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1894 United States House of Representatives election in Wyoming

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← 1892

November 6, 1894 (1894-11-06)

1896 →


Nominee	Frank Wheeler Mondell	Henry A. Coffeen	Shakespeare E. Sealey
Party	Republican	Democratic	Populist
Popular vote	10,068	6,152	2,906
Percentage	52.64%	32.17%	15.19%

County results
Mondell:      40–50%     50–60%      60–70%
Coffeen:      30–40%
     No Data

U.S. Representative before election

Henry A. Coffeen
Democratic

Elected U.S. Representative

Frank Wheeler Mondell
Republican

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The Wyoming United States House election for 1894 was held on November 6, 1894. Republican Frank Wheeler Mondell defeated Democratic incumbent Henry A. Coffeen and Populist Shakespeare E. Sealey with 52.64% of the vote making Coffeen the second incumbent Representative from Wyoming to lose reelection.

Results[edit]

United States House of Representatives election in Wyoming, 1894^[1]
Party		Candidate	Votes	%
	Republican	Frank Wheeler Mondell	10,068	52.64%
	Democratic	Henry A. Coffeen (inc.)	6,152	32.17%
	Populist	Shakespeare E. Sealey	2,906	15.19%
Total votes			19,126	100%

References[edit]

^ "United States House of Representatives election in Wyoming, 1894".

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Retrieved from "https://en.wikipedia.org/w/index.php?title=1894_United_States_House_of_Representatives_election_in_Wyoming&oldid=1216417531" Categories: ●United States House of Representatives elections in Wyoming ●1894 United States House of Representatives elections ●1894 Wyoming elections ●Wyoming election stubs Hidden categories: ●Use mdy dates from September 2023 ●All stub articles ●This page was last edited on 30 March 2024, at 21:48 (UTC). ●Text is available under the Creative Commons Attribution-ShareAlike License 4.0; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. ●Privacy policy ●About Wikipedia ●Disclaimers ●Contact Wikipedia ●Code of Conduct ●Developers ●Statistics ●Cookie statement ●Mobile view