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1872 United States presidential election in Rhode Island

From Wikipedia, the free encyclopedia

United States presidential election in Rhode Island, 1872

← 1868 November 5, 1872 1876 →
 
UlyssesGrant.jpg
Horace Greeley restored (cropped).jpg
Nominee Ulysses S. Grant Horace Greeley
Party Republican Liberal Republican
Home state Ohio New York
Running mate Henry Wilson Benjamin G. Brown
Electoral vote 4 0
Popular vote 13,665 5,329
Percentage 71.94% 28.06%

President before election

Ulysses S. Grant
Republican

Elected President

Ulysses S. Grant
Republican

The 1872 United States presidential election in Rhode Island took place on November 5, 1872. All contemporary 37 states were part of the 1872 United States presidential election. The state voters chose four electors to the Electoral College, which selected the president and vice president.

Rhode Island was won by the Republican nominees, incumbent President Ulysses S. Grant of Illinois and his running mate Senator Henry Wilson of Massachusetts. Grant and Wilson defeated the Liberal Republican and Democratic nominees, former Congressman Horace Greeley of New York and his running mate former Senator and Governor Benjamin Gratz Brown of Missouri by a margin of 43.88%.

With 71.94% of the popular vote, Rhode Island would be Grant's third strongest victory in terms of percentage in the popular vote after Vermont and South Carolina.[1]

Results

United States presidential election in Rhode Island, 1872[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican Ulysses S. Grant of Illinois Henry Wilson of Indiana 13,665 71.94% 4 100.00%
Liberal Republican Horace Greeley of New York Benjamin Gratz Brown of Missouri 5,329 28.06% 0 0.00%
Total 18,994 100.00% 4 100.00%


References

  1. ^ "1872 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
  2. ^ "1872 Presidential General Election Results - Rhode Island".


This page was last edited on 3 May 2019, at 21:55
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