1872 United States presidential election in Rhode Island
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 County Results
Grant 60-70% 70-80%
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| Elections in Rhode Island |
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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] It remains the best Republican performance in Rhode Island's history and the second-best by any candidate since the formation of the Democratic party in 1828 after Lyndon B. Johnson’s 80.87% in 1964.[2]
Results
[edit]| 1872 United States presidential election in Rhode Island[3] | ||||||||
|---|---|---|---|---|---|---|---|---|
| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Republican | Ulysses S. Grant of Illinois | Henry Wilson of Massachusetts | 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% | ||||
See also
[edit]References
[edit]- ^ "1872 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
- ^ "Presidential General Election Results Comparison – Rhode Island". Dave Leip’s U.S. Election Atlas. Retrieved October 25, 2019.
- ^ "1872 Presidential General Election Results - Rhode Island".
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