woodrow Wilson
Democratic

Elected President

Woodrow Wilson
Democratic

Lua error in package.lua at line 80: module 'strict' not found. The 1916 United States Presidential Election in Connecticut took place on November 7, 1916 as part of the 1916 United States Presidential Election which was held throughout all contemporary 48 states. Voters chose five representatives, or electors to the Electoral College, who voted for President and Vice President.

Rhode Island was won by the Republican nominee, U.S. Supreme Court Justice Charles Evans Hughes of New York, and his running mate Senator Charles W. Fairbanks of Indiana. They defeated Democratic nominees, incumbent Democratic President Woodrow Wilson and Vice President Thomas R. Marshall.

Hughes won Rhode Island by a narrow margin of 5.08 percent.

Results

United States presidential election in Rhode Island, 1916[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican Charles Evans Hughes of New York Charles Warren Fairbanks of Indiana 44,858 51.08% 5 100.00%
Democratic Woodrow Wilson of New Jersey Thomas Riley Marshall of Indiana 40,394 46.00% 0 0.00%
Socialist Allan Louis Benson of New York George Ross Kirkpatrick of New Jersey 1,914 2.18% 0 0.00%
Prohibition James Franklin Hanly of Indiana Ira Landrith of Tennessee 470 0.54% 0 0.00%
Socialist Labor Arthur Elmer Reimer of Massachusetts Caleb Harrison of Illinois 180 0.20% 0 0.00%
Total 87,816 100.00% 5 100.00%

References

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