1867 Iowa gubernatorial election

The 1867 Iowa gubernatorial election was held on October 8, 1867. Republican nominee Samuel Merrill defeated Democratic nominee Charles Mason with 62.93% of the vote.

1867 Iowa gubernatorial election

← 1865 October 8, 1867 1869 →
  Samuelmerrill.jpg 3x4.svg
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

General electionEdit

CandidatesEdit

  • Samuel Merrill, Republican
  • Charles Mason, Democratic

ResultsEdit

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

ReferencesEdit

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