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### Example H.1

Posted: **Tue Jul 04, 2017 2:37 am**

by **JD Malana**

In the third step of balancing the equation for the combustion of C_{6}H_{14} + O_{2} > CO_{2} + H_{2}O,

the final balanced equation is C_{6}H_{14} + 19/2 O_{2} > 6 CO_{2} + 7 H_{2}O...

Why/how is the fractional coefficient used for O_{2} 19/2?

Is there a specific method/formula for determining what fractions to use?

### Re: Example H.1

Posted: **Tue Jul 04, 2017 8:58 am**

by **Sarah_Wilen**

To determine what fraction you need, what I would do is count up the oxygens (I need 19). 19 doesn't divide evenly, which is a problem because we have O2. So, I will do 19/2 because I know I need 19 Oxygens and I will "split" up the O2 to get 19 of them. In the end, I just multiply everything by 2 to cancel out the fraction.

A general rule for fractions I think would be, if there is a diatomic element, but you need an uneven number, you just see the number of that element you need and put it over 2.

### Re: Example H.1

Posted: **Mon Oct 02, 2017 4:02 pm**

by **Abby Ellstrom 1I**

Allowing for an improper fraction in your balanced equation makes it easier when initially balancing the equation so you don't have to do it in your head. However, a final balanced reaction should contain no fraction stoichiometric coefficients so simply multiply the entire equation by the denominator for your final answer.

### Re: Example H.1

Posted: **Thu Oct 05, 2017 9:43 am**

by **Madeline Musselman 3H**

When balancing a chemical equation, each side of the reaction needs to have an equal amount of stoichiometric coefficients. Therefore, sometimes the need for fractions is required in order to create the equal amount on both sides. However, fractions are not whole numbers so you would need to multiply the denominator to each coefficient on both sides of the reaction.