1912 United States presidential election in Iowa

The 1912 United States presidential election in Iowa took place on November 5, 1912, as part of the 1912 United States presidential election. Voters chose thirteen representatives, or electors, to the Electoral College, who voted for president and vice president.

Main article: 1912 United States presidential election

1912 United States presidential election in Iowa

November 5, 1912

Nominee Woodrow Wilson Theodore Roosevelt William Howard Taft Party Democratic Progressive Republican Home state New Jersey New York Ohio Running mate Thomas R. Marshall Hiram Johnson Nicholas M. Butler (replacing James S. Sherman) Electoral vote 13 0 0 Popular vote 185,325 161,819 119,805 Percentage 37.64% 32.87% 24.33%

President before election William Howard Taft Republican Elected President Woodrow Wilson Democratic

Iowa was won by Princeton University President Woodrow Wilson (D–Virginia), running with governor of Indiana Thomas R. Marshall, with 37.64% of the popular vote, against the 26th president of the United States Theodore Roosevelt (P–New York), running with governor of California Hiram Johnson, with 32.87% of the popular vote and the 27th president of the United States William Howard Taft (R–Ohio), running with Columbia University President Nicholas Murray Butler, with 24.33% of the popular vote.[1]

Wilson was the first Democrat to win Iowa's electoral votes since Franklin Pierce in 1852. No Democrat would obtain a majority until Franklin Roosevelt in 1932.

Results

1912 United States presidential election in Iowa[1]

Party		Candidate	Votes	%
	Democratic	Woodrow Wilson	185,325	37.64%
	Progressive	Theodore Roosevelt	161,819	32.87%
	Republican	William Howard Taft	119,805	24.33%
	Socialist	Eugene V. Debs	16,967	3.45%
	Prohibition	Stuart Hamblen	8,440	1.71%
Total votes			492,356	100%

References

1. "1912 Presidential Election Results Iowa".