1876 United States presidential election in Rhode Island
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Elections in Rhode Island |
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The 1876 United States presidential election in Rhode Island took place on November 7, 1876, as part of the 1876 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.
Rhode Island voted for the Republican nominee, Rutherford B. Hayes, over the Democratic nominee, Samuel J. Tilden. Hayes won the state by a margin of 19.06%.
With 59.29% of the popular vote, Rhode Island would be Hayes' fourth strongest victory in terms of percentage in the popular vote after Vermont, Nebraska and Kansas[1].
This was the only election between 1860 and 1888 where the Democratic candidate earned more than 40% in at least one Rhode Island county.
Results
United States presidential election in Rhode Island, 1876[2] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Republican | Rutherford B. Hayes of Ohio | William A. Wheeler of New York | 15,787 | 59.29% | 4 | 100.00% | ||
Democratic | Samuel J. Tilden of New York | Thomas A. Hendricks of Indiana | 10,712 | 40.23% | 0 | 0.00% | ||
Greenback | Peter Cooper of New York | Samuel Fenton Cary of Ohio | 68 | 0.26% | 0 | 0.00% | ||
Prohibition | Green Clay Smith of Kentucky | Gideon Tabor Stewart of Ohio | 60 | 0.23% | 0 | 0.00% | ||
Total | 26,627 | 100.00% | 4 | 100.00% |
References
- ^ "1876 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
- ^ "1876 Presidential General Election Results - Rhode Island".