1867 Iowa gubernatorial election
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← 1865 | October 8, 1867 | 1869 → |
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| | | Nominee | Samuel Merrill | Charles Mason | | Party | Republican | Democratic | Popular vote | 90,204 | 62,966 | Percentage | 58.88% | 41.10% | |
Governor before election William M. Stone Republican | Elected Governor Samuel Merrill Republican | |
The 1867 Iowa gubernatorial election was held on October 8, 1867. Republican nominee Samuel Merrill defeated Democratic nominee Charles Mason with 58.88% of the vote.
General election
Candidates
- Samuel Merrill, Republican
- Charles Mason, Democratic
Results
1867 Iowa gubernatorial election[1] Party | Candidate | Votes | % | ±% |
| Republican | Samuel Merrill | 90,204 | 58.88% | |
| Democratic | Charles Mason | 62,966 | 41.10% | |
Majority | 27,238 | | |
Turnout | | | |
| Republican hold | Swing | | |
References
- ^ Kalb, Deborah (December 24, 2015). Guide to U.S. Elections. ISBN 9781483380353. Retrieved June 5, 2021.