United States presidential election in Rhode Island, 1876
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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%.
Results
United States presidential election in Rhode Island, 1876[1] | ||||||||
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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
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