1872 United States presidential election in Vermont
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County Results
Grant 60-70% 70-80% 80-90%
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Elections in Vermont |
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The 1872 United States presidential election in Vermont took place on November 5, 1872. All contemporary 37 states were part of the 1872 United States presidential election. The state voters chose five electors to the Electoral College, which selected the president and vice president.
Benjamin Gratz Brown of Missouri
by a margin of 57.67%.
With 78.29% of the popular vote, Vermont would be Grant's strongest victory in terms of percentage in the popular vote.[1] Grant's performance in the state was the third best for a Republican presidential candidate only after William McKinley's 80.08% in 1896 and Grant's 78.57% from four years earlier.
Results
1872 United States presidential election in Vermont[2] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Republican | Ulysses S. Grant of Illinois | Henry Wilson of Massachusetts | 41,480 | 78.29% | 5 | 100.00% | ||
Liberal Republican | Horace Greeley of New York | Benjamin Gratz Brown of Missouri
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10,926 | 20.62% | 0 | 0.00% | ||
Bourbon Democrat | Charles O'Conor | John Quincy Adams II | 553 | 1.04% | 0 | 0.00% | ||
N/A | Write ins | N/A | 21 | 0.04% | 0 | 0.00% | ||
Total | 52,980 | 100.00% | 5 | 100.00% |
See also
References
- ^ "1872 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
- ^ "1872 Presidential General Election Results - Vermont".