1867 Iowa gubernatorial election

1867 Iowa gubernatorial election
October 8, 1867
 
Nominee Samuel Merrill Charles Mason
Party Republican Democratic
Popular vote 90,204 62,966
Percentage 58.88% 41.10%
County results
Merrill:      50–60%      60–70%      70–80%      80–90%      90–100%
Mason:      50–60%      60–70%
No Data/Votes:      
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

  1. ^ Kalb, Deborah (December 24, 2015). Guide to U.S. Elections. ISBN 9781483380353. Retrieved June 5, 2021.
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