Union, Iowa

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Union is a city in Hardin County, Iowa, United States. The population was 427 at the 2000 census.



Union is located at 42°14′41″N 93°3′51″W / 42.24472°N 93.06417°W / 42.24472; -93.06417 (42.244715, -93.064240)[1].

According to the United States Census Bureau, the city has a total area of 0.6 square miles (1.4 km²), all of it land.


As of the census[2] of 2000, there were 427 people, 183 households, and 124 families residing in the city. The population density was 772.8 people per square mile (299.8/km²). There were 209 housing units at an average density of 378.3/sq mi (146.7/km²). The racial makeup of the city was 99.06% White, 0.23% African American and 0.70% Native American. Hispanic or Latino of any race were 0.70% of the population.

There were 183 households out of which 27.3% had children under the age of 18 living with them, 59.0% were married couples living together, 4.9% had a female householder with no husband present, and 32.2% were non-families. 29.5% of all households were made up of individuals and 19.1% had someone living alone who was 65 years of age or older. The average household size was 2.33 and the average family size was 2.87.

In the city the population was spread out with 24.1% under the age of 18, 7.5% from 18 to 24, 23.7% from 25 to 44, 23.4% from 45 to 64, and 21.3% who were 65 years of age or older. The median age was 41 years. For every 100 females there were 92.3 males. For every 100 females age 18 and over, there were 90.6 males.

The median income for a household in the city was $34,792, and the median income for a family was $39,792. Males had a median income of $25,417 versus $20,000 for females. The per capita income for the city was $16,370. About 8.8% of families and 12.0% of the population were below the poverty line, including 17.0% of those under age 18 and 6.2% of those age 65 or over.


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