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1894 United States House of Representatives election in Wyoming

From Wikipedia, the free encyclopedia

1894 United States House of Representatives election in Wyoming

← 1892 November 6, 1894 (1894-11-06) 1896 →
Henry A. Coffeen.png
Nominee Frank Wheeler Mondell Henry A. Coffeen Shakespeare E. Sealey
Party Republican Democratic People's
Popular vote 10,068 6,152 2,906
Percentage 52.64% 32.17% 15.19%

U.S. Representative before election

Henry A. Coffeen

Elected U.S. Representative

Frank Wheeler Mondell

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.


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 6,152 32.17%
People's Shakespeare E. Sealey 2,906 15.19%
Total votes 19,126 100%


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

This page was last edited on 6 January 2021, at 17:25
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