Jump to content

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 →
 
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

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