1867 Iowa gubernatorial election

1867 Iowa gubernatorial election

←  1865 October 8, 1867 1869  →

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.